Free Math Worksheets

- An intersection lies on both graphs. - Read both the x-coordinate and y-coordinate of each intersection.

1. Locate the two points where the parabola and the line cross. 2. The left intersection has coordinates \((-1,-3)\). 3. The right intersection has coordinates \((3,5)\).

\((-1,-3)\) and \((3,5)\)

- Equal outputs mean the function expressions are equal. - Move all terms to one side. - Factor the resulting quadratic.

1. Set the function expressions equal: \(x^2 = 2x + 8\). 2. Rearrange: \(x^2 - 2x - 8 = 0\). 3. Factor: \((x - 4)(x + 2) = 0\). 4. Therefore, \(x = -2\) or \(x = 4\).

\(x = -2\) and \(x = 4\)

- Move all terms to one side and factor. - An equation \(f(x) = g(x)\) identifies equal outputs. - Equal outputs correspond to graph intersections.

1. Move all terms to one side: \(x^2 - x - 6 = 0\). 2. Factor: \((x - 3)(x + 2) = 0\). 3. Thus, \(x = -2\) or \(x = 3\). 4. These values are the x-coordinates of the intersection points of the graphs of \(f\) and \(g\).

a) \(\{-2, 3\}\) b) The solutions are the x-coordinates of the graphs’ intersection points.

- What do you notice about equation (I) when you consider using substitution? - Use parentheses when substituting an expression into equation (II). - How can you use the value of the first variable to find the second variable?

1. Equation (I) is already solved for \(y\), so substitute \(4x - 3\) for \(y\) in equation (II): \(2x + 3(4x - 3) = 19\). 2. Distribute and combine like terms: \(2x + 12x - 9 = 19\), so \(14x - 9 = 19\). 3. Solve for \(x\): \(14x = 28\), so \(x = 2\). 4. Substitute \(x = 2\) into equation (I): \(y = 4 \cdot 2 - 3 = 5\). 5. Therefore, the solution set is \(\{(2, 5)\}\).

The solution set is \(\{(2, 5)\}\).

- What is the same on the left side of both equations? - If two expressions are both equal to \(x\), what must be true about those expressions? - After finding one variable, how can you find the other?

1. Setting the two expressions equal is most direct because both equations are already solved for \(x\). 2. Set the expressions equal: \(2y + 5 = -3y + 15\). 3. Combine like terms: \(5y + 5 = 15\). 4. Solve for \(y\): \(5y = 10\), so \(y = 2\). 5. Substitute into either original equation: \(x = 2 \cdot 2 + 5 = 9\). 6. The solution is \((9, 2)\).

Setting the two expressions for \(x\) equal is most direct because both equations are already solved for \(x\). The solution set is \(\{(9, 2)\}\).

- Is there a variable that is especially easy to isolate? - Can you express one variable in terms of the other? - Check your result by substituting both values into the original equations.

1. Solve equation (II) for \(y\): \(y = 2x\). 2. Substitute \(2x\) for \(y\) in equation (I): \(x + 3(2x) = 7\). 3. Simplify: \(7x = 7\). 4. Solve for \(x\): \(x = 1\). 5. Substitute \(x = 1\) into \(y = 2x\): \(y = 2\). 6. The solution is \((1, 2)\).

The solution is \((1, 2)\).

- If two expressions are both equal to \(y\), what must be true about those expressions? - Could multiplying by a common denominator help clear the fractions? - What inverse operations isolate \(x\)?

1. Set the expressions for \(y\) equal: \(\frac{2}{5}x + 3 = \frac{4}{5}x - 1\). 2. Subtract \(\frac{2}{5}x\) from both sides: \(3 = \frac{2}{5}x - 1\). 3. Add \(1\): \(4 = \frac{2}{5}x\). 4. Multiply by \(5\): \(20 = 2x\). 5. Divide by \(2\): \(x = 10\).

\(x = 10\)

- Which variable is easiest to isolate in both equations? - If two expressions are both equal to \(y\), what relationship do they have? - Remember to calculate the second variable after finding the first.

1. Solve both equations for \(y\). Equation (I) becomes \(y = 14 - 3x\). Equation (II) becomes \(y = 2x - 1\). 2. Set the expressions equal: \(14 - 3x = 2x - 1\). 3. Solve for \(x\): \(14 = 5x - 1\), so \(15 = 5x\) and \(x = 3\). 4. Substitute into either expression for \(y\): \(y = 2 \cdot 3 - 1 = 5\). 5. The solution set is \(\{(3, 5)\}\).

The solution is \((3, 5)\).

- Can you replace one variable with an expression involving the other variable? - What happens when you substitute the expression for one variable into the other equation? - Pay attention to signs when distributing. - How can you check your result?

1. Substitute the expression for \(y\) from equation (I) into equation (II): \(3x + 2(4x - 5) = 23\). 2. Distribute: \(3x + 8x - 10 = 23\). 3. Combine like terms: \(11x - 10 = 23\). 4. Solve for \(x\): \(11x = 33\), so \(x = 3\). 5. Substitute into equation (I): \(y = 4 \cdot 3 - 5 = 7\).

The solution is \((3, 7)\).

- Which variable in which equation is easiest to isolate? - Put the entire substituted expression in parentheses. - What inverse operation will isolate the chosen variable?

1. Solve equation (I) for \(y\): \(y = 3x - 7\). 2. Substitute into equation (II): \(2x + 4(3x - 7) = 28\). 3. Distribute: \(2x + 12x - 28 = 28\). 4. Combine like terms and solve: \(14x = 56\), so \(x = 4\). 5. Substitute into \(y = 3x - 7\): \(y = 3 \cdot 4 - 7 = 5\).

The solution is \((4, 5)\).

- One variable is already isolated in equation (I). - How can you use that expression in equation (II)? - Replace the corresponding variable in equation (II) with the expression from equation (I). - Use parentheses around a substituted expression with more than one term.

1. Substitute the expression for \(y\) from equation (I) into equation (II): \(3x + 2(2x - 4) = 27\). 2. Distribute: \(3x + 4x - 8 = 27\). 3. Combine like terms: \(7x - 8 = 27\). 4. Add \(8\) to both sides: \(7x = 35\). 5. Divide by \(7\): \(x = 5\). 6. Substitute into equation (I): \(y = 2 \cdot 5 - 4 = 6\). 7. The solution set is \(\{(5, 6)\}\).

The solution set is \(\{(5, 6)\}\).

- What must be true of the right sides when both equations are solved for \(y\)? - How can you combine the two equations into one equation in \(x\)? - After finding \(x\), remember to find \(y\). - Check the ordered pair in both original equations.

1. Set the expressions equal: \(3x - 8 = -2x + 7\). 2. Combine like terms: \(5x = 15\). 3. Divide by \(5\): \(x = 3\). 4. Substitute into equation (I): \(y = 3 \cdot 3 - 8 = 1\). 5. The solution is \((3, 1)\).

The solution is \((3, 1)\).

- Equal outputs mean the function expressions are equal. - Clear the decimal coefficient before factoring. - Substitute each x-value into either function to find the corresponding y-value.

1. Set the function expressions equal: \(x^2 = 1.5x + 1\). 2. Multiply by \(2\): \(2x^2 - 3x - 2 = 0\). 3. Factor: \((2x + 1)(x - 2) = 0\). 4. Thus, \(x = -0.5\) or \(x = 2\). 5. The corresponding outputs are \(f(-0.5) = 0.25\) and \(f(2) = 4\), so the intersection points are \((-0.5, 0.25)\) and \((2, 4)\).

\(x = -0.5\) or \(x = 2\); intersections: \((-0.5, 0.25)\) and \((2, 4)\)

- Equal outputs mean the function expressions are equal. - Clear the decimal coefficient before factoring. - Substitute each candidate value into both sides of the original equation.

1. Set the function expressions equal: \(0.5x^2 = x + 4\). 2. Multiply by \(2\): \(x^2 - 2x - 8 = 0\). 3. Factor: \((x - 4)(x + 2) = 0\), so \(x = -2\) or \(x = 4\). 4. Check \(x = -2\): \(0.5 \cdot (-2)^2 = 2\) and \(-2 + 4 = 2\). 5. Check \(x = 4\): \(0.5 \cdot 4^2 = 8\) and \(4 + 4 = 8\). 6. Both values satisfy the equation.

\(\{-2, 4\}\)

- Set the function expressions equal. - Isolate \(x^2\). - Remember both square roots of a positive number.

1. Set the functions equal: \(x^2 - 2 = -x^2 + 6\). 2. Simplify: \(2x^2 = 8\), so \(x^2 = 4\). 3. Therefore, \(x = -2\) or \(x = 2\). 4. Substituting either value into \(f\) gives \(y = 2\). 5. The intersection points are \((-2, 2)\) and \((2, 2)\).

\((-2, 2)\) and \((2, 2)\)

- What must be true of the right sides when both equations are solved for \(y\)? - Bring all variable terms to one side. - How can you check your result?

1. Set the expressions equal: \(3x + 4.5 = -1.5x + 18\). 2. Add \(1.5x\) and subtract \(4.5\): \(4.5x = 13.5\). 3. Divide by \(4.5\): \(x = 3\). 4. Substitute into either original equation: \(y = 3 \cdot 3 + 4.5 = 13.5\).

The solution is \((3, 13.5)\).

- Can you replace one variable with an expression involving the other variable? - What happens when you substitute the expression for \(x\) into equation (II)? - Distribute the factor to every term inside the parentheses. - How can you check your result?

1. Equation (I) is already solved for \(x\), so substitute \(3y + 2\) for \(x\) in equation (II): \(4(3y + 2) - 5y = 29\). 2. Distribute: \(12y + 8 - 5y = 29\). 3. Combine like terms: \(7y + 8 = 29\). 4. Solve for \(y\): \(7y = 21\), so \(y = 3\). 5. Substitute into equation (I): \(x = 3 \cdot 3 + 2 = 11\). 6. The solution is \((11, 3)\).

The solution is \((11, 3)\).

- Which variable is already isolated in one equation? - Can you substitute that expression into the other equation? - Pay close attention to the negative factor before the parentheses. - After finding one variable, how can you determine the other?

1. Substitute the expression for \(y\) from the first equation into the second equation: \(5x - 2(2x + 1) = 1\). 2. Distribute: \(5x - 4x - 2 = 1\). 3. Combine like terms and solve: \(x - 2 = 1\), so \(x = 3\). 4. Substitute into the first equation: \(y = 2 \cdot 3 + 1 = 7\). 5. The solution is \((3, 7)\).

The solution is \((3, 7)\).

- Which equation is already solved for one variable? - Put parentheses around the entire expression when substituting it into the second equation. - Remember to find both variable values. - Check your result in both original equations.

1. Substitute the expression for \(y\) from equation (I) into equation (II): \(3x - 4(0.5x + 2) = 2\). 2. Distribute: \(3x - 2x - 8 = 2\). 3. Combine like terms: \(x - 8 = 2\). 4. Add \(8\) to both sides: \(x = 10\). 5. Substitute into equation (I): \(y = 0.5 \cdot 10 + 2\). 6. Calculate \(y\): \(y = 7\). 7. The solution is \((10, 7)\).

The solution is \((10, 7)\).

- Which equation is easiest to solve for one variable? - Use parentheses around the expression you substitute into the other equation. - How can you check your result in the original equations?

1. Solve equation (I) for \(x\): \(x = 18 - 4y\). 2. Substitute into equation (II): \(3(18 - 4y) - 2y = -2\). 3. Solve for \(y\): \(54 - 12y - 2y = -2\), so \(-14y = -56\) and \(y = 4\). 4. Substitute into \(x = 18 - 4y\): \(x = 18 - 4 \cdot 4 = 2\). 5. The solution set is \(\{(2, 4)\}\).

The solution set is \(\{(2, 4)\}\).

- Can you rearrange one equation so that one variable is alone on one side? - What happens when you substitute that expression into the other equation? - Put parentheses around the entire substituted expression. - Check your result in both original equations.

1. Solve equation (II) for \(x\): \(x = 2y - 7\). 2. Substitute into equation (I): \(2(2y - 7) + 3y = 7\). 3. Distribute and combine like terms: \(4y - 14 + 3y = 7\), so \(7y - 14 = 7\). 4. Solve for \(y\): \(7y = 21\), so \(y = 3\). 5. Substitute into \(x = 2y - 7\): \(x = 2 \cdot 3 - 7 = -1\). 6. The solution set is \(\{(-1, 3)\}\).

The solution set is \(\{(-1, 3)\}\).

- Which variable can you isolate without introducing fractions? - Substitute the expression into the other equation. - Put parentheses around the entire substituted expression. - Use the first variable value to calculate the second.

1. Solve equation (I) for \(x\): \(x = 4y + 18\). 2. Substitute into equation (II): \(3(4y + 18) + 2y = 12\). 3. Distribute: \(12y + 54 + 2y = 12\). 4. Combine like terms and solve: \(14y + 54 = 12\), so \(14y = -42\) and \(y = -3\). 5. Substitute into \(x = 4y + 18\): \(x = 4(-3) + 18 = 6\). 6. The solution is \((6, -3)\).

The solution is \((6, -3)\).

- Which variable in which equation is easiest to isolate? - Put the entire substituted expression in parentheses. - After finding one variable, substitute it into an original equation to find the other.

1. Solve the first equation for \(y\): \(y = 3x - 5\). 2. Substitute into the second equation: \(2x + 3(3x - 5) = 29\). 3. Solve for \(x\): \(2x + 9x - 15 = 29\), so \(11x = 44\) and \(x = 4\). 4. Substitute into \(y = 3x - 5\): \(y = 3 \cdot 4 - 5 = 7\). 5. The solution is \((4, 7)\).

The solution is \((4, 7)\).

- Can you rearrange one equation so that one variable is alone on one side? - Substitute that expression into the other equation. - Pay close attention to the negative factor before the parentheses. - Check your solution in both original equations.

1. Solve the first equation for \(y\): \(y = 18 - 4x\). 2. Substitute into the second equation: \(3x - 2(18 - 4x) = 8\). 3. Distribute and combine like terms: \(3x - 36 + 8x = 8\), so \(11x - 36 = 8\). 4. Solve for \(x\): \(11x = 44\), so \(x = 4\). 5. Substitute into the expression for \(y\): \(y = 18 - 4 \cdot 4 = 2\).

The solution is \((4, 2)\).

