- Compare the coefficients of the same variable across the two equations. - Opposite coefficients sum to zero. - Add corresponding terms on the two sides without solving the resulting equation.

1. The x-coefficients are opposites, so adding the equations cancels \(x\). 2. The y-terms combine to \(3y\), and the constants combine to \(12\). 3. The resulting equation is \(3y=12\).

\(x\) cancels, leaving \(3y=12\).

- Compare the coefficients of \(y\) in the two equations. - Add matching sides of the equations, term by term. - Which two terms make zero when added?

1. Add the left sides and the right sides: \((x+2y)+(3x-2y)=9+7\). 2. The \(+2y\) and \(-2y\) terms cancel because their sum is \(0\). 3. The resulting equation is \(4x=16\).

The y-terms cancel, leaving \(4x=16\), because \(2y+(-2y)=0\).

- Put parentheses around each full left side before subtracting. - A minus sign before parentheses changes every sign inside. - Be careful with the minus sign in front of \(-2y\).

1. Subtract the equations: \((4x+y)-(4x-2y)=11-5\). 2. The \(4x\) terms cancel, and \(y+2y=3y\). 3. Therefore \(3y=6\), so \(y=2\).

The x-terms cancel. \(3y=6\), so \(y=2\).

- How can you change one equation so that a variable cancels when the equations are added? - What coefficient would be the opposite of the corresponding coefficient in the other equation? - What happens to the remaining variable after you add the equations? - Check the solution in both original equations.

1. Multiply equation (II) by \(2\) so that the \(b\)-coefficients are opposites: \(2a + 4b = 12\). 2. Add this equation to equation (I): \((3a - 4b) + (2a + 4b) = 18 + 12\). 3. Simplify: \(5a = 30\). 4. Divide by \(5\): \(a = 6\). 5. Substitute into equation (II): \(6 + 2b = 6\). 6. Solve for \(b\): \(b = 0\). 7. The solution set is \(\{(6, 0)\}\).

The solution set is \(\{(6, 0)\}\).

- What happens to the \(y\)-terms when you add the equations directly? - How can you make one variable disappear? - Substitute the value you find into either original equation to determine the other variable.

1. Add equations (I) and (II): \(7x = 21\). 2. Solve for \(x\): \(x = 3\). 3. Substitute into equation (I): \(5 \cdot 3 + 3y = 9\). 4. Solve for \(y\): \(15 + 3y = 9\), so \(3y = -6\) and \(y = -2\). 5. The solution set is \(\{(3, -2)\}\).

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

- Can you multiply one equation so that one pair of coefficients becomes opposite? - What common multiple can you use for the coefficients of \(y\)? - Multiply every term in an equation by the same number.

1. Multiply equation (I) by \(2\): \(6x + 4y = 14\). 2. Add this equation to equation (II): \(15x = 15\). 3. Solve for \(x\): \(x = 1\). 4. Substitute into equation (I): \(3 \cdot 1 + 2y = 7\). 5. Solve for \(y\): \(3 + 2y = 7\), so \(y = 2\). 6. The solution set is \(\{(1, 2)\}\).

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

- Can you multiply one equation so that a coefficient becomes the opposite of the corresponding coefficient in the other equation? - How are the \(y\)-coefficients \(9\) and \(-3\) related? - Alternatively, how are the \(x\)-coefficients \(4\) and \(2\) related?

1. Multiply equation (II) by \(3\) to eliminate \(y\): \(6x - 9y = 24\). 2. Add this equation to equation (I): \(10x = 25\), so \(x = 2.5\). 3. Substitute into equation (II): \(2 \cdot 2.5 - 3y = 8\). Then \(-3y = 3\), so \(y = -1\).

One efficient first step is to multiply equation (II) by \(3\), which creates opposite \(y\)-coefficients. The solution is \((2.5, -1)\).

- Multiply one equation so that the variable coefficients match those in the other equation. - What does a false statement such as \(0 = 5\) mean after all variables cancel? - Could the two equations represent distinct parallel lines?

1. Multiply equation (II) by \(2\): \(3x + 4y = 30\). 2. Subtract the transformed equation (II) from equation (I): \((3x + 4y) - (3x + 4y) = 12 - 30\). 3. This gives \(0 = -18\), which is false. 4. Therefore, the system is inconsistent and has no solution. The solution set is \(\emptyset\).

The solution set is \(\emptyset\).

- The elimination process is unchanged when coefficients are decimals. - Can you multiply one equation so that one pair of decimal coefficients becomes opposite? - Choose a multiplier that makes one variable cancel when the equations are added.

1. Multiply equation (II) by \(3\): \(3x - 1.2y = 3\). 2. Add this equation to equation (I): \(3.5x = 7\). 3. Solve for \(x\): \(x = 2\). 4. Substitute into equation (II): \(2 - 0.4y = 1\). 5. Solve for \(y\): \(0.4y = 1\), so \(y = 2.5\). 6. The solution set is \(\{(2, 2.5)\}\).

The solution set is \(\{(2, 2.5)\}\).

- What happens to the \(y\)-terms when you add the equations? - What is the sum of two numbers with equal magnitude and opposite signs? - After finding one variable, how can you find the other?

1. The \(y\)-coefficients, \(-3\) and \(3\), are opposites, so the \(y\)-terms cancel when the equations are added. 2. Add the equations: \(7x = 21\). 3. Solve for \(x\): \(x = 3\). 4. Substitute into equation (II): \(2 \cdot 3 + 3y = 12\), so \(3y = 6\) and \(y = 2\). 5. The solution is \((3, 2)\).

a) The \(y\)-coefficients are opposites, so \(y\) cancels immediately when the equations are added. b) The solution is \((3, 2)\).

- Which pair of coefficients can be made opposites with one simple multiplication? - Pay close attention to signs when adding the equations. - Check the ordered pair in both original equations.

1. Multiply equation (II) by \(-2\): \(-4x - 10y = 2\). 2. Add this equation to equation (I): \(-13y = 13\). 3. Divide by \(-13\): \(y = -1\). 4. Substitute into equation (II): \(2x + 5(-1) = -1\). 5. Solve for \(x\): \(2x - 5 = -1\), so \(x = 2\). 6. The solution set is \(\{(2, -1)\}\).

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

- Can you multiply one equation so that one variable cancels when the equations are added? - Multiplying both equations by \(10\) is another way to remove decimal coefficients. - After finding one variable, substitute it into an original equation to find the other.

1. Multiply equation (II) by \(-2\): \(-0.4x - 1.0y = -4.4\). 2. Add this equation to equation (I): \(-1.3y = -2.6\). 3. Divide by \(-1.3\): \(y = 2\). 4. Substitute into equation (I): \(0.4x - 0.3 \cdot 2 = 1.8\), so \(0.4x = 2.4\) and \(x = 6\).

The solution is \((6, 2)\).

- Can you multiply one equation so that a variable cancels when the equations are added? - Which number should multiply the second equation? - After finding one variable, substitute it into an original equation. - Check the ordered pair in both original equations.

1. Multiply the second equation by \(2\): \(10x - 2y = 6\). 2. Add this equation to the first equation: \(13x = 13\). 3. Divide by \(13\): \(x = 1\). 4. Substitute \(x = 1\) into the second original equation: \(5(1) - y = 3\). 5. Solve for \(y\): \(5 - y = 3\), so \(y = 2\). 6. The solution is \((1, 2)\).

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

- Can you change one equation so that a variable cancels when the equations are added? - Which multiple of a coefficient would be useful? - What happens when the second equation is multiplied by an appropriate number? - Is it easier to eliminate \(x\) or \(y\)?

1. Multiply the second equation by \(3\): \(12x - 3y = 30\). 2. Add this equation to the first equation: \(14x = 42\). 3. Divide by \(14\): \(x = 3\). 4. Substitute \(x = 3\) into the second original equation: \(4(3) - y = 10\). 5. Solve: \(12 - y = 10\), so \(y = 2\).

The solution is \((3, 2)\).

- What happens when the equations are added directly? - Can one variable be eliminated without first multiplying either equation? - Use the first value you find to calculate the other variable.

1. Add the equations to eliminate \(y\): \((\frac{x}{2} + y) + (\frac{x}{2} - y) = 7 + 1\). 2. Simplify: \(x = 8\). 3. Substitute \(x = 8\) into the first equation: \(\frac{8}{2} + y = 7\). 4. Solve: \(4 + y = 7\), so \(y = 3\).

The solution is \((8, 3)\).

- Can one equation be multiplied so that a variable cancels when the equations are added? - You may also clear the decimals and fractions before using elimination. - After finding one variable, substitute to find the other.

1. Multiply the second equation by \(-2\): \(-0.4x + 1.5y = 6\). 2. Add this equation to the first equation: \(2y = 16\). 3. Solve: \(y = 8\). 4. Substitute \(y = 8\) into the first equation: \(0.4x + 4 = 10\). 5. Solve: \(0.4x = 6\), so \(x = 15\). 6. The solution set is \(\{(15, 8)\}\).

The solution set is \(\{(15, 8)\}\).

- Multiply one equation so that a variable cancels when the equations are added. - Which multiplier for the first equation will eliminate \(y\)? - Check the result in both original equations.