- Treat the two sides as separate functions. - Equal outputs give an equation that can be factored. - Solutions of \(f(x) = g(x)\) correspond to graph intersections.

1. Set the function expressions equal: \(x^2 = -2x + 3\). 2. Move all terms to one side: \(x^2 + 2x - 3 = 0\). 3. Factor: \((x + 3)(x - 1) = 0\). 4. Therefore, \(x = -3\) or \(x = 1\). 5. These values are the x-coordinates of the points where the graphs of \(f\) and \(g\) intersect.

1. \(x = -3\) or \(x = 1\) 2. They are the x-coordinates of the graphs’ intersection points.

- Use substitution by setting the two expressions for \(y\) equal. - Clear the denominator before solving the power equation. - Substitute each \(x\)-value into either original equation to find \(y\). - Compare even and odd exponents.

1. Set the two expressions for \(y\) equal: \(x^3=\frac{16}{x}\). Since \(x\ne0\), multiply by \(x\): \(x^4=16\). 2. The real solutions are \(x=\pm2\). 3. Substitute into \(y=x^3\). When \(x=2\), \(y=8\). When \(x=-2\), \(y=-8\). 4. The intersection points are \((2, 8)\) and \((-2, -8)\). 5. In the comparison system, substitution produces \(x^3=8\), which has one real solution because the exponent is odd. Here, \(x^4=16\) has two real solutions because the exponent is even and the right side is positive.

a) \(x=\pm2\) b) \((2, 8)\) and \((-2, -8)\) c) The equation \(x^4=16\) has two real solutions, while \(x^3=8\) has only one.

- Substitute the expression for \(y\) into the other equation. - Combine the powers of \(x\). - Recall how many real solutions \(x^3=a\) has.

1. Substitute \(y=2x\) into \(yx^2=54\): \((2x)x^2=54\). 2. Simplify: \(2x^3=54\), so \(x^3=27\) and \(x=3\). 3. Substitute into \(y=2x\): \(y=6\). The solution is \((3, 6)\). 4. The equation \(x^3=27\) has exactly one real solution because an odd power function takes each real value exactly once.

1. \((3, 6)\) 2. The odd-power equation \(x^3=27\) has exactly one real solution.

- Which system-solving method is most direct when one equation is already solved for \(y\)? - Replace \(y\) in the other equation with the given expression. - Pay attention to the negative factor before the parentheses. - How can you use the value of \(x\) to find \(y\)?

1. Substitute the expression for \(y\) from line \(h\) into the equation for line \(g\): \(3x - 2(2x - 5) = 4\). 2. Distribute: \(3x - 4x + 10 = 4\). 3. Combine like terms: \(-x + 10 = 4\). 4. Solve for \(x\): \(-x = -6\), so \(x = 6\). 5. Substitute into the equation for line \(h\): \(y = 2 \cdot 6 - 5 = 7\). 6. The intersection is \((6, 7)\).

The intersection is \((6, 7)\).

- Solutions of \(f(x) = g(x)\) are the x-coordinates of graph intersections. - Read both intersection points from the coordinate plane. - Substitute each x-value into both function rules.

1. The graphs intersect at exactly two points, so there are two x-values for which \(f(x) = g(x)\). 2. The intersection points are \((0, 0)\) and \((2, 4)\), so the solutions are \(x = 0\) and \(x = 2\). 3. For \(x = 0\), \(f(0) = 0\) and \(g(0) = -(0 - 2)^2 + 4 = 0\). 4. For \(x = 2\), \(f(2) = 4\) and \(g(2) = -(2 - 2)^2 + 4 = 4\).

a) The graphs have two intersection points. b) \(x = 0\) and \(x = 2\) c) Both values give equal outputs for \(f\) and \(g\).

- Identify the equations of the two graphs. - Set the function expressions equal. - Move all terms to one side.

1. The parabola is \(f(x) = x^2 - 2\), and the line is \(g(x) = x\). 2. Set the functions equal: \(x^2 - 2 = x\). 3. Move all terms to one side: \(x^2 - x - 2 = 0\). 4. Its solutions are \(x = -1\) and \(x = 2\), matching the marked intersections.

\(x^2 - x - 2 = 0\)

- Locate the points where the line and parabola meet. - At an intersection, both functions have the same output for the same input. - Move every term to one side of the equation. - The degree is determined by the greatest exponent of the variable.

1. The line and parabola intersect at \(x=1\) and \(x=3\). The corresponding points are \(B=(1, 2)\) and \(C=(3, 2.8)\). 2. Set the function values equal: \(-0.2x^2+1.2x+1=0.4x+1.6\). 3. Move all terms to one side: \(-0.2x^2+0.8x-0.6=0\). 4. The highest exponent of \(x\) is \(2\), so this is a quadratic equation of degree \(2\).

a) \(B=(1, 2)\) and \(C=(3, 2.8)\), in meters b) \(-0.2x^2+1.2x+1=0.4x+1.6\) c) \(-0.2x^2+0.8x-0.6=0\); the equation has degree \(2\).

- Which variable has a coefficient of \(1\)? - Why is a coefficient of \(1\) convenient when isolating a variable? - Remember to include both variable values in the solution set.

1. In equation (I), \(x\) has coefficient \(1\), so it can be isolated without division or fractions: \(x = 2y + 8\). 2. Substitute \(x = 2y + 8\) into equation (II): \(3(2y + 8) + 4y = 4\). 3. Solve for \(y\): \(6y + 24 + 4y = 4\), so \(10y = -20\) and \(y = -2\). 4. Substitute \(y = -2\) into \(x = 2y + 8\): \(x = 2(-2) + 8 = 4\). 5. The solution set is \(\{(4, -2)\}\).

a) Isolate \(x\) in equation (I), giving \(x = 2y + 8\), because \(x\) has coefficient \(1\). b) The solution set is \(\{(4, -2)\}\).

- You may isolate either variable in either equation. Which choice looks simplest here? - Pay close attention to the signs when distributing a negative factor. - How can substitution into both original equations verify your result?

1. Solve equation (I) for \(y\): \(y = 1 - 0.5x\). 2. Substitute the expression for \(y\) into equation (II): \(2x - 3(1 - 0.5x) = 11\). 3. Distribute: \(2x - 3 + 1.5x = 11\). 4. Combine like terms and solve: \(3.5x - 3 = 11\), so \(3.5x = 14\) and \(x = 4\). 5. Substitute \(x = 4\) into the rearranged equation (I): \(y = 1 - 0.5 \cdot 4 = -1\). 6. Check equation (I): \(0.5 \cdot 4 + (-1) = 1\). Check equation (II): \(2 \cdot 4 - 3(-1) = 11\). Both equations are true. 7. The solution set is \(\{(4, -1)\}\).

The solution set is \(\{(4, -1)\}\).

- Identify the two facts given about the invested amounts. - Write each interest rate as a decimal. - Translate the total investment and total interest into two equations.

1. Let \(x\) be the amount invested at \(2\%\) and \(y\) the amount invested at \(3\%\). 2. Write the system: \(x + y = 10{,}000\) and \(0.02x + 0.03y = 260\). 3. Solve the first equation for \(x\): \(x = 10{,}000 - y\). 4. Substitute into the second equation: \(0.02(10{,}000 - y) + 0.03y = 260\). 5. Simplify: \(200 - 0.02y + 0.03y = 260\), so \(0.01y = 60\) and \(y = 6000\). 6. Then \(x = 10{,}000 - 6000 = 4000\).

Ms. Miller invested \(\$4000\) in the \(2\%\) account and \(\$6000\) in the \(3\%\) account.

- Find the slope of \(g_1\) from its two points. - Use the known slope and point to find the equation of \(g_2\). - At an intersection point, what must be true about the two y-values?

1. For \(g_1\), the y-intercept is \(2\), and the slope is \(m_1 = \frac{4 - 2}{4 - 0} = 0.5\). Thus \(g_1: y = 0.5x + 2\). 2. For \(g_2\), substitute \((6, 2)\) into \(y = -x + b_2\): \(2 = -6 + b_2\), so \(b_2 = 8\). Thus \(g_2: y = -x + 8\). 3. At the intersection, the y-values are equal: \(0.5x + 2 = -x + 8\). Solving gives \(x = 4\). 4. Substitute \(x = 4\) into either equation: \(y = 4\). The intersection is \((4, 4)\).

\(g_1: y = 0.5x + 2\), \(g_2: y = -x + 8\), and the intersection is \((4, 4)\).

- What must you do to both equations before you can compare their expressions directly? - Solve both equations for the same variable. - After finding \(x\), how can you find \(y\)? - Check the ordered pair in both original equations.

1. Solve both equations for \(y\). Equation (I) becomes \(3y = 6x + 12\), so \(y = 2x + 4\). Equation (II) becomes \(y = -2x + 20\). 2. Set the expressions equal: \(2x + 4 = -2x + 20\). 3. Solve for \(x\): \(4x = 16\), so \(x = 4\). 4. Substitute into either expression for \(y\): \(y = 2 \cdot 4 + 4 = 12\). 5. The solution set is \(\{(4, 12)\}\).

The solution set is \(\{(4, 12)\}\).

- A candle height can be modeled as initial height minus burn rate times time. - Which variable represents time, and which represents height? - What equation represents the moment when the candles have the same height? - Keep track of the units as you interpret the result.

1. Candle A is modeled by \(h = 25 - 1.5t\). Candle B is modeled by \(h = 20 - 0.5t\). 2. Set the expressions for \(h\) equal: \(25 - 1.5t = 20 - 0.5t\). 3. Solve for \(t\): \(5 - 1.5t = -0.5t\), so \(5 = t\). 4. Substitute \(t = 5\): \(h = 25 - 1.5 \cdot 5 = 17.5\). 5. After \(5\) hours, both candles are \(17.5\,\text{cm}\) tall.

a) Candle A: \(h = 25 - 1.5t\) Candle B: \(h = 20 - 0.5t\) b) After \(5\) hours, both candles are \(17.5\,\text{cm}\) tall.

- Which variable can you isolate most easily before substituting into the other equation? - Pay close attention to the negative sign before the parentheses. - What happens when you square a negative number?

1. Solve equation (I) for \(y\): \(y = 10 - 4x\). 2. Substitute into equation (II): \(x - 2(10 - 4x) = 7\). 3. Distribute and combine like terms: \(x - 20 + 8x = 7\), so \(9x = 27\). 4. Solve for \(x\): \(x = 3\). 5. Find \(y\): \(y = 10 - 4 \cdot 3 = -2\). The solution is \((3, -2)\). 6. Evaluate the expression: \(T = 3^2 + (-2)^2 = 9 + 4 = 13\).

1. The solution is \((3, -2)\). 2. \(T = 13\)

- Would it help to solve an equation for a variable whose coefficient is \(1\)? - Pay close attention to decimal place values. - After expressing one variable in terms of the other, substitute that expression into the remaining equation.

1. Solve equation (II) for \(a\): \(a = 6.5 - 0.5b\). 2. Substitute into equation (I): \(0.4(6.5 - 0.5b) - 0.2b = 1\). 3. Distribute: \(2.6 - 0.2b - 0.2b = 1\). 4. Combine like terms: \(2.6 - 0.4b = 1\). 5. Solve for \(b\): \(-0.4b = -1.6\), so \(b = 4\). 6. Substitute into equation (II): \(a = 6.5 - 0.5 \cdot 4 = 4.5\). 7. The solution is \((a, b) = (4.5, 4)\).

The solution is \((a, b) = (4.5, 4)\).

- Which equation is easiest to solve for one variable? - Substitute the resulting expression into the other equation. - After finding one variable, how can you determine the other?

1. Solve equation (II) for \(r\): \(r = \frac{3}{4}s - 1\). 2. Substitute into equation (I): \(\frac{2}{3}(\frac{3}{4}s - 1) + \frac{1}{2}s = 6\). 3. Distribute: \(\frac{1}{2}s - \frac{2}{3} + \frac{1}{2}s = 6\). 4. Combine like terms: \(s - \frac{2}{3} = 6\). 5. Solve for \(s\): \(s = 6 + \frac{2}{3} = \frac{20}{3}\). 6. Substitute into the expression for \(r\): \(r = \frac{3}{4} \cdot \frac{20}{3} - 1 = 4\). 7. The solution is \((r, s) = (4, \frac{20}{3})\).

The solution is \((r, s) = (4, \frac{20}{3})\).

- What value does a function have at a zero? - At an intersection point, what is true about the two function values? - Compare the results of the first two parts with the equation you solve in part c.

1. For part a, set \(f(x) = 0\): \(\frac{2}{3}x - 4 = 0\). Then \(\frac{2}{3}x = 4\), so \(x = 6\). 2. For part b, set \(g(x) = 0\): \(-\frac{1}{3}x + 2 = 0\). Then \(-\frac{1}{3}x = -2\), so \(x = 6\). 3. For part c, set the function values equal: \(\frac{2}{3}x - 4 = -\frac{1}{3}x + 2\). Adding \(\frac{1}{3}x\) gives \(x - 4 = 2\), so \(x = 6\). 4. Both functions equal \(0\) at \(x = 6\), so their intersection is exactly on the x-axis.

a) \(x = 6\) b) \(x = 6\) c) \(x = 6\). The two functions have the same zero, so they intersect on the x-axis.

- Can you simplify the first equation before choosing a method? - Which method is efficient once one variable is isolated? - Pay close attention to the negative factor before the parentheses. - Include values for both variables in your final answer.

1. Simplify equation (I) and solve for \(y\): \(2x - 6 = y + 2\), so \(y = 2x - 8\). 2. Substitute into equation (II): \(4x - 3(2x - 8) = 2\). 3. Distribute and combine like terms: \(4x - 6x + 24 = 2\), so \(-2x = -22\). 4. Solve for \(x\): \(x = 11\). 5. Substitute into \(y = 2x - 8\): \(y = 2 \cdot 11 - 8 = 14\).

The solution is \((11, 14)\).

- When two expressions give the same value of \(y\), set them equal. - How do you distribute a decimal factor across parentheses? - After finding \(x\), substitute it into either original equation to find \(y\). - Finally, multiply \(x\) and \(y\).

1. Set the expressions for \(y\) equal: \(2.4x - 7 = 0.4(x + 12.5)\). 2. Distribute on the right: \(0.4(x + 12.5) = 0.4x + 5\). 3. Solve \(2.4x - 7 = 0.4x + 5\). 4. Subtract \(0.4x\): \(2x - 7 = 5\). 5. Add \(7\): \(2x = 12\). 6. Divide by \(2\): \(x = 6\). 7. Find \(y\): \(y = 2.4 \cdot 6 - 7 = 7.4\). 8. Calculate the product: \(Z = 6 \cdot 7.4 = 44.4\).

\(x = 6\) and \(Z = 44.4\)

- Can you rearrange one equation so that one variable is alone on one side? - Which choice avoids introducing fractions? - After finding one variable, how can you calculate the other?

1. Solve equation (I) for \(x\): \(x = 4y - 18\). 2. Substitute into equation (II): \(2(4y - 18) + 5y = 16\). 3. Distribute: \(8y - 36 + 5y = 16\). 4. Combine like terms: \(13y - 36 = 16\). 5. Add \(36\) to both sides: \(13y = 52\). 6. Divide by \(13\): \(y = 4\). 7. Substitute into \(x = 4y - 18\): \(x = 4 \cdot 4 - 18 = -2\).

The solution is \((-2, 4)\).

- The substitution process is unchanged when the coefficients are decimals. - When multiplying by \(2\), distribute to every term inside the parentheses. - Combine the \(x\)-terms carefully before continuing to solve.

1. Substitute the expression for \(y\) from equation (I) into equation (II): \(5x + 2(2.4x - 3) = 33.2\). 2. Distribute: \(5x + 4.8x - 6 = 33.2\). 3. Combine like terms: \(9.8x - 6 = 33.2\). 4. Add \(6\): \(9.8x = 39.2\). 5. Divide by \(9.8\): \(x = 4\). 6. Substitute into equation (I): \(y = 2.4 \cdot 4 - 3 = 6.6\).

The solution is \((4, 6.6)\).

- Bring all \(x\)-terms to one side of the equation. - How can you clear or simplify the fractions? - Pay close attention to signs when substituting a negative value of \(x\).

1. Set the expressions equal: \(\frac{3}{4}x + 2 = \frac{1}{4}x - 1\). 2. Subtract \(\frac{1}{4}x\): \(\frac{1}{2}x + 2 = -1\). 3. Subtract \(2\): \(\frac{1}{2}x = -3\). 4. Multiply by \(2\): \(x = -6\). 5. Substitute into equation (II): \(y = \frac{1}{4}(-6) - 1 = -2.5\). 6. The intersection is \((-6, -2.5)\).

The intersection is \((-6, -2.5)\).

- Set the two function expressions equal. - Tangency corresponds to one repeated solution. - Use the discriminant to find \(n\), then substitute to find the point.

1. Set the functions equal: \(-x^2 + 3 = 2x + n\). 2. Rearrange: \(x^2 + 2x + (n - 3) = 0\). 3. Tangency means exactly one intersection, so the discriminant must be \(0\). 4. Calculate \(D = 2^2 - 4 \cdot 1 \cdot (n - 3) = 16 - 4n\). 5. Set \(16 - 4n = 0\), giving \(n = 4\). 6. Then \(x^2 + 2x + 1 = (x + 1)^2 = 0\), so \(x = -1\). The y-coordinate is \(g(-1) = 2\). 7. The point of tangency is \((-1, 2)\).

\(n = 4\); point of tangency \((-1, 2)\)

- Set the function expressions equal to find intersections. - Factor the resulting quadratic. - Test a point by substituting its x-coordinate into each function.

1. Set the functions equal: \(x^2 - 4x + 5 = x + 1\). 2. Rearrange and factor: \(x^2 - 5x + 4 = (x - 1)(x - 4) = 0\). 3. Thus, \(x = 1\) or \(x = 4\). Using \(g(x) = x + 1\), the corresponding y-values are \(2\) and \(5\). The intersections are \((1, 2)\) and \((4, 5)\). 4. For \(Q\), \(g(2.5) = 3.5\), so \(Q\) lies on the line. 5. However, \(p(2.5) = 2.5^2 - 4 \cdot 2.5 + 5 = 1.25\), not \(3.5\). Therefore, \(Q\) does not lie on the parabola.

a) \((1, 2)\) and \((4, 5)\) b) \(Q\) lies on the line only.

- What must be true about the slopes of parallel lines? - How can you use the given point to find the y-intercept? - What is true about the two function values at an intersection point?

1. Because \(h\) is parallel to \(g\), its slope is \(-\frac{2}{3}\). Write \(h(x) = -\frac{2}{3}x + b\). Using \(P(6, 2)\), \(2 = -\frac{2}{3} \cdot 6 + b\), so \(b = 6\). Thus \(h(x) = -\frac{2}{3}x + 6\). 2. At the intersection, the function values are equal: \(-\frac{2}{3}x + 6 = x - 3\). 3. Solving gives \(9 = \frac{5}{3}x\), so \(x = \frac{27}{5} = 5.4\). 4. Substitute into \(k\): \(y = 5.4 - 3 = 2.4\). Therefore, \(S(5.4, 2.4)\).

a) \(h(x) = -\frac{2}{3}x + 6\) b) \(S(5.4, 2.4)\)

- How can you find a line's slope from two points? - Can you read the y-intercept of one line directly from one of its points? - What must be true about the function values at an intersection point?

1. For \(f\), \(m_f = \frac{9 - 1}{3 - (-1)} = 2\). Using \(B(3, 9)\), \(9 = 2 \cdot 3 + b\), so \(b = 3\). Thus \(f(x) = 2x + 3\). 2. For \(g\), \(m_g = \frac{0 - 10}{5 - 0} = -2\). Since \(C(0, 10)\) is the y-intercept, \(g(x) = -2x + 10\). 3. At the intersection, \(2x + 3 = -2x + 10\). Then \(4x = 7\), so \(x = \frac{7}{4} = 1.75\). 4. Substitute into \(f\): \(y = 2 \cdot 1.75 + 3 = 6.5\). Therefore, \(S(1.75, 6.5)\).

a) \(f(x) = 2x + 3\) and \(g(x) = -2x + 10\) b) \(S(1.75, 6.5)\)

- Set the function expressions equal. - Look for a perfect-square trinomial. - A repeated solution corresponds to one common point.

1. Set the functions equal: \(x^2 - 4x + 3 = 2x - 6\). 2. Rearrange: \(x^2 - 6x + 9 = 0\). 3. Factor: \((x - 3)^2 = 0\), so there is one repeated solution, \(x = 3\). 4. The repeated solution means the graphs have exactly one common point, so the line is tangent to the parabola. 5. Calculate the y-coordinate: \(g(3) = 2 \cdot 3 - 6 = 0\). The point of tangency is \((3, 0)\).

The point of tangency is \((3, 0)\). The equation for the intersection has the repeated solution \(x = 3\).

- For each plan, identify the fixed fee and the rate per minute. - When the plans cost the same, what must be true of the two expressions? - What does \(x\) represent, and what unit should your answer use?

1. Plan A is modeled by \(y = 0.20x + 5\). Plan B is modeled by \(y = 0.45x\). 2. Set the expressions equal: \(0.20x + 5 = 0.45x\). 3. Subtract \(0.20x\): \(5 = 0.25x\). 4. Divide by \(0.25\): \(x = 20\). 5. Check the cost: \(y = 0.45 \cdot 20 = 9.00\), so both plans cost \(\$9.00\).

a) Plan A: \(y = 0.20x + 5\) Plan B: \(y = 0.45x\) b) The plans cost the same after \(20\) minutes. Each costs \(\$9.00\).

- Set the two expressions for \(y\) equal. - Rewrite the fractions with a common denominator. - Substitute the value of \(x\) into either original equation.

1. Set the expressions equal: \(\frac{2}{3}x + 5 = \frac{5}{6}x + 2\). 2. Subtract \(2\) and \(\frac{2}{3}x\): \(3 = \frac{5}{6}x - \frac{4}{6}x\). 3. Simplify: \(3 = \frac{1}{6}x\). 4. Multiply by \(6\): \(x = 18\). 5. Substitute into equation (I): \(y = \frac{2}{3} \cdot 18 + 5 = 17\).

The solution is \((18, 17)\).

- Set the function expressions equal. - Clear the decimal coefficient before solving. - Keep irrational coordinates in exact radical form.

1. Set the functions equal: \(0.5x^2 = x + 0.5\). 2. Multiply by \(2\) and rearrange: \(x^2 - 2x - 1 = 0\). 3. Apply the quadratic formula: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}\). 4. Using \(g(x) = x + 0.5\), the y-values are \(\frac{3}{2} + \sqrt{2}\) and \(\frac{3}{2} - \sqrt{2}\). 5. The intersection points are \(\left(1 + \sqrt{2}, \frac{3}{2} + \sqrt{2}\right)\) and \(\left(1 - \sqrt{2}, \frac{3}{2} - \sqrt{2}\right)\).

\(\left(1 + \sqrt{2}, \frac{3}{2} + \sqrt{2}\right)\) and \(\left(1 - \sqrt{2}, \frac{3}{2} - \sqrt{2}\right)\)

- Set the function expressions equal. - Expand the squared binomial and write a quadratic equation. - Use the discriminant to determine whether real intersections exist.

1. Set the functions equal: \((x - 2)^2 + 3 = x\). 2. Expand and rearrange: \(x^2 - 4x + 4 + 3 = x\), so \(x^2 - 5x + 7 = 0\). 3. The discriminant is \(D = (-5)^2 - 4 \cdot 1 \cdot 7 = -3\). 4. Since \(D < 0\), there are no real solutions, so the graphs have no common points.

The graphs have no common points.

- Choose variables for the two planned ticket amounts. - Use the planned total to write the first equation. - Express the percentages as decimal multipliers in the second equation. - After finding the planned amounts, calculate the actual sales.

1. Let \(x\) be the planned number of student tickets and \(y\) the planned number of adult tickets. 2. Write the system \(x + y = 500\) and \(1.05x + 0.90y = 495\). 3. Solve the first equation for \(y\): \(y = 500 - x\). 4. Substitute: \(1.05x + 0.90(500 - x) = 495\). 5. Simplify: \(1.05x + 450 - 0.90x = 495\), so \(0.15x = 45\) and \(x = 300\). 6. Then \(y = 500 - 300 = 200\). 7. Calculate the actual sales: \(1.05 \cdot 300 = 315\) student tickets and \(0.90 \cdot 200 = 180\) adult tickets.

The festival sold \(315\) student tickets and \(180\) adult tickets.

- Determine what the difference between the ending balance and the initial investment represents. - Write one equation for the total principal and another for the total interest. - Convert each percent to a decimal before using it in an equation. - Solve the system using a method such as substitution.

1. Let \(x\) be the amount invested at \(1.5\%\) and \(y\) the amount invested at \(2.5\%\). 2. The total interest earned is \(12{,}230 - 12{,}000 = 230\) dollars. 3. Write the system \(x + y = 12{,}000\) and \(0.015x + 0.025y = 230\). 4. Solve the first equation for \(y\): \(y = 12{,}000 - x\). 5. Substitute: \(0.015x + 0.025(12{,}000 - x) = 230\). 6. Simplify: \(0.015x + 300 - 0.025x = 230\), so \(-0.01x = -70\) and \(x = 7000\). 7. Then \(y = 12{,}000 - 7000 = 5000\).

Ms. Weber invested \(\$7000\) in the \(1.5\%\) account and \(\$5000\) in the \(2.5\%\) account.

- Which equation is easiest to solve for one variable? - Pay close attention to the negative factor before the parentheses. - After finding one variable, where can you substitute it most easily to find the other? - Check both values in the original equations.

1. Solve equation (I) for \(y\): \(y = 7 - 2x\). 2. Substitute into equation (II): \(3x - 2(7 - 2x) = 21\). 3. Distribute carefully: \(3x - 14 + 4x = 21\). 4. Combine like terms: \(7x - 14 = 21\). 5. Solve for \(x\): \(7x = 35\), so \(x = 5\). 6. Substitute into \(y = 7 - 2x\): \(y = 7 - 2 \cdot 5 = -3\). 7. The solution set is \(\{(5, -3)\}\).

The solution set is \(\{(5, -3)\}\).

- In which equation is one variable easiest to isolate? - Look for a variable with coefficient \(1\). - Substitute the resulting expression into the equation you have not changed. - Check both values in the original equations.

1. Solve the first equation for \(y\), since its coefficient is \(1\): \(y = 10 - 3x\). 2. Substitute into the second equation: \(2x + 3(10 - 3x) = 2\). 3. Distribute: \(2x + 30 - 9x = 2\). 4. Combine like terms and solve: \(-7x + 30 = 2\), so \(-7x = -28\) and \(x = 4\). 5. Substitute into \(y = 10 - 3x\): \(y = 10 - 3 \cdot 4 = -2\). 6. The solution set is \(\{(4, -2)\}\).

The solution set is \(\{(4, -2)\}\).

- Look for a variable with coefficient \(1\) to make isolating it easier. - Pay close attention to signs when distributing. - After finding one variable, substitute it into an original equation to find the other.

1. Solve equation (II) for \(x\), since its coefficient is \(1\): \(x = 3y + 5\). 2. Substitute into equation (I): \(2(3y + 5) + 5y = -1\). 3. Distribute: \(6y + 10 + 5y = -1\). 4. Combine like terms: \(11y + 10 = -1\). 5. Solve for \(y\): \(11y = -11\), so \(y = -1\). 6. Substitute into \(x = 3y + 5\): \(x = 3(-1) + 5 = 2\). 7. The solution is \((2, -1)\).

The solution is \((2, -1)\).

- Can you isolate a multiple of a variable without introducing fractions? - Can an expression for \(2y\) be used directly to replace \(4y\)? - Remember to determine both \(x\) and \(y\).

1. Solve equation (I) for \(2y\): \(2y = 5x - 19\). 2. Since \(4y = 2(2y)\), substitute \(2(5x - 19)\) for \(4y\) in equation (II): \(3x + 2(5x - 19) = 1\). 3. Solve for \(x\): \(3x + 10x - 38 = 1\), so \(13x = 39\) and \(x = 3\). 4. Substitute into \(2y = 5x - 19\): \(2y = 5 \cdot 3 - 19 = -4\), so \(y = -2\). 5. The solution is \((3, -2)\).

The solution set is \(\{(3, -2)\}\).