1. Multiply the first equation by \(3\): \(12x - 3y = 21\). 2. Add it to the second equation: \(14x = 28\). 3. Solve: \(x = 2\). 4. Substitute into \(4x - y = 7\): \(8 - y = 7\), so \(y = 1\). 5. The solution is \((2, 1)\).

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

- Multiply one equation so that a variable cancels when the equations are added. - What multiplier for the second equation creates opposite \(y\)-coefficients? - Substitute the first value you find to determine the other. - Check the ordered pair in both equations.

1. Multiply the second equation by \(-2\): \(-30x - 8y + 14 = 0\). 2. Add it to the first equation: \(-25x + 25 = 0\). 3. Solve: \(x = 1\). 4. Substitute into the first equation: \(5(1) + 8y + 11 = 0\). 5. Solve: \(16 + 8y = 0\), so \(y = -2\). 6. The solution is \((1, -2)\).

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

- What happens when one equation is subtracted from the other? - Use a common denominator to combine the fractional coefficients. - Can a variable be eliminated without first scaling either equation? - Check the result in both original equations.

1. Subtract the second equation from the first to eliminate \(x\): \(\frac{1}{3}y - (-\frac{1}{6}y) = 7 - 1\). 2. Combine the fractions: \(\frac{1}{2}y = 6\), so \(y = 12\). 3. Substitute \(y = 12\) into the first equation: \(\frac{1}{2}x + 4 = 7\). 4. Solve: \(\frac{1}{2}x = 3\), so \(x = 6\). 5. The solution is \((6, 12)\).

The solution is \((6, 12)\).

- Which terms already have matching coefficients in the two equations? - After one variable is removed, how can the remaining equation determine one container weight? - Check whether the two weights reproduce both shipment totals.

1. Let \(a\) be the weight of one artifact case and \(p\) the weight of one pedestal crate. The system is (I) \(4a+3p=650\) (II) \(7a+3p=890\). 2. Subtract equation (I) from equation (II): \(3a=240\), so \(a=80\). 3. Substitute into equation (I): \(4\cdot 80+3p=650\), so \(3p=330\) and \(p=110\).

Each artifact case weighs \(80\,\text{lb}\), and each pedestal crate weighs \(110\,\text{lb}\).

- Compare the requested expression with the coefficients in the two given equations. - Consider whether a direct combination can produce the requested coefficients. - There is no need to determine either coordinate individually.

1. Add equations (I) and (II): \((4x+y)+(-2x+3y)=17+1\). 2. The left side simplifies to \(2x+4y\), so \(2x+4y=18\).

\(2x+4y=18\).

- Look for coefficients that are already opposites. - Combine the equations before doing any rearrangement. - After one variable is known, use either original equation for the other coordinate.

1. Add the equations. The y-terms cancel, giving \(2x=6\). 2. Thus \(x=3\). 3. Substitute into either original equation to get \(y=2\).

\((3,2)\)

- Both equations begin with the same \(2x\) term. - Subtraction can remove that term. - After finding \(y\), use one original equation to find \(x\).

1. Subtract (I) from (II): \((2x+3y)-(2x+y)=15-9\). 2. This gives \(2y=6\), so \(y=3\). 3. Substitute into (I): \(2x+3=9\), so \(x=3\). 4. The solution is \((3,3)\).

After subtraction: \(2y=6\). Solution: \((3,3)\).

- Align terms with the same variable before combining the equations. - Add the left sides and the right sides separately. - Put the known x-value into either original equation.

1. Add (I) and (II): \(4x=16\). 2. Then \(x=4\). 3. Use (I): \(4+2y=10\), so \(2y=6\) and \(y=3\). 4. Check (II): \(3\cdot4-2\cdot3=6\).

After addition: \(4x=16\). Solution: \((4,3)\).

- Compare the coefficients before choosing whether to add or subtract. - Combine like terms after the variable cancels. - Check the ordered pair in both original equations before answering.

1. Add the equations. The x-terms cancel, leaving \(3y=12\). 2. Thus \(y=4\). 3. In (I), \(-3x+4=2\), so \(-3x=-2\) and \(x=\frac{2}{3}\). 4. The solution is \(\left(\frac{2}{3},4\right)\).

After addition: \(3y=12\). Solution: \(\left(\frac{2}{3},4\right)\).

- Use one variable for each ticket price. - Ticket counts are coefficients; ticket prices are the unknowns. - Compare the left sides: which terms would disappear if you subtracted them?

1. Let \(a\) be the adult price and \(c\) the child price. Then \(2a+c=19\) and \(2a+3c=29\). 2. Subtract the first equation from the second: \(2c=10\), so \(c=5\). 3. Substitute into \(2a+c=19\): \(2a+5=19\), so \(a=7\). 4. Adult tickets cost \(\$7\); child tickets cost \(\$5\).

\(2a+c=19\) and \(2a+3c=29\). After subtraction: \(2c=10\). Adult ticket: \(\$7\). Child ticket: \(\$5\).

- Look for coefficients that could become opposites after scaling just one equation. - Multiply the entire equation, not only one term. - Substitute the value found after elimination into an original equation.

1. Multiply (II) by \(2\): \(4x-2y=6\). 2. Add to (I): \(5x=10\), so \(x=2\). 3. From (II), \(4-y=3\), so \(y=1\). 4. Both originals hold for \((2,1)\).

Double (II): \(4x-2y=6\). After addition: \(5x=10\). Solution: \((2,1)\).

- Find a common multiple of the coefficients of \(x\). - Use opposite signs so the \(x\)-terms cancel when added. - After finding one coordinate, return to an original equation for the other.

1. Multiply (I) by \(3\): \(6x+9y=39\). 2. Multiply (II) by \(-2\): \(-6x-4y=-24\). 3. Add: \(5y=15\), so \(y=3\). 4. Substitute into (I): \(2x+9=13\), so \(x=2\). 5. Check in (II): \(6+6=12\).

Multiply (I) by \(3\): \(6x+9y=39\). Multiply (II) by \(-2\): \(-6x-4y=-24\). After addition: \(5y=15\). Solution: \((2,3)\).

- Add the two \(x\)-terms mentally. Do they make zero? - What operation removes two identical terms? - Check your final ordered pair in both equations.

1. Adding \(2x+2x\) gives \(4x\), not zero. To cancel equal \(x\)-terms, subtract the equations. 2. Compute (I) minus (II): \(4y=4\), so \(y=1\). 3. From (II), \(2x-1=3\), so \(x=2\). 4. The solution \((2,1)\) satisfies both equations.

The \(2x\)-terms have the same sign, so adding does not cancel them. Subtract to get \(4y=4\); the solution is \((2,1)\).

- Compare the \(y\)-coefficients before choosing an operation. - What happens when you add equalities with opposite \(y\)-terms? - Think about how many rearranging steps substitution would need.

1. The \(y\)-coefficients are opposites already, so add the equations: \(6x=18\). 2. Thus \(x=3\). In (I), \(6+3y=12\), so \(y=2\). 3. Both equations are satisfied by \((3,2)\). 4. Elimination cancels \(y\) immediately, while substitution would first require isolating a variable.

\((3,2)\). Add the equations: the \(y\)-terms cancel without rearranging or scaling.

- Which terms cancel if you add the first two equations? - A point lies on a line when its coordinates satisfy the line equation. - Substitute the intersection into the third equation to test it. - Substitute the same coordinates into \(x - 2y = c\) to find \(c\).

1. Add the equations \(3x - y = 5\) and \(x + y = 7\): \(4x = 12\), so \(x = 3\). Substitute into \(x + y = 7\) to get \(y = 4\). The intersection is \((3, 4)\). 2. Check the third equation: \(2 \cdot 3 + 3 \cdot 4 = 18\). Therefore, the intersection satisfies the equation. 3. Substitute \(x = 3\) and \(y = 4\) into \(x - 2y = c\): \(c = 3 - 2 \cdot 4 = -5\).

a) The intersection is \((3, 4)\). b) Yes. Substitution gives \(2 \cdot 3 + 3 \cdot 4 = 18\). c) \(c = -5\)

- When neither pair of coefficients is already equal or opposite, what multiples can make one pair cancel? - Find a common multiple of the coefficients of \(u\) or \(v\). - Choose signs so that one variable cancels when the equations are added.

1. Multiply equation (I) by \(3\) and equation (II) by \(-5\) to eliminate \(u\): (I′) \(15u + 21v = 33\) (II′) \(-15u - 20v = -30\) 2. Add the new equations: \(v = 3\). 3. Substitute \(v = 3\) into equation (II): \(3u + 4 \cdot 3 = 6\). 4. Solve for \(u\): \(3u + 12 = 6\), so \(3u = -6\) and \(u = -2\). 5. The solution set is \(\{(-2, 3)\}\).

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

- Expand the parentheses to simplify each equation. - Arrange both equations with variable terms on one side and constants on the other. - Which variable is easiest to eliminate after rewriting? - The solution may contain decimals.