- When no coefficient is \(1\), isolating a variable may introduce a fraction. - Which variable is most convenient to isolate? - How can you clear denominators after substituting? - Pay attention to signs when substituting a negative value.

1. Solve equation (I) for \(x\): \(3x = 11 + 5y\), so \(x = \frac{11 + 5y}{3}\). 2. Substitute into equation (II): \(4 \cdot \frac{11 + 5y}{3} + 3y = 5\). 3. Multiply by \(3\) to clear the denominator: \(4(11 + 5y) + 9y = 15\). 4. Distribute and combine like terms: \(44 + 20y + 9y = 15\), so \(44 + 29y = 15\). 5. Solve for \(y\): \(29y = -29\), so \(y = -1\). 6. Substitute into the expression for \(x\): \(x = \frac{11 + 5(-1)}{3} = 2\). 7. The solution set is \(\{(2, -1)\}\).

The solution set is \(\{(2, -1)\}\).

- Solve one equation for one variable. Decide whether fractions or decimals are more convenient. - Substitute the expression into the other equation and distribute carefully. - Check the final ordered pair in both original equations.

1. Solve equation (I) for \(y\): \(2y = 5 - 3x\), so \(y = 2.5 - 1.5x\). 2. Substitute into equation (II): \(5x + 3(2.5 - 1.5x) = 8\). 3. Distribute: \(5x + 7.5 - 4.5x = 8\). 4. Combine like terms and solve: \(0.5x + 7.5 = 8\), so \(0.5x = 0.5\) and \(x = 1\). 5. Substitute into the expression for \(y\): \(y = 2.5 - 1.5 \cdot 1 = 1\). 6. Check equation (I): \(3 \cdot 1 + 2 \cdot 1 = 5\). Check equation (II): \(5 \cdot 1 + 3 \cdot 1 = 8\). Both equations are true. 7. The solution is \((1, 1)\).

The solution is \((1, 1)\).

- Which variable is easiest to isolate? - Treat \(p\) as a constant while solving for \(x\) and \(y\). - Put parentheses around the substituted expression. - Your final ordered pair should be written in terms of \(p\).

1. Solve equation (II) for \(y\): \(y = 2x - 3p\). 2. Substitute into equation (I): \(x + 2(2x - 3p) = 4p\). 3. Solve for \(x\): \(x + 4x - 6p = 4p\), so \(5x = 10p\) and \(x = 2p\). 4. Substitute into the expression for \(y\): \(y = 2(2p) - 3p = p\).

The solution is \((x, y) = (2p, p)\).

- No variable is isolated, so begin by rearranging one equation. - Dividing by a coefficient may produce manageable decimals. - Check your result in both original equations.

1. Solve the second equation for \(y\): \(-2y = 14 - 5x\), so \(y = 2.5x - 7\). 2. Substitute into the first equation: \(4x + 3(2.5x - 7) = 2\). 3. Distribute and solve for \(x\): \(4x + 7.5x - 21 = 2\), so \(11.5x = 23\). 4. Therefore, \(x = 2\). 5. Substitute into the expression for \(y\): \(y = 2.5 \cdot 2 - 7 = -2\). 6. The solution is \((2, -2)\).

The solution set is \(\{(2, -2)\}\).

- How can you clear the fraction in the first equation? - Rewrite both equations in the form \(Aa + Bb = C\). - Which variable is easiest to isolate? - Check both values in the original equations.

1. Simplify the first equation: \(\frac{1}{2}a + 2 = b + 1\), so \(\frac{1}{2}a - b = -1\). Multiply by \(2\) to get \(a - 2b = -2\). 2. Simplify the second equation: \(3a - 3b = 2b + 11\), so \(3a - 5b = 11\). 3. From the first simplified equation, \(a = 2b - 2\). 4. Substitute into the second simplified equation: \(3(2b - 2) - 5b = 11\). 5. Solve for \(b\): \(6b - 6 - 5b = 11\), so \(b = 17\). 6. Find \(a\): \(a = 2 \cdot 17 - 2 = 32\).

The solution is \((a, b) = (32, 17)\).

- Begin by expanding and combining like terms in each equation. - Could the expressions \(x + y\) and \(x - y\) also be treated as temporary variables? - Which system-solving method is most efficient after simplification?

1. Expand and simplify each equation. The first becomes \(2x + 2y + 3x - 3y = 14\), so \(5x - y = 14\). The second becomes \(5x + 5y - 2x + 2y = 16\), so \(3x + 7y = 16\). 2. Solve the first simplified equation for \(y\): \(y = 5x - 14\). 3. Substitute into the second simplified equation: \(3x + 7(5x - 14) = 16\). 4. Distribute and combine like terms: \(3x + 35x - 98 = 16\), so \(38x = 114\). 5. Solve for \(x\): \(x = 3\). 6. Substitute into \(y = 5x - 14\): \(y = 1\). 7. The solution is \((3, 1)\).

The solution is \((3, 1)\).

- Choose variables for last year’s membership in each program. - Use the combined membership from last year to write one equation. - Represent the percentage increase and decrease with decimal multipliers. - Express one variable in terms of the other and substitute.

1. Let \(x\) be last year’s soccer membership and \(y\) last year’s tennis membership. 2. Write the system \(x + y = 200\) and \(1.1x + 0.9y = 202\). 3. Solve the first equation for \(y\): \(y = 200 - x\). 4. Substitute: \(1.1x + 0.9(200 - x) = 202\). 5. Simplify: \(1.1x + 180 - 0.9x = 202\), so \(0.2x = 22\) and \(x = 110\). 6. Then \(y = 200 - 110 = 90\).

Last year, the soccer program had \(110\) members, and the tennis program had \(90\) members.

- Identify whether the unknown variables should represent last month or this month. - Express a \(5\%\) decrease and a \(20\%\) increase with decimal multipliers. - Pay attention to which month the question asks about. - A table comparing last month and this month may help organize the information.

1. Let \(b\) be last month’s number of Basic members and \(p\) last month’s number of Premium members. 2. Write the system \(b + p = 600\) and \(0.95b + 1.2p = 630\). 3. Solve the first equation for \(p\): \(p = 600 - b\). 4. Substitute: \(0.95b + 1.2(600 - b) = 630\). 5. Simplify: \(0.95b + 720 - 1.2b = 630\), so \(-0.25b = -90\) and \(b = 360\). 6. Then \(p = 600 - 360 = 240\). 7. Calculate this month’s membership: \(0.95 \cdot 360 = 342\) Basic members and \(1.2 \cdot 240 = 288\) Premium members.

This month, \(342\) members use the Basic plan and \(288\) members use the Premium plan.

- First determine the technicians’ combined hourly rate. - Express the \(180\) completed components as work done during the individual and joint time periods. - Write a system with the two unknown individual rates. - Use the known combined rate in the second equation.

1. Let \(x\) be Mr. Allen’s rate and \(y\) Ms. Baker’s rate, in components per hour. 2. Their combined work gives \(8(x + y) = 240\), so \(x + y = 30\). 3. The second situation gives \(3x + 4(x + y) = 180\). 4. Substitute \(x + y = 30\): \(3x + 4 \cdot 30 = 180\). 5. Solve: \(3x + 120 = 180\), so \(x = 20\). 6. Then \(20 + y = 30\), so \(y = 10\).

Mr. Allen processes \(20\) components per hour, and Ms. Baker processes \(10\) components per hour.

- Interpret flow rate as the volume moved per unit of time. - Write one equation for each operating schedule. - Once you know both rates, add them to find the combined rate. - Divide the tank’s capacity by the combined rate to find the filling time.

1. Let \(a\) and \(b\) be the flow rates of Pumps A and B, in gallons per minute. 2. Write the system \(3a + 2b = 110\) and \(2a + 4b = 140\). 3. Divide the second equation by \(2\): \(a + 2b = 70\), so \(a = 70 - 2b\). 4. Substitute into the first equation: \(3(70 - 2b) + 2b = 110\). 5. Simplify: \(210 - 6b + 2b = 110\), so \(-4b = -100\) and \(b = 25\). 6. Then \(a = 70 - 2 \cdot 25 = 20\). 7. Together, the pumps move \(20 + 25 = 45\,\text{gal}\) per minute. Filling \(225\,\text{gal}\) takes \(225 \div 45 = 5\) minutes.

a) Pump A moves \(20\,\text{gal}\) per minute, and Pump B moves \(25\,\text{gal}\) per minute. b) Together, the pumps would fill the tank in \(5\) minutes.

- Set the two expressions for \(y\) equal. - Use the discriminant to test for intersections. - Write the parabola in vertex form to interpret the result geometrically.

1. Set the two expressions for \(y\) equal: \(x^2 - 4x + 6 = 1\). 2. Rearrange: \(x^2 - 4x + 5 = 0\). 3. The discriminant is \(D = (-4)^2 - 4 \cdot 1 \cdot 5 = -4\). Since \(D < 0\), there are no real intersection points. 4. Complete the square: \(y = x^2 - 4x + 6 = (x - 2)^2 + 2\). The vertex is \((2, 2)\). 5. The parabola opens upward and has minimum value \(2\), so it lies entirely above the line \(y = 1\).

The system has no solution. The parabola has vertex \((2, 2)\) and lies entirely above \(y = 1\).

- Set the two function expressions equal. - Clear decimals before calculating the discriminant. - A negative discriminant means the line and parabola do not intersect.

1. Set the functions equal: \(-0.5x^2 + 2x - 4 = 0.5x - 1\). 2. Rearrange: \(-0.5x^2 + 1.5x - 3 = 0\). 3. Multiply by \(-2\): \(x^2 - 3x + 6 = 0\). 4. The discriminant is \(D = (-3)^2 - 4 \cdot 1 \cdot 6 = -15\). 5. Since \(D < 0\), the graphs have no common points. The line does not intersect the parabola.

The graphs have no common points; the line does not intersect the parabola.

- Both equations already give \(y\) in terms of \(x\). - Set the right sides equal and factor. - Substitute each x-value to find its paired y-value.

1. Both equations are solved for \(y\), so their right sides must be equal at an intersection. 2. Set them equal: \(x^2 - 4x + 1 = 2x - 4\). 3. Rearrange and factor: \(x^2 - 6x + 5 = (x - 1)(x - 5) = 0\). 4. Thus, \(x = 1\) or \(x = 5\). 5. Substitute into \(y = 2x - 4\): the corresponding y-values are \(-2\) and \(6\). 6. The solutions are \((1, -2)\) and \((5, 6)\).

\(\{(1, -2), (5, 6)\}\)

- Solve the linear equation for one variable. - Substitute that expression into the squared equation. - Remember both cases when taking a square root.

1. Solve the linear equation for \(x\): \(x = y + 1\). 2. Substitute into the first equation: \((y + 1 + y)^2 = 25\), so \((2y + 1)^2 = 25\). 3. Take both square roots: \(2y + 1 = 5\) or \(2y + 1 = -5\). 4. The first case gives \(y = 2\) and \(x = 3\). The second case gives \(y = -3\) and \(x = -2\). 5. Therefore, the system has exactly two ordered-pair solutions.

\(\{(3, 2), (-2, -3)\}\)

- Solve the linear equation for one variable. - Substitute into the quadratic equation and keep the parentheses. - Factor the resulting quadratic, then find each matching value of the other variable.

1. Solve the linear equation for \(y\): \(y = 3 - x\). 2. Substitute into the quadratic equation: \(3x^2 - (3 - x)^2 = 11\). 3. Expand and simplify: \(3x^2 - (9 - 6x + x^2) = 11\), so \(2x^2 + 6x - 20 = 0\). 4. Divide by \(2\): \(x^2 + 3x - 10 = 0\). 5. Factor: \((x - 2)(x + 5) = 0\). Thus, \(x = 2\) or \(x = -5\). 6. Using \(y = 3 - x\), the corresponding values are \(y = 1\) and \(y = 8\).

\(\{(2, 1), (-5, 8)\}\)

- Isolate \(y\) in the linear equation. - Substitute carefully into every occurrence of \(y\). - Combine like terms carefully before solving the resulting equation.

1. Solve the second equation for \(y\): \(y = x + 2\). 2. Substitute into the first equation: \(x^2 - 2x(x + 2) + (x + 2)^2 + 2x + 2(x + 2) = 8\). 3. Expand and combine like terms: \(4x + 8 = 8\). 4. Solve: \(x = 0\). 5. Substitute into \(y = x + 2\): \(y = 2\). 6. The ordered pair \((0, 2)\) satisfies both original equations.

\(\{(0, 2)\}\)

- Solve the linear equation for one variable. - Substitute into the quadratic equation and expand carefully. - Factor the resulting quadratic, then find the matching value of the other variable.

1. Solve the linear equation for \(x\): \(x = 3 - y\). 2. Substitute into the quadratic equation: \((3 - y)^2 + 2y^2 = 33\). 3. Expand and simplify: \(9 - 6y + 3y^2 = 33\), so \(y^2 - 2y - 8 = 0\). 4. Factor: \((y - 4)(y + 2) = 0\). Thus, \(y = 4\) or \(y = -2\). 5. Using \(x = 3 - y\), the corresponding values are \(x = -1\) and \(x = 5\).

\(\{(-1, 4), (5, -2)\}\)

- Set the function expressions equal. - For part b), express the discriminant in terms of \(m\). - A tangent gives one repeated intersection x-value.

1. When \(m = -2\), set \(x^2 - 4x + 3 = -2x + 3\). 2. This simplifies to \(x^2 - 2x = x(x - 2) = 0\), so \(x = 0\) or \(x = 2\). 3. Using \(y = -2x + 3\), the intersection points are \((0, 3)\) and \((2, -1)\). 4. In general, setting the functions equal gives \(x^2 - (4 + m)x = 0\). 5. Its discriminant is \(D = (4 + m)^2\). Tangency requires \(D = 0\), so \(m = -4\). 6. Then the repeated solution is \(x = 0\), and \(y = 3\). The point of tangency is \((0, 3)\).

a) \((0, 3)\) and \((2, -1)\) b) \(m = -4\); point of tangency \((0, 3)\)

- Set the line equal to the parabola. - Require the resulting quadratic to have discriminant \(0\). - Use the repeated x-value to find the y-coordinate.