1. Rewrite equation (I): \(3x - 6 = 4y + 1\), so \(3x - 4y = 7\). 2. Rewrite equation (II): \(5x = 2y + 6x + 7\), so \(-x - 2y = 7\). 3. Multiply the simplified equation (II) by \(-2\): \(2x + 4y = -14\). 4. Add it to the simplified equation (I): \((3x - 4y) + (2x + 4y) = 7 - 14\). 5. Simplify: \(5x = -7\), so \(x = -1.4\). 6. Substitute into \(-x - 2y = 7\): \(1.4 - 2y = 7\), so \(-2y = 5.6\) and \(y = -2.8\). 7. The solution set is \(\{(-1.4, -2.8)\}\).

The solution set is \(\{(-1.4, -2.8)\}\).

- Multiply the equations so that one variable can be eliminated. - Find a common multiple of one pair of coefficients. - Choose signs that make those coefficients opposites.

1. Multiply equation (I) by \(2\) and equation (II) by \(5\): (I′) \(8x - 10y = 26\) (II′) \(15x + 10y = 20\) 2. Add the new equations: \(23x = 46\). 3. Solve for \(x\): \(x = 2\). 4. Substitute into equation (II): \(3 \cdot 2 + 2y = 4\). 5. Solve for \(y\): \(6 + 2y = 4\), so \(y = -1\). 6. The solution set is \(\{(2, -1)\}\).

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

- What is the goal of elimination? - Which strategy requires fewer equation transformations? - Which strategy produces smaller coefficients?

1. Strategy A is more efficient because only one equation must be multiplied, and the resulting coefficients are smaller. It creates opposite \(y\)-coefficients, \(4y\) and \(-4y\). 2. Multiply equation (I) by \(2\): \(10x + 4y = 32\). 3. Add this equation to equation (II): \(13x = 39\). 4. Solve for \(x\): \(x = 3\). 5. Substitute into equation (I): \(5 \cdot 3 + 2y = 16\), so \(2y = 1\) and \(y = 0.5\). 6. The solution set is \(\{(3, 0.5)\}\).

a) Strategy A is more efficient because it changes only one equation and keeps the coefficients smaller. b) The solution set is \(\{(3, 0.5)\}\).

- Compare the coefficients \(1.2\) and \(-0.6\). - What multiplication would remove all decimal points? - After finding \(x\), how can you find \(y\)?

1. Multiply equation (II) by \(2\), because \(-0.6 \cdot 2 = -1.2\), the opposite of \(1.2\). 2. Multiplying both equations by \(10\) changes all decimal coefficients into integers, which can make the arithmetic easier and reduce errors. 3. Using equation (II) multiplied by \(2\) gives \(4x - 1.2y = 4.4\). 4. Add this equation to equation (I): \(4.5x = 9\). 5. Solve for \(x\): \(x = 2\). 6. Substitute into equation (II): \(2 \cdot 2 - 0.6y = 2.2\), so \(1.8 = 0.6y\) and \(y = 3\). 7. The solution set is \(\{(2, 3)\}\).

a) Multiply equation (II) by \(2\). b) Multiplying by \(10\) removes the decimal coefficients and allows integer arithmetic. c) The solution set is \(\{(2, 3)\}\).

- When do two terms cancel when equations are added? - Compare the coefficients directly. - After finding one variable, how can you determine the other?

1. For part a), choose equations (1) and (2), because the \(y\)-coefficients \(4\) and \(-4\) are opposites. 2. Add the equations: \(8x = 16\), so \(x = 2\). 3. Substitute into equation (1): \(3 \cdot 2 + 4y = 10\), so \(y = 1\). The solution is \((2, 1)\). 4. For part b), choose equations (1) and (3), because the \(x\)-coefficients \(3\) and \(-3\) are opposites. 5. Add the equations: \(6y = 18\), so \(y = 3\). 6. Substitute into equation (3): \(-3x + 2 \cdot 3 = 8\), so \(-3x = 2\) and \(x = -\frac{2}{3}\). The solution is \((-\frac{2}{3}, 3)\).

a) Use equations (1) and (2). The solution is \((2, 1)\). b) Use equations (1) and (3). The solution is \((-\frac{2}{3}, 3)\).

- Substitute the found value of \(x\) into either original equation to recover \(y\). - For two \(x\)-terms to cancel by addition, how must their coefficients be related? - If every variable term disappears after addition, interpret the remaining statement before deciding the solution type.

1. The addition is correct: the \(y\)-terms cancel, and \(9x=45\) gives \(x=5\). 2. Substitute \(x=5\) into equation (II): \(3\cdot5+5y=31\). Then \(5y=16\), so \(y=3.2\). 3. To eliminate \(x\) by addition as well, the new coefficient in equation (II) must be \(-6\), giving \(-6x+5y=31\). 4. Adding the modified equation to equation (I) gives \(0=45\), a contradiction. Therefore, the modified system has no solution.

a) The work for \(x\) is correct, and \(y=3.2\). b) Change the coefficient to \(-6\). The resulting equation is \(0=45\), so the modified system has no solution.

- First rewrite the system with variable terms on one side and constants on the other. - Find a common multiple of one pair of coefficients. - Substitute the first variable value into an original equation to find the second.

1. Rewrite equation (II) in standard form: \(5a + 4b = 23\). 2. Multiply equation (I) by \(4\) and the rewritten equation (II) by \(3\): (I′) \(8a - 12b = 0\) (II′) \(15a + 12b = 69\) 3. Add the new equations: \(23a = 69\). 4. Divide by \(23\): \(a = 3\). 5. Substitute into equation (I): \(2 \cdot 3 - 3b = 0\), so \(b = 2\). 6. The solution set is \(\{(3, 2)\}\).

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

- Compare the coefficients of \(x\) and \(y\). Is one pair related by a simple multiple? - Can you change one equation so that a variable cancels when the equations are added? - Is multiplying one equation simpler than isolating a variable with fractions?

1. Elimination is efficient because multiplying equation (I) by \(2\) makes the \(y\)-coefficients opposites. 2. Multiply equation (I) by \(2\): \(10x + 4y = 2\). 3. Add this equation to equation (II): \(13x = 13\). 4. Solve for \(x\): \(x = 1\). 5. Substitute into equation (I): \(5 \cdot 1 + 2y = 1\), so \(2y = -4\) and \(y = -2\). 6. The solution is \((1, -2)\).

Elimination is efficient because multiplying equation (I) by \(2\) creates opposite \(y\)-coefficients. The solution set is \(\{(1, -2)\}\).

- Compare the decimal coefficient of \(x\) in equation (I) with the \(x\)-coefficient in equation (II). - Look for a multiplier that makes one pair of corresponding coefficients equal or opposite. - Once one variable has been eliminated, use the remaining equation before substituting back.

1. Multiply equation (I) by \(5\): \(2x+1.5y=9\). This is useful because the \(x\)-coefficient now matches the one in equation (II). 2. Subtract equation (II) from the transformed equation (I): \((2x-2x)+(1.5y-(-5y))=9-9\). 3. Simplify: \(6.5y=0\), so \(y=0\). 4. Substitute into equation (II): \(2x-5\cdot0=9\), so \(x=4.5\). 5. The solution set is \(\{(4.5,0)\}\).

Multiplying equation (I) by \(5\) makes the \(x\)-coefficients equal, so subtraction eliminates \(x\). The solution set is \(\{(4.5,0)\}\).

- What multiplier clears the denominators? - What number is a common multiple of \(2\) and \(5\)? - To eliminate a variable, how should its coefficients compare in the two equations?

1. Multiply both equations by \(10\): (I′) \(4x - 5y = -10\) (II′) \(2x + 15y = 170\). 2. Multiply equation (I′) by \(3\): \(12x - 15y = -30\). 3. Add this equation to equation (II′): \(14x = 140\), so \(x = 10\). 4. Substitute into equation (II′): \(2 \cdot 10 + 15y = 170\), so \(15y = 150\) and \(y = 10\). 5. The solution set is \(\{(10, 10)\}\).

The solution set is \(\{(10, 10)\}\).

- What power of \(10\) moves each decimal point one place to the right? - How can you make one variable cancel when the equations are added? - Multiply every term, including the constant, by the same factor.

1. Multiply both equations by \(10\): (I′) \(3x + 4y = 25\) (II′) \(7x - 2y = 13\). 2. Multiply equation (II′) by \(2\): \(14x - 4y = 26\). 3. Add this equation to equation (I′): \(17x = 51\), so \(x = 3\). 4. Substitute into equation (I′): \(3 \cdot 3 + 4y = 25\), so \(4y = 16\) and \(y = 4\). 5. The solution set is \(\{(3, 4)\}\).

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

- Write one equation for each order. - Represent each percentage increase with a multiplier. - Simplify the equations before applying elimination.

1. Let \(x\) be the original price of one jersey and \(y\) the original price of one pair of shorts, in dollars. 2. Write the system \(20x + 30y = 900\) and \(20(1.1x) + 30(1.2y) = 1020\). 3. Simplify the equations: \(2x + 3y = 90\) and \(22x + 36y = 1020\). 4. Multiply the first simplified equation by \(11\): \(22x + 33y = 990\). 5. Subtract: \(3y = 30\), so \(y = 10\). 6. Substitute into \(2x + 3y = 90\): \(2x + 30 = 90\), so \(x = 30\).

A jersey originally cost \(\$30\), and a pair of shorts originally cost \(\$10\).

- Write one revenue equation for each lunch period. - Use multipliers to represent the discount and price increase. - Simplify both equations before using elimination.