1. Set the functions equal: \(x^2 - 4x + 5 = 2x + c\). 2. Rearrange: \(x^2 - 6x + (5 - c) = 0\). 3. The discriminant is \(D = (-6)^2 - 4 \cdot 1 \cdot (5 - c) = 16 + 4c\). 4. Tangency requires \(D = 0\), so \(16 + 4c = 0\) and \(c = -4\). 5. Then \(x^2 - 6x + 9 = (x - 3)^2 = 0\), so \(x = 3\). 6. The y-coordinate is \(2 \cdot 3 - 4 = 2\). The point of tangency is \((3, 2)\).

\(c = -4\); point of tangency \((3, 2)\)

- Solve the sum equation for one variable. - Substitute into the product equation. - Factor the resulting quadratic and find each corresponding ordered pair.

1. Solve the first equation for \(y\): \(y = -2 - x\). 2. Substitute into the product equation: \(x(-2 - x) = -15\). 3. Rewrite in standard form: \(x^2 + 2x - 15 = 0\). 4. Factor: \((x + 5)(x - 3) = 0\). Thus, \(x = -5\) or \(x = 3\). 5. The corresponding values of \(y\) are \(3\) and \(-5\).

\(\{(-5, 3), (3, -5)\}\)

- Apply the zero-product property to the second equation. - Substitute each resulting fixed value into the circle equation. - List each distinct ordered pair only once.

1. By the zero-product property, \(x - 3 = 0\) or \(y - 4 = 0\). Thus, consider the cases \(x = 3\) and \(y = 4\). 2. If \(x = 3\), then \(9 + y^2 = 25\), so \(y = 4\) or \(y = -4\). This gives \((3, 4)\) and \((3, -4)\). 3. If \(y = 4\), then \(x^2 + 16 = 25\), so \(x = 3\) or \(x = -3\). This gives \((3, 4)\) and \((-3, 4)\). 4. Remove the repeated ordered pair \((3, 4)\).

\(\{(3, 4), (3, -4), (-3, 4)\}\)

- Solve the linear equation for one variable. - Substitute into the quadratic equation and expand. - Factor the resulting quadratic, then find each matching coordinate.

1. Solve the linear equation for \(x\): \(x = y + 5\). 2. Substitute into the quadratic equation: \((y + 5)^2 + y^2 = 13\). 3. Expand and simplify: \(2y^2 + 10y + 12 = 0\). Divide by \(2\): \(y^2 + 5y + 6 = 0\). 4. Factor: \((y + 2)(y + 3) = 0\). Thus, \(y = -2\) or \(y = -3\). 5. Using \(x = y + 5\), the corresponding values are \(x = 3\) and \(x = 2\).

\(\{(3, -2), (2, -3)\}\)

- Set the two expressions for \(y\) equal. - Factor the resulting quadratic. - Substitute each x-value into either equation to find the paired y-value.

1. Set the function expressions equal: \(x^2 - 3 = 2x\). 2. Rearrange and factor: \(x^2 - 2x - 3 = (x - 3)(x + 1) = 0\). 3. Thus, \(x = -1\) or \(x = 3\). 4. Substituting into \(y = 2x\) gives \(y = -2\) and \(y = 6\). 5. The solutions are \((-1, -2)\) and \((3, 6)\); these are the graphs’ intersection points.

\((-1, -2)\) and \((3, 6)\)

- State the value excluded by the denominator. - Solve the rational equation for one variable. - Substitute into the quadratic equation, then verify the domain restriction.

1. The denominator requires \(y \ne 1\). 2. Solve the first equation for \(x\): \(x + 2 = 2(y - 1)\), so \(x = 2y - 4\). 3. Substitute into the second equation: \((2y - 4)^2 + y^2 = 13\). 4. Expand and simplify: \(5y^2 - 16y + 3 = 0\). 5. Factor: \((5y - 1)(y - 3) = 0\). Thus, \(y = \frac{1}{5}\) or \(y = 3\). 6. Using \(x = 2y - 4\), the corresponding values are \(x = -\frac{18}{5}\) and \(x = 2\). Both y-values satisfy \(y \ne 1\).

\(\left\{\left(-\frac{18}{5}, \frac{1}{5}\right), (2, 3)\right\}\)

- Solve the linear equation for one variable. - Substitute into the sum-of-squares equation. - Factor the resulting quadratic and find each corresponding coordinate.

1. Solve the linear equation for \(y\): \(y = 1 - x\). 2. Substitute into the second equation: \(x^2 + (1 - x)^2 = 13\). 3. Expand and simplify: \(2x^2 - 2x - 12 = 0\). Divide by \(2\): \(x^2 - x - 6 = 0\). 4. Factor: \((x - 3)(x + 2) = 0\). Thus, \(x = 3\) or \(x = -2\). 5. Using \(y = 1 - x\), the corresponding values are \(y = -2\) and \(y = 3\).

\(\{(3, -2), (-2, 3)\}\)

- What is the general shape of a graph whose variable is in the denominator? - Use a graphing tool to display both equations on the same coordinate plane. - Make sure the reciprocal graph is viewed for both positive and negative inputs. - What do common points of the two graphs represent?

1. Enter \(y = \frac{4}{x}\) and \(y = x + 3\) in a graphing tool and display both graphs on the same coordinate plane. 2. The reciprocal graph includes points such as \((1, 4)\), \((2, 2)\), \((4, 1)\), \((-1, -4)\), \((-2, -2)\), and \((-4, -1)\). The line has slope \(1\) and \(y\)-intercept \(3\). 3. The graphs intersect at \((1, 4)\) and \((-4, -1)\). 4. These intersection points are the solutions of the system.

The graphs intersect at \((1, 4)\) and \((-4, -1)\), so these are the solutions.

- Split the equation so that each side becomes a function rule. - Remember that \(x = 0\) is not in the domain of the reciprocal function. - Use a graphing tool to find the intersection points, then identify the requested \(x\)-coordinates. - How can substitution confirm that a proposed value satisfies the original equation?

1. Use \(f(x) = \frac{6}{x}\) and \(g(x) = x + 1\). 2. Enter both functions in a graphing tool and display them on the same coordinate plane. 3. The graphs intersect at \((2, 3)\) and \((-3, -2)\), so the solutions are \(x = 2\) and \(x = -3\). 4. Check \(x = 2\): \(\frac{6}{2} = 3\) and \(2 + 1 = 3\). 5. Check \(x = -3\): \(\frac{6}{-3} = -2\) and \(-3 + 1 = -2\). Both values satisfy the original equation.

a) \(f(x) = \frac{6}{x}\) and \(g(x) = x + 1\) b) The graphs intersect at \((2, 3)\) and \((-3, -2)\), so \(x = 2\) and \(x = -3\). c) \(\frac{6}{2} = 2 + 1 = 3\), and \(\frac{6}{-3} = -3 + 1 = -2\).

- Read the y-intercept and choose another grid point on each line. - Use the two points to calculate each slope. - To get integer coefficients, clear any fractions before rearranging. - At the intersection, the two expressions for \(y\) are equal.

1. Line \(f\) crosses the y-axis at \((0, -2)\) and passes through \((2, 1)\). Its slope is \(\frac{1 - (-2)}{2 - 0} = \frac{3}{2}\), so \(f\) has equation \(y = \frac{3}{2}x - 2\). 2. Line \(g\) crosses the y-axis at \((0, 3)\) and passes through \((2, 1)\). Its slope is \(-1\), so \(g\) has equation \(y = -x + 3\). 3. Rewrite the first equation with integer coefficients: \(2y = 3x - 4\), so \(3x - 2y = 4\). The second becomes \(x + y = 3\). 4. At the intersection, \(\frac{3}{2}x - 2 = -x + 3\). Solving gives \(\frac{5}{2}x = 5\), so \(x = 2\). Then \(y = 1\), so \(S(2, 1)\).

a) \(f: y = \frac{3}{2}x - 2\); \(g: y = -x + 3\) b) \(f: 3x - 2y = 4\); \(g: x + y = 3\) c) \(S(2, 1)\)

- Read the marked coordinates carefully. - Use the vertex to identify each parabola. - Use two points to determine each line, then set the functions equal.

1. Graph a): The intersections are \((-1, 1)\) and \((3, 9)\), so the x-values are \(-1\) and \(3\). 2. The parabola has vertex \((0, 0)\), so \(f(x) = x^2\). The line has slope \(2\) and y-intercept \(3\), so \(g(x) = 2x + 3\). 3. Setting the functions equal gives \(x^2 = 2x + 3\), or \(x^2 - 2x - 3 = 0\). 4. Graph b): The intersections are \((-2, 0)\) and \((1, -3)\), so the x-values are \(-2\) and \(1\). 5. The parabola has vertex \((0, -4)\), so \(f(x) = x^2 - 4\). The line has slope \(-1\) and y-intercept \(-2\), so \(g(x) = -x - 2\). 6. Setting the functions equal gives \(x^2 - 4 = -x - 2\), or \(x^2 + x - 2 = 0\).

a) Intersections: \((-1, 1)\), \((3, 9)\); \(f(x) = x^2\); \(g(x) = 2x + 3\); equation \(x^2 - 2x - 3 = 0\) b) Intersections: \((-2, 0)\), \((1, -3)\); \(f(x) = x^2 - 4\); \(g(x) = -x - 2\); equation \(x^2 + x - 2 = 0\)

- Interpret each side of an equation as a function. - Match vertex positions and horizontal or slanted lines. - Solutions are the x-coordinates of intersections.

1. Graph a) shows \(y = x^2 - 4x + 3\) and \(y = 0\), so it matches equation (1). 2. Graph b) shows \(y = x^2\) and \(y = 4x - 3\), so it matches equation (2). 3. Graph c) shows \(y = x^2 - 4x\) and \(y = -3\), so it matches equation (3). 4. Graph d) shows \(y = (x - 2)^2\) and \(y = 1\), so it matches equation (4). 5. The intersections are: a) \((1, 0)\) and \((3, 0)\); b) \((1, 1)\) and \((3, 9)\); c) \((1, -3)\) and \((3, -3)\); d) \((1, 1)\) and \((3, 1)\). 6. Every graph gives the same solution x-values, \(1\) and \(3\), because the equations are equivalent.

a) (1) → a), (2) → b), (3) → c), (4) → d) b) The solution set for every equation is \(\{1, 3\}\). The y-coordinates differ, but the intersection x-coordinates are the same.

- Read the intersection coordinates from the axes. - Use point-slope information for the line and vertex form for the parabola. - Set the equations equal and factor.

1. From the graph, the intersection points are \((-1, 0)\) and \((4, 5)\). 2. Write \(g(x) = x + b\). Substituting \((-1, 0)\) gives \(b = 1\), so \(g(x) = x + 1\). 3. Write \(f(x) = a(x - 1)^2 - 4\). Substituting \((3, 0)\) gives \(0 = 4a - 4\), so \(a = 1\). 4. Therefore, \(f(x) = (x - 1)^2 - 4 = x^2 - 2x - 3\). 5. Set the functions equal: \(x^2 - 2x - 3 = x + 1\), so \(x^2 - 3x - 4 = 0\). 6. Factor: \((x - 4)(x + 1) = 0\), giving \(x = -1\) or \(x = 4\). 7. Substitution into \(g\) gives the points \((-1, 0)\) and \((4, 5)\), matching the graph.

1. \((-1, 0)\) and \((4, 5)\) 2. \(g(x) = x + 1\), \(f(x) = x^2 - 2x - 3\) 3. The calculated intersections are \((-1, 0)\) and \((4, 5)\).

- Identify which two graphs meet at each labeled point. - Set the corresponding function expressions equal. - Factor the quadratic equations.

1. From the graph, \(g(x) = -x + 2\) and \(h(x) = x\). 2. Points \(A\) and \(D\) are intersections of \(p\) and \(g\), so solve \(x^2 - 2x - 4 = -x + 2\). 3. This becomes \(x^2 - x - 6 = (x - 3)(x + 2) = 0\), so the x-coordinates are \(-2\) and \(3\). 4. Points \(B\) and \(E\) are intersections of \(p\) and \(h\), so solve \(x^2 - 2x - 4 = x\). 5. This becomes \(x^2 - 3x - 4 = (x - 4)(x + 1) = 0\), so the x-coordinates are \(-1\) and \(4\). 6. Point \(C\) is the intersection of \(g\) and \(h\), so solve \(-x + 2 = x\), giving \(x = 1\).

a) \(A,D\): \(x^2 - 2x - 4 = -x + 2\) \(B,E\): \(x^2 - 2x - 4 = x\) \(C\): \(-x + 2 = x\) b) \(A,D\): \(x = -2, 3\); \(B,E\): \(x = -1, 4\); \(C\): \(x = 1\)

- Use the vertex and one point for the parabola. - Use slope and y-intercept for the line. - Set the functions equal and factor.

1. The parabola has vertex \((1, 4)\) and passes through \((2, 3)\), so \(f(x) = -(x - 1)^2 + 4 = -x^2 + 2x + 3\). 2. The line has slope \(1\) and y-intercept \(1\), so \(g(x) = x + 1\). 3. Set the functions equal: \(-x^2 + 2x + 3 = x + 1\). 4. Rearrange: \(x^2 - x - 2 = 0\). 5. Factor: \((x - 2)(x + 1) = 0\), so \(x = -1\) or \(x = 2\). 6. Substituting into \(g\) gives \((-1, 0)\) and \((2, 3)\), which match the graph.

a) \(f(x) = -x^2 + 2x + 3\), \(g(x) = x + 1\) b) \(x^2 - x - 2 = 0\) c) \((-1, 0)\) and \((2, 3)\)

- Isolate \(x^2\). - Read the marked coordinates carefully. - Factor the quadratic for the exact check.

1. Rewrite the equation as \(x^2 = x + 2\), so \(g(x) = x + 2\). 2. The graph shows intersections \(A(-1, 1)\) and \(B(2, 4)\). 3. Algebraically, factor \(x^2 - x - 2 = (x - 2)(x + 1)\). 4. Thus, \(x = -1\) or \(x = 2\), matching the graph.

a) \(g(x) = x + 2\) b) \(A(-1, 1)\), \(B(2, 4)\) c) \(x = -1\) and \(x = 2\)

- Read a point whose coordinates lie exactly on grid lines. - Substitute that point into \(f(x)=ax^4\). - Write the equation of the horizontal line and set the two function values equal. - An even power equal to a positive number gives two real solutions.