1. Let \(p\) be the original price of one pretzel and \(j\) the original price of one bottle of juice, in dollars. 2. Write the system \(40p + 50j = 115\) and \(30(0.8p) + 40(1.5j) = 102\). 3. Simplify: \(4p + 5j = 11.5\) and \(4p + 10j = 17\). 4. Subtract the first simplified equation from the second: \(5j = 5.5\), so \(j = 1.10\). 5. Substitute into \(4p + 5j = 11.5\): \(4p + 5.5 = 11.5\), so \(p = 1.50\).

A pretzel originally cost \(\$1.50\), and a bottle of juice originally cost \(\$1.10\).

- When no variable is ready to cancel, multiply both equations. - Find a least common multiple of one pair of coefficients. - Check the final ordered pair in both original equations.

1. Multiply equation (I) by \(3\) and equation (II) by \(4\): (I′) \(9x + 12y = 15\) (II′) \(20x - 12y = 72\). 2. Add the transformed equations: \(29x = 87\). 3. Divide by \(29\): \(x = 3\). 4. Substitute into equation (I): \(3 \cdot 3 + 4y = 5\). 5. Solve for \(y\): \(9 + 4y = 5\), so \(y = -1\). 6. The solution set is \(\{(3, -1)\}\).

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

- What must happen to the \(y\)-coefficients so that they cancel when the equations are added? - Find the least common multiple of the coefficients. - Multiply every term, including the constant, by the chosen factor. - Check the solution in both original equations.

1. A common multiple of the \(y\)-coefficients \(3\) and \(4\) is \(12\). 2. Multiply equation (I) by \(4\): \(20x + 12y = 28\). 3. Multiply equation (II) by \(3\): \(6x - 12y = 24\). 4. Add the transformed equations: \(26x = 52\), so \(x = 2\). 5. Substitute into equation (I): \(5 \cdot 2 + 3y = 7\), so \(3y = -3\) and \(y = -1\). 6. The solution set is \(\{(2, -1)\}\). 7. Neither \(3\) nor \(4\) is an integer multiple of the other, so both equations must be scaled to create opposite integer coefficients, such as \(12\) and \(-12\).

a) \(20x + 12y = 28\) and \(6x - 12y = 24\) b) The solution set is \(\{(2, -1)\}\). c) Both equations must be multiplied because \(3\) and \(4\) are not integer multiples of one another, so both must be scaled to create opposite integer coefficients.

- It may help to write all coefficients in the same form, either fractions or decimals. - Can you multiply one equation so that a coefficient matches the corresponding coefficient in the other equation? - Which variable is easier to eliminate?

1. Multiply equation (II) by \(2\): \(0.5x - y = -2\). Since \(\frac{1}{2} = 0.5\), the \(x\)-coefficients now match. 2. Subtract the transformed equation (II) from equation (I): \((0.5x - 0.5x) + (\frac{2}{3}y - (-y)) = 4 - (-2)\). 3. Simplify: \(\frac{5}{3}y = 6\), so \(y = \frac{18}{5} = 3.6\). 4. Substitute into equation (II): \(0.25x - 0.5 \cdot 3.6 = -1\). Then \(0.25x = 0.8\), so \(x = 3.2\). 5. The solution set is \(\{(3.2, 3.6)\}\). 6. Test \((3, 3.6)\) in equation (I): \(\frac{1}{2} \cdot 3 + \frac{2}{3} \cdot 3.6 = 3.9 \ne 4\). Therefore, the given ordered pair is not a solution.

The solution set is \(\{(3.2, 3.6)\}\). The ordered pair \((3, 3.6)\) is not a solution.

- Compare the coefficients of one variable across the two equations in each system. - What happens when you add terms with opposite coefficients? What happens when you subtract equal coefficients? - After one variable cancels, use an original equation to find the other.

1. For system 1, add the equations: \(7x=21\), so \(x=3\). Substitute into \(2x-3y=0\): \(6-3y=0\), giving \(y=2\). Adding first cancels \(y\). 2. For system 2, subtract the first equation from the second: \(9b=-9\), so \(b=-1\). Substitute into \(6a-4b=10\): \(6a+4=10\), giving \(a=1\). Subtracting first cancels \(a\).

1) Add first; \((x, y)=(3, 2)\). 2) Subtract first; \((a, b)=(1, -1)\).

- Rewrite each equation with the variable terms on the left and the constant on the right. - How can you create equal or opposite coefficients for one variable? - Be especially careful with signs when subtracting equations. - Write the two values as an ordered pair in the solution set.

1. Rewrite equation (II) by subtracting \(2y\) from both sides: \(3x - 2y = 8\). 2. Multiply equation (I) by \(3\): \(6x + 15y = -3\). 3. Multiply the rewritten equation (II) by \(2\): \(6x - 4y = 16\). 4. Subtract the second transformed equation from the first: \((6x + 15y) - (6x - 4y) = -3 - 16\), so \(19y = -19\) and \(y = -1\). 5. Substitute \(y = -1\) into equation (II): \(3x = 2(-1) + 8 = 6\), so \(x = 2\). 6. The solution set is \(\{(2, -1)\}\).

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

- Look for a common multiple of a pair of coefficients. - You may multiply the two equations by different numbers. - Choose multipliers that create opposite coefficients for one variable. - Make sure you find both variables.

1. Multiply the first equation by \(2\): \(8x + 10y = 4\). 2. Multiply the second equation by \(5\): \(15x - 10y = 65\). 3. Add the transformed equations: \(23x = 69\), so \(x = 3\). 4. Substitute \(x = 3\) into the first original equation: \(4(3) + 5y = 2\). 5. Solve: \(12 + 5y = 2\), so \(5y = -10\) and \(y = -2\). 6. The solution is \((3, -2)\).

The solution is \((3, -2)\).

- Find a common multiple of one pair of coefficients. - What signs must the coefficients have so that they cancel when the equations are added? - You may need to multiply both equations by different numbers. - What is the least common multiple of \(2\) and \(3\)?

1. Multiply the first equation by \(2\): \(8x + 6y = 4\). 2. Multiply the second equation by \(-3\): \(-15x - 6y = 3\). 3. Add the transformed equations: \(-7x = 7\), so \(x = -1\). 4. Substitute \(x = -1\) into the first original equation: \(4(-1) + 3y = 2\). 5. Solve: \(-4 + 3y = 2\), so \(3y = 6\) and \(y = 2\).

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

- What multipliers will create opposite coefficients for one variable? - What happens when the transformed equations are added? - You may also solve one equation for a variable, but elimination is efficient here. - After finding one variable, substitute to find the other.

1. Multiply the first equation by \(2\): \(8x - 10y = -4\). 2. Multiply the second equation by \(5\): \(15x + 10y = 50\). 3. Add the transformed equations: \(23x = 46\), so \(x = 2\). 4. Substitute \(x = 2\) into the second original equation: \(3(2) + 2y = 10\). 5. Solve: \(6 + 2y = 10\), so \(y = 2\). 6. The solution is \((2, 2)\).

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

- What happens when the two equations are added? - Can subtraction eliminate the other variable? - Treat \(u\) and \(v\) as fixed numbers while you solve for \(x\) and \(y\). - Check by substituting both expressions into the original equations.

1. Add the equations to eliminate \(y\): \((x + y) + (x - y) = (5u - v) + (u + 5v)\). 2. Simplify: \(2x = 6u + 4v\), so \(x = 3u + 2v\). 3. Subtract the second equation from the first to eliminate \(x\): \((x + y) - (x - y) = (5u - v) - (u + 5v)\). 4. Simplify: \(2y = 4u - 6v\), so \(y = 2u - 3v\).

\(x = 3u + 2v\) and \(y = 2u - 3v\)

- Compare the coefficients of \(y\). Which operation eliminates that variable immediately? - After finding one variable, how can you find the other? - The parameter may remain in the final expressions. - Be careful with signs when substituting and simplifying.

1. Add the equations to eliminate \(y\): \((4x + 3y) + (2x - 3y) = 10a + 2a\). 2. Simplify: \(6x = 12a\), so \(x = 2a\). 3. Substitute \(x = 2a\) into the first equation: \(4(2a) + 3y = 10a\). 4. Solve: \(8a + 3y = 10a\), so \(3y = 2a\) and \(y = \frac{2}{3}a\).

\(x = 2a\) and \(y = \frac{2}{3}a\)

- What happens when the two equations are added? - Can subtraction eliminate the other variable? - Treat \(k\) as a fixed number while solving for \(x\) and \(y\). - How can the expression for \(x\) help you verify the expression for \(y\)?

1. Add the equations: \((4x + 3y) + (4x - 3y) = 11k + 5k\). 2. Simplify: \(8x = 16k\), so \(x = 2k\). 3. Subtract equation (2) from equation (1): \((4x + 3y) - (4x - 3y) = 11k - 5k\). 4. Simplify: \(6y = 6k\), so \(y = k\).

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

- First move the constant terms so that each equation is in standard form. - What multipliers will make one pair of coefficients equal? - After finding one variable, substitute to find the other. - Check the ordered pair in the original equations.

1. Rewrite the equations in standard form: \(4x + 3y = 1\) and \(5x + 2y = 3\). 2. Multiply the first equation by \(2\): \(8x + 6y = 2\). 3. Multiply the second equation by \(3\): \(15x + 6y = 9\). 4. Subtract the first transformed equation from the second: \(7x = 7\), so \(x = 1\). 5. Substitute \(x = 1\) into \(4x + 3y = 1\): \(4 + 3y = 1\), so \(y = -1\).

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

- Expand the expressions and put both equations in standard form first. - What factor will make one variable cancel when the equations are added? - Pay close attention to signs when distributing. - Check the result in both original equations.

1. Expand and rewrite the equations in standard form: \(2x - 3y = -5\) and \(4x - 5y = -3\). 2. Multiply the first equation by \(-2\): \(-4x + 6y = 10\). 3. Add this equation to the second equation: \(y = 7\). 4. Substitute \(y = 7\) into \(2x - 3y = -5\): \(2x - 21 = -5\). 5. Solve: \(2x = 16\), so \(x = 8\). 6. The solution is \((8, 7)\).

The solution is \((8, 7)\).

- Multiply each equation by its least common denominator to remove fractions. - What is the least common multiple of the denominators in each equation? - After removing fractions, which variable can be eliminated immediately? - Check the solution in the original equations.

1. Multiply the first equation by \(4\) to remove fractions: \(2x + 3y = 24\). 2. Multiply the second equation by \(6\): \(4x - 3y = 12\). 3. Add the new equations: \(6x = 36\), so \(x = 6\). 4. Substitute \(x = 6\) into \(2x + 3y = 24\): \(12 + 3y = 24\). 5. Solve: \(3y = 12\), so \(y = 4\). 6. The solution set is \(\{(6, 4)\}\).

The solution set is \(\{(6, 4)\}\).

- Multiply each equation by a common denominator to remove fractions. - After removing fractions, look for opposite coefficients. - What is the least common denominator in the first equation?

1. Multiply the first equation by \(6\): \(3x - 2y = 6\). 2. Multiply the second equation by \(4\): \(x + 2y = 10\). 3. Add the new equations: \(4x = 16\), so \(x = 4\). 4. Substitute \(x = 4\) into \(x + 2y = 10\): \(4 + 2y = 10\). 5. Solve: \(2y = 6\), so \(y = 3\).

The solution is \((4, 3)\).

- Multiply each equation to remove all denominators. - What number clears both denominators in the first equation? - Compare the \(y\)-terms after the fractions are removed. - Check the solution in the original equations.

1. Multiply the first equation by \(6\): \(3x - 2y = 12\). 2. Multiply the second equation by \(12\): \(3x + 2y = 24\). 3. Add the new equations: \(6x = 36\), so \(x = 6\). 4. Substitute \(x = 6\) into \(3x + 2y = 24\): \(18 + 2y = 24\). 5. Solve: \(2y = 6\), so \(y = 3\). 6. The solution set is \(\{(6, 3)\}\).

The solution set is \(\{(6, 3)\}\).

- What multipliers will create opposite coefficients for one variable? - Elimination is useful when adding equations can remove a variable. - Find a common multiple of one pair of coefficients. - Check the result in both equations.

1. Multiply the first equation by \(2\): \(8x - 6y = 10\). 2. Multiply the second equation by \(3\): \(9x + 6y = 24\). 3. Add the transformed equations: \(17x = 34\), so \(x = 2\). 4. Substitute \(x = 2\) into the first equation: \(4(2) - 3y = 5\). 5. Solve: \(8 - 3y = 5\), so \(y = 1\).

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

- Change the equations so that one variable cancels when they are added. - What multipliers make a pair of coefficients opposites? - Check the ordered pair in both original equations.

1. Multiply the first equation by \(5\): \(25x + 10y = 5\). 2. Multiply the second equation by \(-2\): \(-6x - 10y = -24\). 3. Add the transformed equations: \(19x = -19\), so \(x = -1\). 4. Substitute \(x = -1\) into the first equation: \(5(-1) + 2y = 1\). 5. Solve: \(-5 + 2y = 1\), so \(y = 3\).

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

- First remove the decimal coefficients by multiplying each equation by \(10\). - Can one equation then be scaled to create equal coefficients? - Subtracting equations with equal coefficients will eliminate that variable. - Check the solution in both original equations.

1. Multiply both equations by \(10\): (I) \(4x + 3y = 50\) (II) \(2x + 5y = 60\). 2. Multiply equation (II) by \(2\): \(4x + 10y = 120\). 3. Subtract equation (I): \(7y = 70\), so \(y = 10\). 4. Substitute \(y = 10\) into equation (I): \(4x + 30 = 50\), so \(x = 5\). 5. The solution set is \(\{(5, 10)\}\).

The solution set is \(\{(5, 10)\}\).

- Rewrite the mixed number as an improper fraction. - Multiply each equation by a common denominator to remove fractions. - Which algebraic method is simplest after the fractions are removed?

1. Rewrite the mixed number: \(2\frac{1}{12} = \frac{25}{12}\). 2. Multiply the first equation by \(12\): \(3x + 4y = 25\). 3. Multiply the second equation by \(6\): \(3x - y = 5\). 4. Subtract the second equation from the first: \(5y = 20\), so \(y = 4\). 5. Substitute \(y = 4\) into \(3x - y = 5\): \(3x - 4 = 5\), so \(x = 3\). 6. The solution set is \(\{(3, 4)\}\).

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

- Write one total-cost equation for each day. - Use multipliers such as \(1.1\) and \(1.2\) for the price increases. - Eliminate one variable after simplifying the equations. - For part b, divide the sandwich cost by the total cost.

1. Let \(s\) be the original sandwich price and \(a\) the original apple price, in dollars. 2. Monday’s cost gives \(30s + 20a = 90\), or \(3s + 2a = 9\). 3. Tuesday’s cost gives \(30(1.1s) + 20(1.2a) = 102\), or \(33s + 24a = 102\). 4. Multiply \(3s + 2a = 9\) by \(12\): \(36s + 24a = 108\). 5. Subtract the Tuesday equation: \(3s = 6\), so \(s = 2\). 6. Substitute into \(3s + 2a = 9\): \(6 + 2a = 9\), so \(a = 1.50\). 7. Monday’s sandwich cost was \(30 \cdot 2 = 60\) dollars. The percentage was \(\frac{60}{90} \cdot 100\% = 66.\overline{6}\% \approx 66.7\%\).

a) A sandwich originally cost \(\$2.00\), and an apple originally cost \(\$1.50\). b) Sandwiches accounted for approximately \(66.7\%\) of Monday’s total cost.

- Which variable has the same coefficient in both equations? - Subtract the equations so that variable disappears. - Use a common denominator to combine the remaining fractions.

1. Subtract equation (II) from equation (I) to cancel \(b\): \(\frac{a}{4}-\frac{a}{20}=12\). 2. Rewrite with denominator \(20\): \(\frac{5a-a}{20}=12\), so \(\frac{a}{5}=12\) and \(a=60\). 3. Substitute into equation (I): \(15+b=20\), giving \(b=5\). 4. Check: \(60\div4+5=20\) and \(60\div20+5=8\).

\((a, b)=(60, 5)\). Check (I): \(\frac{60}{4}+5=20\). Check (II): \(\frac{60}{20}+5=8\).

- You may need to multiply both equations before adding them. - Find the least common multiple of one pair of coefficients. - Decide whether eliminating \(x\) or \(y\) is more efficient.

1. Multiply the first equation by \(3\): \(9x + 12y = 15\). 2. Multiply the second equation by \(4\): \(8x - 12y = 36\). 3. Add the equations: \(17x = 51\), so \(x = 3\). 4. Substitute into the first equation: \(3(3) + 4y = 5\). 5. Solve: \(9 + 4y = 5\), so \(y = -1\). 6. The solution set is \(\{(3, -1)\}\).

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

- Scale both equations so that one pair of coefficients has equal magnitude and opposite signs. - Find the least common multiple of the \(x\)-coefficients or the \(y\)-coefficients. - A fractional coordinate is possible. - Simplify before multiplying when substituting a fraction.

1. A common multiple of \(15\) and \(20\) is \(60\). 2. Multiply the first equation by \(4\): \(88x - 60y + 4 = 0\). 3. Multiply the second equation by \(-3\): \(-99x + 60y - 12 = 0\). 4. Add the transformed equations: \(-11x - 8 = 0\), so \(x = -\frac{8}{11}\). 5. Substitute into the first original equation: \(22\left(-\frac{8}{11}\right) - 15y + 1 = 0\). 6. Simplify: \(-16 - 15y + 1 = 0\), so \(-15y = 15\) and \(y = -1\).

The solution is \(\left(-\frac{8}{11}, -1\right)\).

- Can one variable be eliminated by adding suitably scaled equations? - What multiplier makes the \(y\)-coefficients opposites? - Treat the expressions involving \(u\) and \(v\) as constants while solving for \(x\) and \(y\). - Simplify the final expressions.

1. Multiply the second equation by \(2\): \(4x - 2y = 2u + 6v\). 2. Add it to the first equation: \(5x = 5u + 5v\). 3. Divide by \(5\): \(x = u + v\). 4. Substitute into the first equation: \((u + v) + 2y = 3u - v\). 5. Simplify: \(2y = 2u - 2v\), so \(y = u - v\).