1. The graph of \(f\) passes through \((2, 8)\). Substitute into \(f(x)=ax^4\): \(8=a(2)^4=16a\), so \(a=0.5\). Thus \(f(x)=0.5x^4\). 2. The line is \(g(x)=3\). Set the functions equal: \(0.5x^4=3\). 3. From the graph, the intersections occur at approximately \(x=-1.6\) and \(x=1.6\). 4. Algebraically, \(x^4=6\), so \(x=\pm\sqrt[4]{6}\approx\pm1.565\).

a) \(a=0.5\), so \(f(x)=0.5x^4\) b) \(0.5x^4=3\) c) \(x\approx-1.6\) and \(x\approx1.6\) d) \(x=\pm\sqrt[4]{6}\approx\pm1.565\)

- Fixed costs are the costs when production is \(0\). - One process is cheaper when its graph lies below the other cost graph. - A break-even point is an intersection of a revenue graph and a cost graph.

1. Fixed cost is the y-intercept of each cost graph. Process A has a fixed cost of \(\$200\), and Process B has a fixed cost of \(\$50\). Therefore, Process A has the greater fixed cost. 2. The cost graphs intersect at \((150,275)\). For \(x>150\), the graph of \(K_A\) lies below the graph of \(K_B\). Because production quantities are whole numbers, Process A is less expensive starting at \(151\) cases. 3. The graph of \(R\) intersects \(K_B\) at \((50,125)\), so Process B breaks even at \(50\) cases. The graph of \(R\) intersects \(K_A\) at \((100,250)\), so Process A breaks even at \(100\) cases.

a) Process A; Process A: \(\$200\), Process B: \(\$50\) b) For \(x>150\), so starting at \(151\) cases c) Process B: \(50\) cases; Process A: \(100\) cases

- Read both coordinates at each crossing. - Set the function expressions equal. - Factor out \(x\).

1. From the graph, the intersections are \((0, 0)\) and \((3, 3)\). 2. Set the functions equal: \(-x^2 + 4x = x\). 3. Rearrange and factor: \(x^2 - 3x = x(x - 3) = 0\). 4. Thus, \(x = 0\) or \(x = 3\). 5. Since \(g(x) = x\), the corresponding y-values are \(0\) and \(3\), confirming the graph.

a) \((0, 0)\) and \((3, 3)\) b) The calculation gives the same points.

- Find the line’s slope from two points. - Set the line equal to the parabola. - A tangent produces a repeated solution.

1. The parabola is \(f(x) = x^2\). 2. The line passes through \((1, 0)\) and \((2, 4)\), so its slope is \(4\) and its equation is \(g(x) = 4x - 4\). 3. Set the functions equal: \(x^2 = 4x - 4\). 4. Move all terms to one side: \(x^2 - 4x + 4 = 0\). 5. This is \((x - 2)^2 = 0\), so it has the single repeated solution \(x = 2\).

\(x^2 - 4x + 4 = 0\)

- Identify which graphs meet at each labeled point. - A point on the x-axis has a function value of \(0\). - Use the vertex and intercepts to determine the function equations. - Set the appropriate function expressions equal to find an intersection's x-coordinate.

1. Point \(A\) lies on all three graphs, so its x-coordinate satisfies \(f(x)=g(x)\), \(f(x)=h(x)\), or \(g(x)=h(x)\). Point \(B\) satisfies \(f(x)=h(x)\). Point \(C\) satisfies \(f(x)=g(x)\), and because it lies on the x-axis, it also satisfies \(f(x)=0\). Point \(D\) satisfies \(f(x)=0\). 2. The horizontal line is \(h(x)=3\). The line \(g\) passes through \((0,3)\) and \((3,0)\), so its slope is \(-1\) and \(g(x)=-x+3\). 3. The parabola has vertex \((2,-1)\), so write \(f(x)=a(x-2)^2-1\). Since it passes through \((0,3)\), \(3=4a-1\), which gives \(a=1\). Thus \(f(x)=(x-2)^2-1=x^2-4x+3\). 4. For \(B\), solve \(f(x)=h(x)\): \(x^2-4x+3=3\), so \(x(x-4)=0\). The solution \(x=0\) belongs to \(A\), so the x-coordinate of \(B\) is \(4\). 5. For \(C\), solve \(f(x)=g(x)\): \(x^2-4x+3=-x+3\), so \(x(x-3)=0\). The solution \(x=0\) belongs to \(A\), so the x-coordinate of \(C\) is \(3\).

\(A\): \(f(x)=g(x)\), \(f(x)=h(x)\), or \(g(x)=h(x)\) \(B\): \(f(x)=h(x)\) \(C\): \(f(x)=g(x)\) or \(f(x)=0\) \(D\): \(f(x)=0\) The functions are \(f(x)=x^2-4x+3\), \(g(x)=-x+3\), and \(h(x)=3\). The x-coordinate of \(B\) is \(4\), and the x-coordinate of \(C\) is \(3\).

- Look for the x-values where the parabola and line meet. - Consider what happens when two equal values are subtracted. - Recall the name for an x-value where a function's output is \(0\).

1. From the graph, the functions intersect at \(x_1=2\) and \(x_2=4\). 2. Compute the difference: \(h(x)=f(x)-g(x)=0.5x^2-3x+4\). 3. At each intersection, \(f(x)=g(x)\). Therefore, \(h(2)=f(2)-g(2)=0\) and \(h(4)=f(4)-g(4)=0\). Thus, \(2\) and \(4\) are the zeros of \(h\). 4. In general, \(f(x)=g(x)\) is equivalent to \(f(x)-g(x)=0\), so the x-coordinates of the intersections of two graphs are the zeros of their difference function.

a) \(x_1=2\) and \(x_2=4\) b) The values \(2\) and \(4\) are the zeros of \(h\). c) At an intersection, \(f(x)=g(x)\), which is equivalent to \(f(x)-g(x)=0\).

- Use the vertex and another visible point to determine the parabola. - Use two points to find the line's slope and y-intercept. - At an intersection, the function outputs are equal. - Rearrange the quadratic equation and look for a factorization.

1. The parabola has vertex \((0,4)\), opens downward, and passes through \((2,0)\). Therefore, \(f(x)=-x^2+4\). 2. The line has y-intercept \(2\) and slope \(-1\), so \(g(x)=-x+2\). 3. Set the functions equal: \(-x^2+4=-x+2\). 4. Rearrange to obtain \(x^2-x-2=0\), and factor: \((x-2)(x+1)=0\). Thus, \(x=2\) or \(x=-1\). 5. Substitute into \(g\). The corresponding points are \((2,0)\) and \((-1,3)\), which agree with the graph.

a) \(f(x)=-x^2+4\); \(g(x)=-x+2\) b) \(-x^2+4=-x+2\) c) \((-1,3)\) and \((2,0)\)

- At an intersection, the two functions have equal outputs. - Move all terms to one side and factor the quadratic expression. - An intersection point includes both an x-coordinate and a y-coordinate.

1. At an intersection, \(f(x)=g(x)\), so \(x^2-4=x-2\). 2. Move all terms to one side: \(x^2-x-2=0\). 3. Factor: \((x-2)(x+1)=0\). Therefore, \(x=2\) or \(x=-1\). 4. Substitute into \(g(x)=x-2\). When \(x=2\), \(y=0\). When \(x=-1\), \(y=-3\). 5. The intersections are \((2,0)\) and \((-1,-3)\), which match the graph.

a) \(x^2-4=x-2\) b) \((-1,-3)\) and \((2,0)\) c) The calculated points agree with the graph.

- Use the vertex to write the equation of the parabola. - Determine the slope and y-intercept of the line from the graph. - Set the function expressions equal to find intersection x-coordinates. - Look for a common factor in the resulting quadratic equation.

1. The vertex of the parabola is \((2,-4)\), so \(p(x)=(x-2)^2-4=x^2-4x\). 2. The line passes through the origin and has slope \(-1\), so \(g(x)=-x\). 3. Set the functions equal: \(x^2-4x=-x\). This gives \(x^2-3x=0\). 4. Factor: \(x(x-3)=0\), so \(x=0\) or \(x=3\). 5. Substitute into \(g\): \(g(0)=0\) and \(g(3)=-3\). Therefore, the intersections are \((0,0)\) and \((3,-3)\).

a) \(p(x)=(x-2)^2-4\) and \(g(x)=-x\) b) \((0,0)\) and \((3,-3)\)

- Use the vertex and one additional point to determine the parabola. - Use two points to determine the slope and y-intercept of the line. - Set the function expressions equal and solve for \(x\). - Substitute each x-value into one function to find the corresponding y-value.

1. The parabola has vertex \((2,-1)\), so write \(p(x)=a(x-2)^2-1\). Since it passes through \((0,3)\), \(3=4a-1\), which gives \(a=1\). Thus, \(p(x)=(x-2)^2-1=x^2-4x+3\). 2. The line passes through \((1,0)\) and \((3,2)\). Its slope is \(\frac{2-0}{3-1}=1\), so \(g(x)=x-1\). 3. Set the functions equal: \(x^2-4x+3=x-1\). Rearranging gives \(x^2-5x+4=0\). 4. Factor: \((x-1)(x-4)=0\), so \(x=1\) or \(x=4\). 5. Substitute into \(g\): \(g(1)=0\) and \(g(4)=3\). Therefore, the intersections are \((1,0)\) and \((4,3)\).

The equations are \(p(x)=x^2-4x+3\) and \(g(x)=x-1\). The intersections are \((1,0)\) and \((4,3)\).

- Connect the zeros of a function to an equation of the form \(f(x)=0\). - In the second graph, interpret what it means when the two function outputs are equal. - Decide whether x-coordinates or y-coordinates represent the equation solutions. - Find two integers whose product is \(-2\) and whose sum is \(1\).

1. In graph a), the function \(f(x)=x^2+x-2\) is shown. The solutions are the x-intercepts of the parabola because they satisfy \(f(x)=0\). 2. In graph b), the equation is rewritten as \(x^2=-x+2\). The solutions are the x-coordinates of the intersections of \(y=x^2\) and \(y=-x+2\). 3. Both graphs show the solutions \(x=-2\) and \(x=1\). 4. To verify, factor the original equation: \(x^2+x-2=(x+2)(x-1)=0\). Therefore, \(x=-2\) or \(x=1\), confirming the graphical results.

a) In graph a), use the x-intercepts of \(f\). In graph b), use the x-coordinates of the intersections of the parabola and line. b) \(x=-2\) and \(x=1\) c) \((x+2)(x-1)=0\), so \(x=-2\) or \(x=1\).

- Use a point with convenient coordinates to determine \(a\). - Read the y-value of the horizontal line. - Set the function values equal and take a cube root.

1. The graph of \(f\) passes through \((2, 4)\). Substitution gives \(4=a(2)^3=8a\), so \(a=0.5\). Thus \(f(x)=0.5x^3\). The horizontal line is \(g(x)=-2\). 2. Set the functions equal: \(0.5x^3=-2\), so \(x^3=-4\). 3. The graph suggests \(x\approx-1.6\). Algebraically, \(x=-\sqrt[3]{4}\approx-1.587\).

a) \(a=0.5\) and \(g(x)=-2\) b) Graphically, \(x\approx-1.6\); algebraically, \(x=-\sqrt[3]{4}\approx-1.587\)

- Use the point \((2, -4)\) to determine \(a\). - Read the constant y-value of the line. - Set the two functions equal and include both real fourth roots.

1. The graph of \(f\) passes through \((2, -4)\). Thus \(-4=a(2)^4=16a\), so \(a=-0.25\). The horizontal line is \(g(x)=-2\), so \(c=-2\). 2. Set the functions equal: \(-0.25x^4=-2\). Therefore, \(x^4=8\). 3. The real solutions are \(x=\pm\sqrt[4]{8}\approx\pm1.682\).

a) \(a=-0.25\) and \(c=-2\) b) \(x\approx-1.682\) and \(x\approx1.682\)

- At an intersection, the two functions have equal outputs. - Move all terms to one side so the other side is \(0\). - Look for two factors of the quadratic expression. - Find the y-coordinate for each solution.

1. Set the functions equal: \(x^2-2x-1=x-3\). 2. Rearrange: \(x^2-3x+2=0\). 3. Factor: \((x-1)(x-2)=0\), so \(x=1\) or \(x=2\). 4. Substitute into \(g\): \(g(1)=-2\) and \(g(2)=-1\). 5. Therefore, the intersections are \((1,-2)\) and \((2,-1)\), which agree with the graph.

\((1,-2)\) and \((2,-1)\)

- Use vertex form for the parabola. - Use the y-intercept and slope of the line. - Set the function expressions equal to find their common points.

1. The parabola has vertex \((1,0)\) and passes through \((0,1)\), so \(p(x)=(x-1)^2\). 2. The line passes through \((0,-1)\) and \((1,0)\), so its slope is \(1\) and \(g(x)=x-1\). 3. Set the functions equal: \((x-1)^2=x-1\). 4. Move all terms to one side and factor: \((x-1)^2-(x-1)=0\), so \((x-1)(x-2)=0\). Therefore, \(x=1\) or \(x=2\). 5. Substituting into \(g\) gives the intersections \((1,0)\) and \((2,1)\).

\(p(x)=(x-1)^2\); \(g(x)=x-1\) Intersections: \((1,0)\) and \((2,1)\)

- Express each vertex in terms of \(c\). - Which two vertices lie on the horizontal line \(y = c\)? - What happens to the square of a quantity when the quantity is doubled?

1. The lines \(f\) and \(g\) intersect where \(x = -x\), so \(x = 0\) and \(y = 0\). One vertex is \((0, 0)\). 2. The intersection of \(f\) and \(h\) satisfies \(x = c\), giving \((c, c)\). The intersection of \(g\) and \(h\) satisfies \(-x = c\), giving \((-c, c)\). 3. The horizontal base has length \(c - (-c) = 2c\), and the height from the origin to \(y = c\) is \(c\). 4. Therefore, \(A = \frac{1}{2} \cdot 2c \cdot c = c^2\). 5. If \(c\) is doubled, the new area is \((2c)^2 = 4c^2\), so the area is multiplied by \(4\).

a) The vertices are \((0, 0)\), \((c, c)\), and \((-c, c)\). The base is \(2c\) and the height is \(c\), so \(A = c^2\). b) The area quadruples because \((2c)^2 = 4c^2\).