\(x = u + v\) and \(y = u - v\)

- Scale the equations so that one variable cancels when they are added. - Create coefficients with equal magnitude and opposite signs. - Substitute the first value you find into an original equation.

1. Multiply equation (I) by \(3\): \(18x - 15y = 33\). 2. Multiply equation (II) by \(5\): \(20x + 15y = 195\). 3. Add the equations: \(38x = 228\), so \(x = 6\). 4. Substitute into equation (II): \(4(6) + 3y = 39\). 5. Solve: \(24 + 3y = 39\), so \(y = 5\). 6. The solution set is \(\{(6, 5)\}\).

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

- First move each constant term to the right side. - You may multiply by powers of \(10\) to replace decimal coefficients with integers. - Which variable can be eliminated efficiently?

1. Rewrite the equations in standard form: (I) \(0.4x - 0.3y = 0.5\) (II) \(0.5x + 0.2y = 1.2\). 2. Multiply equation (I) by \(2\): \(0.8x - 0.6y = 1.0\). 3. Multiply equation (II) by \(3\): \(1.5x + 0.6y = 3.6\). 4. Add the equations: \(2.3x = 4.6\), so \(x = 2\). 5. Substitute into equation (II): \(0.5(2) + 0.2y = 1.2\), so \(y = 1\). 6. The solution set is \(\{(2, 1)\}\).

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

- What variable value is determined by the surviving equation? - Use the equation whose constant is known before recovering the erased constant. - Verify that combining the completed equations gives the stated elimination result.

1. From \(10x=50\), \(x=5\). 2. Substitute into equation (I): \(7\cdot 5+4y=39\), so \(4y=4\) and \(y=1\). 3. Use equation (II): \(c=3\cdot 5-4\cdot 1=11\).

\(c=11\), and the completed system has solution set \(\{(5, 1)\}\).

- Track the subtraction sign across every term of the second equation. - Which way of combining the equations makes a pair of opposite terms disappear? - Check the ordered pair in both original equations.

1. In the subtraction, the y-terms are \(-7y-(7y)=-14y\), not \(0\), so subtraction does not eliminate \(y\). 2. Add the equations instead: \(8x=32\), so \(x=4\). 3. Substitute into equation (II): \(2\cdot4+7y=22\), so \(7y=14\) and \(y=2\).

a) Elena treated \(-7y-7y\) as \(0\), but it equals \(-14y\). b) The solution set is \(\{(4,2)\}\).

- Look for the smallest common magnitude the two x-coefficients can reach. - The transformed x-terms must have opposite signs. - After one variable remains, use either original equation to recover the other.

1. The least common multiple of \(6\) and \(9\) is \(18\), so multiply equation (I) by \(3\) and equation (II) by \(-2\). 2. The transformed equations are \(18x+15y=21\) and \(-18x+8y=-44\). Adding gives \(23y=-23\), so \(y=-1\). 3. Substitute into equation (I): \(6x+5(-1)=7\), so \(6x=12\) and \(x=2\).

One pair of smallest multipliers is \(3\) and \(-2\). The solution set is \(\{(2, -1)\}\).

- First determine the ordered pair forced by the equations with known constants. - Choose a multiplier that creates opposite coefficients for one variable. - The final constant must equal the third left side at the common solution.

1. Multiply equation (I) by \(2\): \(4x-2y=14\). 2. Add the result to equation (II): \(7x=18\), so \(x=\frac{18}{7}\). 3. Substitute into equation (I): \(2\cdot\frac{18}{7}-y=7\), so \(y=-\frac{13}{7}\). 4. Evaluate the left side of equation (III): \(5x+y=5\cdot\frac{18}{7}-\frac{13}{7}=11\). Thus, \(k=11\).

\(k=11\).

- Assign one variable to each type of time block. - Which equation can be scaled so one pair of terms becomes opposites? - Use the recovered unit costs only after the system has been solved.

1. Let \(s\) be the studio-block cost and \(e\) the editing-block cost. The system is (I) \(3s+4e=247\) (II) \(5s+2e=281\). 2. Multiply equation (II) by \(-2\): \(-10s-4e=-562\). Add the result to equation (I): \(-7s=-315\), so \(s=45\). 3. Substitute into equation (II): \(5\cdot45+2e=281\), so \(e=28\). 4. The requested cost is \(2\cdot45+3\cdot28=174\).

a) A studio block costs \(\$45\), and an editing block costs \(\$28\). b) The cost is \(\$174\).

- Compare the transformed coefficients before carrying out the arithmetic. - Keep the same operation on the constants as on the left sides. - Either surviving one-variable equation should lead to the same ordered pair.

1. Plan A gives \(3(8x+3y)-2(12x-5y)=3\cdot10-2\cdot34\), so \(19y=-38\). It eliminates \(x\). 2. Plan B gives \(5(8x+3y)+3(12x-5y)=5\cdot10+3\cdot34\), so \(76x=152\). It eliminates \(y\). 3. Either result gives \(y=-2\) or \(x=2\). Substitution into an original equation gives the other value, so the solution set is \(\{(2, -2)\}\).

a) Plan A eliminates \(x\); Plan B eliminates \(y\). b) Plan A gives \(19y=-38\); Plan B gives \(76x=152\). c) The solution set is \(\{(2, -2)\}\).

- Express the combined coefficient of the variable that should disappear. - The disappearing coefficient must equal zero. - After finding the multiplier, apply it to every term, including the constant.

1. The y-coefficient in the combination is \(6-2q\). Setting it equal to \(0\) gives \(q=3\). 2. Add \(3\) times equation (I) to equation (II): \((12x-6y)+(5x+6y)=30+4\), so \(17x=34\). 3. Thus, \(x=2\). Substitute into equation (I): \(4\cdot2-2y=10\), so \(y=-1\).

a) \(q=3\). b) The resulting equation is \(17x=34\). c) The solution set is \(\{(2,-1)\}\).

- The original equations already contain one useful pair of opposite terms. - For the second coordinate, create a new pair of opposite coefficients for the other variable. - Keep the two elimination combinations separate so neither coordinate is found by substitution.

1. Add equations (I) and (II): \(12x=36\), so \(x=3\). 2. To eliminate \(x\), multiply equation (I) by \(5\) and equation (II) by \(-7\): \(35x+20y=145\) and \(-35x+28y=-49\). 3. Add the transformed equations: \(48y=96\), so \(y=2\).

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

- Make the variable terms match before subtracting. - Think about what a true statement with no variables means. - What if the constant sides do not match after cancellation?

1. Double (I): \(2x+4y=10\). 2. Subtract this from (II): \(0=k-10\). 3. If \(k=10\), the equations represent the same line, so there are infinitely many solutions. 4. If \(k\ne10\), the equation says \(0\) equals a nonzero number, so the system has no solution.

After elimination: \(0=k-10\). \(k=10\): infinitely many solutions. \(k\ne10\): no solution.

- Compare each original equation with its transformed version. - Which terms cancel when the transformed equations are added? - Which terms cancel when one transformed equation is subtracted from the other?

1. Equation (I) was multiplied by \(2\), and equation (II) was multiplied by \(3\). 2. Adding equations (I′) and (II′) eliminates \(y\). Subtracting equation (I′) from equation (II′) eliminates \(x\). 3. Add the transformed equations: \(36x = 96\), so \(x = \frac{8}{3}\). 4. Substitute into equation (II): \(6 \cdot \frac{8}{3} + 4y = 24\), so \(16 + 4y = 24\) and \(y = 2\). 5. The solution set is \(\{(\frac{8}{3}, 2)\}\).

a) Equation (I) was multiplied by \(2\), and equation (II) was multiplied by \(3\). b) Addition eliminates \(y\); subtraction eliminates \(x\). c) The solution set is \(\{(\frac{8}{3}, 2)\}\).

- Decide what relationship the two \(x\)-coefficients need so addition removes \(x\). - After choosing a \(y\)-coefficient, check whether the new equation would accidentally be equivalent to the first equation. - Use the required solution point to determine the constant term of your equation.

1. To eliminate \(x\), the coefficient \(a\) in equation (II) must be the opposite of \(1\), so \(a=-1\). 2. Start with \(-x+by=c\). Choose any \(b\ne-1\); for example, let \(b=2\). If \(b=-1\), the second equation would be the negative of equation (I), and the system would not have a unique solution. 3. Substitute \((5,2)\) to find \(c\): \(-5+2\cdot2=c\), so \(c=-1\). One possible equation is \(-x+2y=-1\). 4. Add the equations: \(3y=6\), so \(y=2\). 5. Substitute into equation (I): \(x+2=7\), so \(x=5\).

One possible equation is (II) \(-x+2y=-1\). More generally, any equation \(-x+by=-5+2b\) with \(b\ne-1\) works. Solving the system confirms the solution \((5,2)\).

- Choose two nonzero \(x\)-coefficients for which neither is an integer multiple of the other. - Make sure the two equations are not equivalent and that eliminating \(x\) will not eliminate \(y\) at the same time. - Use the required solution point to determine each constant after choosing your coefficients.

1. Choose \(x\)-coefficients that are not integer multiples of one another, such as \(2\) and \(3\). Choose \(y\)-coefficients that will not also cancel after scaling, such as \(3\) and \(2\). 2. Use \((3,4)\) to calculate the constants: \(2\cdot3+3\cdot4=18\), and \(3\cdot3+2\cdot4=17\). 3. One possible system is (I) \(2x+3y=18\) (II) \(3x+2y=17\). 4. Multiply equation (I) by \(3\) and equation (II) by \(-2\): (I′) \(6x+9y=54\) (II′) \(-6x-4y=-34\). 5. Add the transformed equations: \(5y=20\), so \(y=4\). Substitution gives \(x=3\).

One possible system is (I) \(2x+3y=18\) (II) \(3x+2y=17\). Multiplying equation (I) by \(3\) and equation (II) by \(-2\) creates \(6x\) and \(-6x\). Adding eliminates \(x\) and gives \(5y=20\), so \(y=4\), and then \(x=3\).

- Multiply each entire equation by its least common denominator. - Pay close attention to signs when a subtraction applies to a numerator in parentheses. - Arrange the simplified equations with the variable terms aligned before applying elimination.

1. Multiply equation (I) by \(6\): \(3(x - 2) + 2(y + 1) = 12\). Expanding and simplifying gives \(3x + 2y = 16\). 2. Multiply equation (II) by \(6\): \(2(x + 4) - 3(y - 1) = 0\). Expanding and simplifying gives \(2x - 3y = -11\). 3. Multiply the first simplified equation by \(3\) and the second by \(2\): \(9x + 6y = 48\) and \(4x - 6y = -22\). 4. Add the equations: \(13x = 26\), so \(x = 2\). 5. Substitute into \(3x + 2y = 16\): \(6 + 2y = 16\), so \(y = 5\). 6. The solution set is \(\{(2, 5)\}\).

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

- Represent each printer’s hourly rate as a fraction of one complete order. - Doubling the number of printers of one model doubles that model’s hourly contribution. - Write one equation for each staffing arrangement. - Relate a machine’s hourly rate to the time it needs to complete one order.

1. Let \(r_X\) and \(r_Y\) be the fractions of the order completed per hour by one printer of each model. 2. Write the system: \(r_X + r_Y = \frac{1}{4}\) and \(2r_X + r_Y = \frac{1}{3}\). 3. Subtract the first equation from the second: \(r_X = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}\). 4. Substitute into the first equation: \(\frac{1}{12} + r_Y = \frac{1}{4}\), so \(r_Y = \frac{1}{6}\). 5. Time is the reciprocal of rate. Model X takes \(12\) hours, and Model Y takes \(6\) hours.

One Model X printer would take \(12\) hours, and one Model Y printer would take \(6\) hours.

- Write an equation for the fraction of the park completed in each situation. - Remember that the first situation accounts for only half of the park. - After solving the system, compare the rates by forming their quotient.

1. Let \(g\) be Mr. Green’s work rate and \(b\) be Ms. Bloom’s work rate, measured in parks completed per hour. 2. Write the system: \(4g + 2b = \frac{1}{2}\) and \(2g + 5b = 1\). 3. Multiply the second equation by \(2\): \(4g + 10b = 2\). 4. Subtract the first equation: \(8b = \frac{3}{2}\), so \(b = \frac{3}{16}\). 5. Substitute into \(2g + 5b = 1\): \(2g + \frac{15}{16} = 1\), so \(g = \frac{1}{32}\). 6. Since \(\frac{3}{16} = \frac{6}{32}\), Ms. Bloom’s rate is greater. The ratio of her rate to Mr. Green’s rate is \(6:1\).

Ms. Bloom has the greater work rate. Her rate is \(6\) times Mr. Green’s rate, so the ratio is \(6:1\).

- Determine what fraction of the load is completed during the first \(4\) hours. - The remaining work is completed entirely by Belt B, which helps determine its rate. - Doubling Belt B’s speed doubles only Belt B’s hourly contribution. - Compare the new combined time with half of the original time.

1. Let \(r_A\) and \(r_B\) be the fractions of one full load completed per hour by Belts A and B. 2. From their combined time, \(r_A + r_B = \frac{1}{12}\). 3. The test gives \(4r_A + 24r_B = 1\), because both belts run for \(4\) hours and Belt B runs for \(20\) additional hours. 4. Multiply the first equation by \(4\): \(4r_A + 4r_B = \frac{1}{3}\). Subtract this equation from the test equation: \(20r_B = \frac{2}{3}\), so \(r_B = \frac{1}{30}\). 5. Then \(r_A = \frac{1}{12} - \frac{1}{30} = \frac{1}{20}\). Therefore, Belt A takes \(20\) hours alone, and Belt B takes \(30\) hours alone. 6. If Belt B’s rate doubles, the new combined rate is \(\frac{1}{20} + 2 \cdot \frac{1}{30} = \frac{7}{60}\). 7. The new time is \(1 \div \frac{7}{60} = \frac{60}{7} \approx 8.57\) hours. Half of \(12\) hours is \(6\) hours, so the technician’s claim is false.

a) Belt A would take \(20\) hours, and Belt B would take \(30\) hours. b) The claim is false. The new combined time is \(\frac{60}{7} \approx 8.57\) hours, not \(6\) hours.

- What happens when the corresponding coefficients are added? - Can you divide the resulting equation by a common factor? - This method works especially well when the coefficients are swapped between the two equations.

1. Add equations (I) and (II): \(24x + 24y = 48\). Divide by \(24\) to get \(x + y = 2\). 2. Subtract equation (II) from equation (I): \(-2x + 2y = 2\). Divide by \(2\) to get \(-x + y = 1\). 3. Solve the simpler system: (A) \(x + y = 2\) (B) \(-x + y = 1\). Adding gives \(2y = 3\), so \(y = 1.5\). Then \(x = 0.5\). 4. Adding and subtracting exploits the swapped coefficients and immediately produces small coefficients, avoiding products such as \(11 \cdot 13 = 143\).

a) \(x + y = 2\) b) \(-x + y = 1\) c) The solution set is \(\{(0.5, 1.5)\}\). d) Addition and subtraction immediately produce equations with small coefficients, avoiding arithmetic with large coefficients such as \(143\).

- Compare the coefficients of \(y\). Which operation eliminates that variable immediately? - After finding \(x\), substitute its expression into an original equation to find \(y\). - In part b, what equation represents the condition \(x = y\)? - Set up an equation containing only the parameter \(a\).

1. Add equations (1) and (2): \(4x = 8a - 4\). 2. Divide by \(4\): \(x = 2a - 1\). 3. Substitute into equation (2): \((2a - 1) - 2y = a - 5\). 4. Solve for \(y\): \(-2y = -a - 4\), so \(y = \frac{a + 4}{2} = 0.5a + 2\). 5. For part b, set the two expressions equal: \(2a - 1 = 0.5a + 2\). 6. Solve: \(1.5a = 3\), so \(a = 2\).

a) \(x = 2a - 1\) and \(y = \frac{a + 4}{2}\) b) \(a = 2\)

- Multiply each equation by the least common denominator to remove fractions. - Distribute carefully, especially when a minus sign precedes a fraction. - Rewrite the resulting equations in standard form. - Then use elimination to solve the simpler system.

1. Multiply the first equation by \(12\): \(4(x + 2) + 3(y - 1) = 24\), which simplifies to \(4x + 3y = 19\). 2. Multiply the second equation by \(6\): \(3(x - 1) - 2(y + 2) = 3\), which simplifies to \(3x - 2y = 10\). 3. Multiply \(4x + 3y = 19\) by \(2\): \(8x + 6y = 38\). 4. Multiply \(3x - 2y = 10\) by \(3\): \(9x - 6y = 30\). 5. Add the transformed equations: \(17x = 68\), so \(x = 4\). 6. Substitute \(x = 4\) into \(3x - 2y = 10\): \(12 - 2y = 10\), so \(y = 1\). 7. The solution set is \(\{(4, 1)\}\).

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

- Multiply each equation by the least common denominator. - Distribute carefully, especially after a subtraction sign. - Once the equations are simplified, choose the most efficient solving method. - Can suitable multipliers create opposite coefficients for one variable?

1. Multiply the first equation by \(6\): \(3(a + 1) + 2(b - 2) = 18\), which simplifies to \(3a + 2b = 19\). 2. Multiply the second equation by \(12\): \(4(a - 2) - 3(b + 2) = 0\), which simplifies to \(4a - 3b = 14\). 3. Multiply \(3a + 2b = 19\) by \(3\): \(9a + 6b = 57\). 4. Multiply \(4a - 3b = 14\) by \(2\): \(8a - 6b = 28\). 5. Add the transformed equations: \(17a = 85\), so \(a = 5\). 6. Substitute \(a = 5\) into \(3a + 2b = 19\): \(15 + 2b = 19\), so \(b = 2\).

The solution is \((a, b) = (5, 2)\).

- Clear the fractions by multiplying each equation by its least common denominator. - Multiply every term on both sides of each equation. - Pay attention to the subtraction sign before the second fraction in the first equation. - After simplifying, which variable is easiest to eliminate?