- Use the difference in Plan A’s costs and data amounts to find its price per gigabyte. - Substitute the per-gigabyte price into one Plan A equation to find the base fee. - Use the stated differences to determine Plan B’s fees. - Equal monthly costs mean the two cost expressions are equal.

1. Let \(G_A\) be Plan A’s base fee and \(p_A\) its price per gigabyte. 2. Write the system \(G_A + 3p_A = 14\) and \(G_A + 5p_A = 20\). 3. Subtract the first equation from the second: \(2p_A = 6\), so \(p_A = 3\). 4. Substitute into \(G_A + 3p_A = 14\): \(G_A + 9 = 14\), so \(G_A = 5\). 5. Plan B therefore has base fee \(G_B = 5 - 2 = 3\) and price \(p_B = 3 + 0.50 = 3.50\) per gigabyte. 6. For \(x\) gigabytes, set the costs equal: \(5 + 3x = 3 + 3.5x\). 7. Solve: \(2 = 0.5x\), so \(x = 4\).

Plan A has a \(\$5\) base fee and costs \(\$3.00\) per gigabyte. Plan B has a \(\$3\) base fee and costs \(\$3.50\) per gigabyte. The plans have the same cost at \(4\,\text{GB}\).

- Set the function expressions equal. - Use the discriminant to test for common points. - For part b), require the discriminant to be nonnegative.

1. Set \(f(x) = g(x)\): \(x^2 + 1 = x - 1\), so \(x^2 - x + 2 = 0\). 2. Its discriminant is \(D = (-1)^2 - 4 \cdot 1 \cdot 2 = -7\). Since \(D < 0\), the graphs do not intersect. 3. For the shifted line, set \(x^2 + 1 = x - 1 + d\), giving \(x^2 - x + (2 - d) = 0\). 4. Touching or intersecting requires \(D \ge 0\): \(1 - 4(2 - d) = 4d - 7 \ge 0\). 5. Therefore, \(d \ge \frac{7}{4}\), so the minimum upward shift is \(\frac{7}{4}\) units.

a) The discriminant is \(-7\), so there are no intersections. b) The minimum shift is \(d = \frac{7}{4}\).

- Treat \(p\) and \(q\) as fixed constants. - Which system-solving method is most efficient here? - Use parentheses when substituting an expression. - Isolate one variable at a time.

1. Solve equation (I) for \(x\): \(x = p - 2y\). 2. Substitute into equation (II): \(2(p - 2y) + y = q\). 3. Distribute and combine like terms: \(2p - 4y + y = q\), so \(2p - 3y = q\). 4. Solve for \(y\): \(3y = 2p - q\), so \(y = \frac{2p - q}{3}\). 5. Substitute into the expression for \(x\): \(x = p - 2 \cdot \frac{2p - q}{3} = \frac{2q - p}{3}\). 6. The solution is \((x, y) = (\frac{2q - p}{3}, \frac{2p - q}{3})\).

The solution is \((x, y) = (\frac{2q - p}{3}, \frac{2p - q}{3})\).

- A minus sign before a fraction affects the entire numerator. - Multiply each equation by the least common multiple of its denominators. - After simplifying, solve the resulting linear system using a method of your choice. - Check your result in the original equations.

1. Multiply the first equation by \(4\): \((3x - 2y + 2) - 2(x + y - 5) = 4\). Expand and simplify to get \(x - 4y = -8\). 2. Multiply the second equation by \(20\): \(4(2x + y - 1) + 5(x - 2y + 6) = 60\). Expand and simplify to get \(13x - 6y = 34\). 3. Solve the first simplified equation for \(x\): \(x = 4y - 8\). 4. Substitute into the second simplified equation: \(13(4y - 8) - 6y = 34\). This gives \(52y - 104 - 6y = 34\), so \(46y = 138\) and \(y = 3\). 5. Substitute into \(x = 4y - 8\): \(x = 4 \cdot 3 - 8 = 4\).

The solution set is \(\{(4, 3)\}\).

- Expand both sides of equation (I). - What happens to the \(xy\)-terms when you simplify equation (I)? - How can you clear the fraction in equation (II)? - Which system-solving method is most efficient after simplification?

1. Expand equation (I): \(xy - 2x + 3y - 6 = xy + 4x - y - 4\). 2. Subtract \(xy\) from both sides: \(-2x + 3y - 6 = 4x - y - 4\). 3. Rearrange equation (I): \(-6x + 4y = 2\), or equivalently \(3x - 2y = -1\). 4. Multiply equation (II) by \(2\): \(x + y = 6\). 5. From equation (II), \(y = 6 - x\). Substitute into the simplified equation (I): \(3x - 2(6 - x) = -1\). Then \(5x = 11\), so \(x = \frac{11}{5}\). 6. Find \(y\): \(y = 6 - \frac{11}{5} = \frac{19}{5}\). 7. The term \(xy\) has the same coefficient on both sides of equation (I), so it cancels. Only first-degree terms in \(x\) and \(y\) remain.

a) The solution set is \(\{(\frac{11}{5}, \frac{19}{5})\}\). b) The \(xy\)-terms cancel during simplification, leaving only linear terms.

- Choose variables for the two original ball counts. - Write one equation for the total purchase cost. - Use each percentage to represent the value of the worn-out part of the inventory. - Simplify the percentage equation before solving the system.

1. Let \(x\) be the number of basketballs and \(y\) the number of volleyballs. 2. The total purchase cost gives \(15x + 20y = 1225\). 3. The value of the worn-out balls gives \(0.20 \cdot 15x + 0.30 \cdot 20y = 285\). 4. Simplify the second equation: \(3x + 6y = 285\), or \(x + 2y = 95\). 5. Solve for \(x\): \(x = 95 - 2y\). 6. Substitute into the cost equation: \(15(95 - 2y) + 20y = 1225\). 7. Simplify: \(1425 - 30y + 20y = 1225\), so \(-10y = -200\) and \(y = 20\). 8. Then \(x = 95 - 2 \cdot 20 = 55\).

The club originally bought \(55\) basketballs and \(20\) volleyballs.

- Notice that the question asks for the inventory remaining after the sale, not the original inventory. - Write one equation for the original inventory value. - Write another equation for the revenue from the percentages sold. - After finding the original counts, calculate the percentages that remain.

1. Let \(x\) be the original number of tablets and \(y\) the original number of laptops. 2. The original inventory value gives \(300x + 800y = 110{,}000\). 3. The sales revenue gives \(0.40 \cdot 300x + 0.60 \cdot 800y = 60{,}000\). 4. Simplify the second equation: \(120x + 480y = 60{,}000\), or \(x + 4y = 500\). 5. Solve for \(x\): \(x = 500 - 4y\). 6. Substitute into the inventory equation: \(300(500 - 4y) + 800y = 110{,}000\). 7. Simplify: \(150{,}000 - 1200y + 800y = 110{,}000\), so \(-400y = -40{,}000\) and \(y = 100\). 8. Then \(x = 500 - 4 \cdot 100 = 100\). 9. The store keeps \(60\%\) of the tablets and \(40\%\) of the laptops, so \(60\) tablets and \(40\) laptops remain.

After the sale, \(60\) tablets and \(40\) laptops remain in stock.

- Interpret worker-hours as the number of workers multiplied by the hours per worker. - Choose variables for the two unknown planned quantities. - Express the same total work for each staffing scenario. - Expand the equations so the common product cancels. - Verify that both alternative staffing plans produce the same total worker-hours.

1. Let \(n\) be the planned number of workers and \(t\) the planned hours per worker. The total work is \(nt\) worker-hours. 2. Equal total work gives \(nt = (n + 4)(t - 1)\) and \(nt = (n - 3)(t + 1.5)\). 3. Expand and simplify the first equation: \(nt = nt - n + 4t - 4\), so \(-n + 4t = 4\). 4. Expand and simplify the second equation: \(nt = nt + 1.5n - 3t - 4.5\), so \(1.5n - 3t = 4.5\). 5. From the first equation, \(n = 4t - 4\). Substitute into the second: \(1.5(4t - 4) - 3t = 4.5\). 6. Simplify: \(6t - 6 - 3t = 4.5\), so \(3t = 10.5\) and \(t = 3.5\). 7. Then \(n = 4 \cdot 3.5 - 4 = 10\). 8. The total work is \(nt = 10 \cdot 3.5 = 35\) worker-hours.

The plan called for \(10\) workers at \(3.5\) hours each, for a total of \(35\) worker-hours.

- Represent each pump’s hourly rate as a fraction of the full reservoir. - Translate each operating schedule into an equation. - Write \(40\%\) as a decimal or fraction. - Once you know a pump’s hourly rate, use its reciprocal to find the total filling time.

1. Let \(x\) and \(y\) be the fractions of the reservoir filled per hour by Pumps 1 and 2. 2. Write the system \(4(x + y) = \frac{1}{3}\) and \(6x + 3y = 0.4\). 3. Simplify the first equation: \(x + y = \frac{1}{12}\). 4. Solve for \(y\): \(y = \frac{1}{12} - x\). Substitute into the second equation: \(6x + 3\left(\frac{1}{12} - x\right) = \frac{2}{5}\). 5. Simplify: \(3x + \frac{1}{4} = \frac{2}{5}\), so \(3x = \frac{3}{20}\) and \(x = \frac{1}{20}\). 6. Then \(y = \frac{1}{12} - \frac{1}{20} = \frac{1}{30}\). 7. Time is the reciprocal of rate, so Pump 1 takes \(20\) hours and Pump 2 takes \(30\) hours.

Pump 1 would take \(20\) hours, and Pump 2 would take \(30\) hours.

- Write a system using each crew’s fraction of the job completed per hour. - Translate one-third and one-half into the equations. - For part b, determine the time required at the combined rate. - Compare the combined time with the time of the faster crew.

1. Let \(x\) and \(y\) be the fractions of the job completed per hour by Crews 1 and 2. 2. Write the system \(2(x + y) = \frac{1}{3}\) and \(4x + y = \frac{1}{2}\). 3. The first equation gives \(x + y = \frac{1}{6}\), so \(y = \frac{1}{6} - x\). 4. Substitute into the second equation: \(4x + \frac{1}{6} - x = \frac{1}{2}\). 5. Simplify: \(3x = \frac{1}{3}\), so \(x = \frac{1}{9}\). Then \(y = \frac{1}{18}\). 6. Crew 1 takes \(9\) hours alone, and Crew 2 takes \(18\) hours alone. 7. Their combined rate is \(\frac{1}{9} + \frac{1}{18} = \frac{1}{6}\), so together they take \(6\) hours. 8. Compared with the faster crew’s \(9\) hours, the time saved is \(9 - 6 = 3\) hours.

a) Crew 1 would take \(9\) hours, and Crew 2 would take \(18\) hours. b) Working together saves \(3\) hours.

- Write an equation for the salt in the initial mixture. - Write a second salt-balance equation for the mixture after the additional solution is added. - For part b, compare the concentration being added with the concentration of the initial mixture.

1. Let \(x\) be the volume of the \(5\%\) solution and \(y\) the volume of the \(20\%\) solution, in liters. 2. The initial mixture gives \(0.05x + 0.20y = 0.10(x + y)\). Simplifying gives \(0.10y = 0.05x\), so \(x = 2y\). 3. After adding \(3\,\text{L}\) of the \(20\%\) solution, the salt balance is \(0.10(x + y) + 0.20 \cdot 3 = 0.15(x + y + 3)\). 4. Substitute \(x = 2y\): \(0.10(3y) + 0.60 = 0.15(3y + 3)\). 5. Simplify: \(0.30y + 0.60 = 0.45y + 0.45\), so \(0.15 = 0.15y\) and \(y = 1\). 6. Then \(x = 2y = 2\). 7. For part b, the added solution is less concentrated than the \(10\%\) mixture. Adding it dilutes the mixture, so the concentration falls below \(10\%\).

a) The initial mixture used \(2\,\text{L}\) of the \(5\%\) solution and \(1\,\text{L}\) of the \(20\%\) solution. b) The concentration would decrease below \(10\%\).

- Choose variables for the original amount of each drink. - Write an equation for the amount of apple juice in the original mixture. - Use the first equation to relate the two amounts. - Write a second equation after increasing both amounts by \(3\,\text{gal}\).

1. Let \(x\) be the volume of Drink A and \(y\) the volume of Drink B, in gallons. 2. The original mixture gives \(0.10x + 0.40y = 0.20(x + y)\). Simplifying gives \(0.20y = 0.10x\), so \(x = 2y\). 3. The larger mixture gives \(0.10(x + 3) + 0.40(y + 3) = 0.225(x + y + 6)\). 4. Substitute \(x = 2y\): \(0.10(2y + 3) + 0.40(y + 3) = 0.225(3y + 6)\). 5. Simplify: \(0.60y + 1.50 = 0.675y + 1.35\). 6. Then \(0.15 = 0.075y\), so \(y = 2\). 7. Therefore, \(x = 2y = 4\).

The original plan used \(4\,\text{gal}\) of Drink A and \(2\,\text{gal}\) of Drink B.

- Determine what percent of each full dispenser is removed to fill the pitcher. - Express \(A\) and \(B\) in terms of \(C\). - Translate “\(20\) more than four times” into an equation. - Substitute and solve.

1. Filling \(C\) from \(A\) removes \(40\%\) of \(A\), so \(C = 0.40A\), or \(A = 2.5C\). 2. Filling \(C\) from \(B\) removes \(25\%\) of \(B\), so \(C = 0.25B\), or \(B = 4C\). 3. The total-capacity condition is \(A + B = 4C + 20\). 4. Substitute the expressions for \(A\) and \(B\): \(2.5C + 4C = 4C + 20\). 5. Solving gives \(2.5C = 20\), so \(C = 8\). 6. Then \(A = 2.5(8) = 20\) and \(B = 4(8) = 32\).

Dispenser \(A\) holds \(20\) gallons, dispenser \(B\) holds \(32\) gallons, and pitcher \(C\) holds \(8\) gallons.

- Solve the linear equation for one variable. - Substitute to obtain a quadratic equation in one variable. - Require its discriminant to equal \(0\).

1. Solve the second equation for \(y\): \(y = 10 - x\). 2. Substitute into \(xy = c\): \(x(10 - x) = c\). 3. Rearrange: \(x^2 - 10x + c = 0\). 4. The discriminant is \(D = (-10)^2 - 4 \cdot 1 \cdot c = 100 - 4c\). 5. For exactly one value of \(x\), set \(D = 0\): \(100 - 4c = 0\), so \(c = 25\). 6. Then \((x - 5)^2 = 0\), so \(x = 5\), and \(y = 10 - 5 = 5\).

When \(c = 25\), the system has the single solution \((5, 5)\).

- Isolate one variable in the linear equation. - Substitute into the product equation and factor the resulting quadratic. - Compare the two ordered pairs when \(s = 0\) and when \(s \ne 0\).

1. Solve the first equation for \(y\): \(y = 7s - x\). 2. Substitute into the second equation: \(x(7s - x) = 10s^2\), so \(x^2 - 7sx + 10s^2 = 0\). 3. Factor: \((x - 2s)(x - 5s) = 0\). Thus, \(x = 2s\) or \(x = 5s\). 4. The corresponding values of \(y\) are \(5s\) and \(2s\), giving \((2s, 5s)\) and \((5s, 2s)\). 5. When \(s = 0\), both ordered pairs become \((0, 0)\). When \(s \ne 0\), \(2s \ne 5s\), so the ordered pairs are distinct.

a) \(\{(2s, 5s), (5s, 2s)\}\) b) For \(s = 0\), both expressions give \((0, 0)\). For \(s \ne 0\), they give two distinct ordered pairs.

- Solve the linear equation for \(x\). - Substitute and expand the squared binomial. - Factor the resulting quadratic in \(y\), then find the matching \(x\)-values.

1. Solve the linear equation for \(x\): \(x = y + c\). 2. Substitute into the first equation: \((y + c)^2 + y^2 = 25c^2\). 3. Expand and simplify: \(2y^2 + 2cy - 24c^2 = 0\). Divide by \(2\): \(y^2 + cy - 12c^2 = 0\). 4. Factor: \((y - 3c)(y + 4c) = 0\). Thus, \(y = 3c\) or \(y = -4c\). 5. Using \(x = y + c\), the ordered pairs are \((4c, 3c)\) and \((-3c, -4c)\). 6. When \(c = 0\), these expressions represent the same ordered pair, \((0, 0)\).

a) \(y^2 + cy - 12c^2 = 0\) b) For \(c \ne 0\), \(\{(4c, 3c), (-3c, -4c)\}\). For \(c = 0\), \(\{(0, 0)\}\).

- Set the two expressions for \(y\) equal. - Require the resulting quadratic to have a zero discriminant. - Substitute the repeated x-value to find \(y\).

1. Set the expressions for \(y\) equal: \(x^2 - 4x + k = 2x - 5\). 2. Rearrange: \(x^2 - 6x + (k + 5) = 0\). 3. The discriminant is \(D = (-6)^2 - 4 \cdot 1 \cdot (k + 5) = 16 - 4k\). 4. For exactly one solution, set \(D = 0\): \(16 - 4k = 0\), so \(k = 4\). 5. Then \(x^2 - 6x + 9 = (x - 3)^2 = 0\), so \(x = 3\). 6. Substitute into the line: \(y = 2 \cdot 3 - 5 = 1\).

a) \(k = 4\) b) \((3, 1)\)

- Write the circle equation and identify the parabola's vertex. - Use the condition that the vertex lies on the circle. - Rewrite one equation so that \(x^2\) can be substituted into the other.

1. The circle has equation \(x^2 + y^2 = 100\), and the parabola has vertex \((0, c)\). 2. For the vertex to lie on the circle, \(c^2 = 100\). Since \(c < 0\), \(c = -10\). 3. The system is \(x^2 + y^2 = 100\) and \(y = \frac{1}{4}x^2 - 10\). Rewrite the second equation as \(x^2 = 4y + 40\). 4. Substitute into the circle equation: \(4y + 40 + y^2 = 100\), so \(y^2 + 4y - 60 = 0\). 5. Factor: \((y + 10)(y - 6) = 0\). Thus, \(y = -10\) or \(y = 6\). 6. When \(y = -10\), \(x^2 = 0\), giving \((0, -10)\). When \(y = 6\), \(x^2 = 64\), giving \((8, 6)\) and \((-8, 6)\).

a) \(c = -10\) b) \(\{(0, -10), (8, 6), (-8, 6)\}\)

- Solve the linear equation for one variable. - Substitute and simplify the resulting quadratic. - Factor, then check whether the two ordered-pair expressions coincide for a special value of \(k\).

1. Solve the first equation for \(y\): \(y = 5k - x\). 2. Substitute into the second equation: \(x^2 + (5k - x)^2 = 13k^2\). 3. Expand and simplify: \(2x^2 - 10kx + 12k^2 = 0\). Divide by \(2\): \(x^2 - 5kx + 6k^2 = 0\). 4. Factor: \((x - 2k)(x - 3k) = 0\). Thus, \(x = 2k\) or \(x = 3k\). 5. Using \(y = 5k - x\), the ordered pairs are \((2k, 3k)\) and \((3k, 2k)\). 6. When \(k = 0\), these expressions represent the same ordered pair, \((0, 0)\).

For \(k \ne 0\), \(\{(2k, 3k), (3k, 2k)\}\). For \(k = 0\), \(\{(0, 0)\}\).

- Solve the sum equation for one variable. - Substitute into the product equation and factor the resulting quadratic. - Determine when the two ordered-pair expressions are identical.

1. Solve the first equation for \(y\): \(y = 2a + b - x\). 2. Substitute into the product equation: \(x(2a + b - x) = a^2 + ab\). 3. Rewrite in standard form: \(x^2 - (2a + b)x + a^2 + ab = 0\). 4. Factor: \((x - a)(x - a - b) = 0\). Thus, \(x = a\) or \(x = a + b\). 5. The corresponding values of \(y\) are \(a + b\) and \(a\). When \(b = 0\), the two ordered-pair expressions coincide.

For \(b \ne 0\), \(\{(a, a + b), (a + b, a)\}\). For \(b = 0\), \(\{(a, a)\}\).

- Solve the linear equation for one variable. - Substitute and simplify the resulting quadratic. - After factoring, determine when the two ordered-pair expressions coincide.

1. Solve the linear equation for \(y\): \(y = 5m - x\). 2. Substitute into the quadratic equation: \(x^2 + (5m - x)^2 = 17m^2\). 3. Expand and simplify: \(2x^2 - 10mx + 8m^2 = 0\). Divide by \(2\): \(x^2 - 5mx + 4m^2 = 0\). 4. Factor: \((x - 4m)(x - m) = 0\). Thus, \(x = 4m\) or \(x = m\). 5. The corresponding values of \(y\) are \(m\) and \(4m\). When \(m = 0\), the two ordered-pair expressions coincide.

For \(m \ne 0\), \(\{(4m, m), (m, 4m)\}\). For \(m = 0\), \(\{(0, 0)\}\).

- Solve the system while treating \(k\) as a constant. - Write the conditions \(x > 0\) and \(y > 0\) as inequalities in \(k\). - Find the overlap of the two resulting intervals.

1. Solve equation (II) for \(x\): \(x = k - 2y\). 2. Substitute into equation (I): \(4(k - 2y) + 3y = 24\), so \(y = \frac{4k - 24}{5}\). 3. Then \(x = k - 2y = \frac{48 - 3k}{5}\). 4. Require \(y > 0\): \(4k - 24 > 0\), so \(k > 6\). 5. Require \(x > 0\): \(48 - 3k > 0\), so \(k < 16\). 6. Both conditions hold when \(6 < k < 16\).

\(6 < k < 16\)

- Set the line and parabola equal. - Use the discriminant to count intersections. - Tangency means a zero discriminant.

1. Set the expressions equal: \(x^2 + 2x + 2 = -2x - 3\), so \(x^2 + 4x + 5 = 0\). 2. The discriminant is \(D = 4^2 - 4 \cdot 1 \cdot 5 = -4\), so the original system has no real solutions. 3. For the shifted line, setting the functions equal gives \(x^2 + 4x + (2 - c) = 0\). 4. Its discriminant is \(D = 4^2 - 4 \cdot 1 \cdot (2 - c) = 8 + 4c\). 5. Tangency requires \(D = 0\), so \(8 + 4c = 0\) and \(c = -2\). 6. If \(c > -2\), then \(D > 0\), so the line and parabola have two intersection points.

1) No real solutions 2) \(c = -2\) 3) Two intersections when \(c > -2\)

- Use each parabola's vertex and one additional point to determine its equation. - At an intersection, the two functions have the same output. - After simplifying the resulting quadratic equation, look for two binomials whose product is zero. - Substitute each x-value into either function to find the corresponding y-value.

1. The graph of \(f\) has vertex \((0,-4)\) and passes through \((2,0)\). Therefore, \(f(x)=x^2-4\). 2. The graph of \(g\) has vertex \((1,1)\) and opens downward. Write \(g(x)=a(x-1)^2+1\). Since it passes through \((0,0)\), \(0=a+1\), so \(a=-1\). Thus \(g(x)=-(x-1)^2+1=-x^2+2x\). 3. Set the functions equal: \(x^2-4=-x^2+2x\). This simplifies to \(x^2-x-2=0\). 4. Factor: \((x-2)(x+1)=0\), so \(x=2\) or \(x=-1\). 5. Substitute into either function. When \(x=-1\), \(y=-3\). When \(x=2\), \(y=0\). Therefore, the intersections are \((-1,-3)\) and \((2,0)\).

a) \(f(x)=x^2-4\) and \(g(x)=-x^2+2x\) b) \((-1,-3)\) and \((2,0)\)

- Recall the vertex form of a translated parent quadratic. - Read the lowest point of each parabola from the graph. - Notice what happens to the \(x^2\)-terms when the functions are set equal. - For part c), identify the type of equation that remains after cancellation.

1. The vertex of \(f\) is \((-2,-1)\), so \(f(x)=(x+2)^2-1\). The vertex of \(g\) is \((1,-4)\), so \(g(x)=(x-1)^2-4\). 2. Set the functions equal: \((x+2)^2-1=(x-1)^2-4\). 3. Expand and simplify: \(x^2+4x+3=x^2-2x-3\), so \(6x=-6\) and \(x=-1\). 4. Substitute into either function: \(f(-1)=(-1+2)^2-1=0\). Therefore, the intersection is \((-1,0)\). 5. In general, two such parabolas can be written as \(x^2+b_1x+c_1\) and \(x^2+b_2x+c_2\). Setting them equal cancels the \(x^2\)-terms and leaves \((b_1-b_2)x=c_2-c_1\). Different vertex x-coordinates imply \(b_1\ne b_2\), so the remaining linear equation has exactly one solution.

a) \(f(x)=(x+2)^2-1\) and \(g(x)=(x-1)^2-4\) b) \((-1,0)\) c) The \(x^2\)-terms cancel when the equations are set equal, leaving a linear equation with one solution because the vertices have different x-coordinates.

- Write one equation for the number of vehicles and one for the number of wheels. - Treat \(n\) and \(r\) as fixed quantities while solving for \(m\) and \(a\). - Isolate one of the vehicle variables and substitute. - Consider the minimum and maximum possible number of wheels for \(n\) vehicles, and determine when the formulas are integers.

1. Write the system \(m + a = n\) and \(2m + 4a = r\). 2. Solve the first equation for \(m\): \(m = n - a\). 3. Substitute into the second equation: \(2(n - a) + 4a = r\). 4. Simplify: \(2n + 2a = r\), so \(a = \frac{r - 2n}{2}\). 5. Substitute into \(m = n - a\): \(m = n - \frac{r - 2n}{2} = \frac{4n - r}{2}\). 6. For nonnegative counts, \(a \ge 0\) requires \(r \ge 2n\), and \(m \ge 0\) requires \(r \le 4n\). 7. For whole-number counts, \(n\) and \(r\) must be nonnegative integers and \(r\) must be even. Thus the conditions are \(n, r \in \mathbb{Z}_{\ge 0}\), \(2n \le r \le 4n\), and even \(r\).

The formulas are \(a = \frac{r - 2n}{2}\) and \(m = \frac{4n - r}{2}\). They give nonnegative whole numbers when \(n, r \in \mathbb{Z}_{\ge 0}\), \(2n \le r \le 4n\), and \(r\) is even.

- Write one equation for total mass and one for the mass of salt. - Convert each percent concentration into a fraction over \(100\). - Use the total-mass equation to replace one unknown. - Treat \(a\), \(b\), \(k\), and \(M\) as constants while solving for \(x\) and \(y\).

1. Let \(x\) be the mass of Solution A and \(y\) the mass of Solution B, in grams. 2. The total-mass equation is \(x + y = M\). 3. The salt-mass equation is \(\frac{a}{100}x + \frac{b}{100}y = \frac{k}{100}M\). Multiplying by \(100\) gives \(ax + by = kM\). 4. Solve the first equation for \(y\): \(y = M - x\). 5. Substitute into the salt equation: \(ax + b(M - x) = kM\). 6. Simplify: \((a - b)x = M(k - b)\), so \(x = \frac{M(k - b)}{a - b}\). 7. Use \(y = M - x\): \(y = M - \frac{M(k - b)}{a - b} = \frac{M(a - k)}{a - b}\). 8. Because \(a > k > b\), both calculated masses are positive.

The required masses are \(x = \frac{M(k - b)}{a - b}\,\text{g}\) of Solution A and \(y = \frac{M(a - k)}{a - b}\,\text{g}\) of Solution B.

- Express \(x\) and \(y\) in terms of \(a\). - A Quadrant II point must have a negative x-coordinate and a positive y-coordinate. - Determine when each rational expression has the required sign. - Remember to exclude parameter values that make a denominator zero.

1. Solve the first equation for \(x\): \(x=5-y\). 2. Substitute into the second equation: \(2(5-y)+ay=12\). 3. Simplify: \(10-2y+ay=12\), so \((a-2)y=2\). Thus \(y=\frac{2}{a-2}\), where \(a\neq2\). 4. Substitute into \(x=5-y\): \(x=5-\frac{2}{a-2}=\frac{5a-12}{a-2}\). 5. Since \(y>0\), \(\frac{2}{a-2}>0\), which requires \(a>2\). 6. With \(a>2\), the denominator of \(x\) is positive. Therefore, \(x<0\) requires \(5a-12<0\), so \(a<\frac{12}{5}\). 7. Combining the conditions gives \(2<a<\frac{12}{5}\).

\(2<a<\frac{12}{5}\)