1. Multiply the first equation by \(12\): \(3(3x - 1) - 4(2y + 1) = 2(x - y)\). 2. Simplify: \(9x - 3 - 8y - 4 = 2x - 2y\), so \(7x - 6y = 7\). 3. Multiply the second equation by \(10\): \(5(x + y) + 2(x - y) = 20\). 4. Simplify: \(7x + 3y = 20\). 5. Subtract \(7x - 6y = 7\) from \(7x + 3y = 20\): \(9y = 13\), so \(y = \frac{13}{9}\). 6. Substitute into \(7x + 3y = 20\): \(7x + \frac{13}{3} = 20\), so \(7x = \frac{47}{3}\) and \(x = \frac{47}{21}\).

The solution is \(\left(\frac{47}{21}, \frac{13}{9}\right)\).

- Expand both sides of each equation first. - Combine like terms carefully and track all signs. - What happens to the \(xy\)-term when it appears on both sides? - Rewrite each result in the form \(Ax + By = C\). - Which method is efficient once the \(x\)-coefficients are opposites?

1. Expand the first equation: \(xy - 2x + y - 2 = xy + 3x - y - 3\). Cancel \(xy\) and simplify to get \(-5x + 2y = -1\). 2. Expand the second equation: \(xy + 4x - 3y - 12 = xy - x + 2y - 2\). Cancel \(xy\) and simplify to get \(5x - 5y = 10\). 3. Add the two simplified equations: \(-3y = 9\), so \(y = -3\). 4. Substitute \(y = -3\) into \(5x - 5y = 10\): \(5x + 15 = 10\), so \(x = -1\). 5. The solution is \((-1, -3)\).

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

- Write one equation for the original production time and one for the improved production time. - Represent a \(20\%\) reduction with a multiplier of \(0.8\) and a \(10\%\) reduction with a multiplier of \(0.9\). - For part b, consider which model accounted for more of the original total labor time and how that affects a weighted average.

1. Let \(x\) be the original labor time for one Standard unit and \(y\) the original labor time for one Deluxe unit, in hours. 2. The original production time gives \(50x + 40y = 130\), or \(5x + 4y = 13\). 3. After the reductions, the production time gives \(50(0.8x) + 40(0.9y) = 111\), or \(40x + 36y = 111\). 4. Multiply \(5x + 4y = 13\) by \(8\): \(40x + 32y = 104\). 5. Subtract this equation from \(40x + 36y = 111\): \(4y = 7\), so \(y = 1.75\). 6. Substitute into \(5x + 4y = 13\): \(5x + 7 = 13\), so \(x = 1.2\). 7. One Standard unit originally required \(1.2\) hours, or \(72\) minutes. One Deluxe unit originally required \(1.75\) hours, or \(105\) minutes. 8. The overall savings is a weighted average. The Deluxe units account for more of the original labor time, so their smaller \(10\%\) reduction has more influence than the Standard units’ \(20\%\) reduction.

a) A Standard unit originally required \(1.2\) hours, and a Deluxe unit originally required \(1.75\) hours. b) The overall savings is below \(15\%\) because more of the original labor time was spent on Deluxe units, which had the smaller percentage reduction.

- Identify the quantity that remains constant in every scenario. - Express the total work as the number of machines times the number of days. - Expand the equations and notice what happens to the product of the two unknowns. - Solve the resulting linear system.

1. Let \(m\) be the planned number of combines and \(d\) the planned number of days. The total work is proportional to \(md\). 2. Equal amounts of work give \((m + 2)(d - 2) = md\) and \((m - 3)(d + 6) = md\). 3. Expand the first equation: \(md - 2m + 2d - 4 = md\), so \(-m + d = 2\). 4. Expand the second equation: \(md + 6m - 3d - 18 = md\), so \(2m - d = 6\). 5. Add the simplified equations: \(m = 8\). 6. Substitute into \(-m + d = 2\): \(-8 + d = 2\), so \(d = 10\). 7. The planned work was \(8 \cdot 10 = 80\) combine-days. The alternatives also give \(10 \cdot 8 = 80\) and \(5 \cdot 16 = 80\) combine-days.

The plan called for \(8\) combines working for \(10\) days.

- Identify the total quantity that stays constant in every scenario. - Express total cost as the number of students times the cost per student. - Write an equation for each change in group size and price. - Expand the products to obtain a linear system.

1. Let \(n\) be the planned number of students and \(p\) the planned cost per student. The total bus cost is \(np\). 2. The first scenario gives \((n - 5)(p + 2.4) = np\). 3. Expand and simplify: \(np + 2.4n - 5p - 12 = np\), so \(2.4n - 5p = 12\). 4. The second scenario gives \((n + 10)(p - 3.2) = np\). 5. Expand and simplify: \(np - 3.2n + 10p - 32 = np\), so \(-3.2n + 10p = 32\). 6. Multiply the first simplified equation by \(2\): \(4.8n - 10p = 24\). 7. Add the simplified equations: \(1.6n = 56\), so \(n = 35\). 8. Substitute into \(2.4n - 5p = 12\): \(84 - 5p = 12\), so \(p = 14.4\).

The plan included \(35\) students at \(\$14.40\) per student.

- Choose variables for the two printing rates. - Use the \(50\)-minute completion time to find a simple relationship between the rates. - Determine how using multiple printers of the same model changes the combined rate. - Convert the fractional part of a minute to seconds.

1. Let \(a\) and \(b\) be the printing speeds of Models Alpha and Beta, in pages per minute. 2. Their combined work gives \(50(a + b) = 5000\), so \(a + b = 100\). 3. The test gives \(20a + 30b = 2600\). 4. Multiply \(a + b = 100\) by \(20\): \(20a + 20b = 2000\). 5. Subtract this equation from the test equation: \(10b = 600\), so \(b = 60\). Then \(a = 40\). 6. Three Alpha printers have a combined rate of \(3 \cdot 40 = 120\) pages per minute. 7. The time is \(5000 \div 120 = \frac{125}{3}\) minutes, which is \(41\) minutes and \(40\) seconds.

a) Model Alpha prints \(40\) pages per minute, and Model Beta prints \(60\) pages per minute. b) Three Alpha printers would take \(41\) minutes and \(40\) seconds.

- If a printer’s full-job time is unknown, express the time for a fraction of the job as the same fraction of that time. - Write one equation for each work plan. - Multiply the equations to clear the fractions. - Solve the resulting system of two linear equations.

1. Let \(t_A\) be the time Printer A needs for the entire job and \(t_B\) the time Printer B needs, in minutes. 2. Time is proportional to the fraction of the job completed, and the printer times for the two portions add. Thus, \(\frac{2}{5}t_A + \frac{3}{5}t_B = 100\) and \(\frac{3}{5}t_A + \frac{2}{5}t_B = 80\). 3. Multiply both equations by \(5\): \(2t_A + 3t_B = 500\) and \(3t_A + 2t_B = 400\). 4. Multiply the first equation by \(3\): \(6t_A + 9t_B = 1500\). 5. Multiply the second equation by \(2\): \(6t_A + 4t_B = 800\). 6. Subtract: \(5t_B = 700\), so \(t_B = 140\). 7. Substitute into \(2t_A + 3t_B = 500\): \(2t_A + 420 = 500\), so \(t_A = 40\).

Printer A would take \(40\) minutes, and Printer B would take \(140\) minutes, or \(2\) hours \(20\) minutes.

- Expand and simplify each equation before choosing a solving method. - Rewrite \(0.25\) as \(\frac{1}{4}\) when clearing denominators. - Put both equations in the form \(Ax + By = C\). - Reduce any fractions in the final answer.

1. Expand the first equation: \(0.8x - 2y + 1.2 = 0.2x\). 2. Rewrite it as \(0.6x - 2y = -1.2\), then multiply by \(10\): \(6x - 20y = -12\). 3. Multiply the second equation by \(4\): \(2(x - 1) = 3y + 1 - 1\). 4. Simplify: \(2x - 3y = 2\). 5. Multiply \(2x - 3y = 2\) by \(-3\): \(-6x + 9y = -6\). 6. Add the equations: \(-11y = -18\), so \(y = \frac{18}{11}\). 7. Substitute into \(2x - 3y = 2\): \(2x - \frac{54}{11} = 2\), so \(2x = \frac{76}{11}\) and \(x = \frac{38}{11}\).

The solution set is \(\left\{\left(\frac{38}{11}, \frac{18}{11}\right)\right\}\).

- Combine corresponding coefficients and constants in the same way on both sides. - Choose a multiplier that makes one pair of variable terms opposites. - To justify equivalence, ask whether the original equation can be recovered from the replacement.

1. Twice equation (I) is \(4x+10y=38\). Adding it to equation (II) gives \(x+12y=38\). 2. Multiply the replacement equation by \(-2\) and add it to equation (I): \(-19y=-57\), so \(y=3\). 3. Substitute into \(x+12y=38\): \(x+36=38\), so \(x=2\). 4. Every solution of the original system satisfies the replacement equation. Conversely, subtracting twice the unchanged equation (I) from the replacement equation recovers equation (II), so no solutions are gained or lost.

a) The replacement equation is \(x+12y=38\). b) The solution set is \(\{(2, 3)\}\). c) The operation is reversible, so the original and transformed systems have the same solution set.

- Undo the stated equation operation in reverse order. - Apply the reverse operation to both sides of the replacement equation. - Once the original pair is restored, choose coefficients that can be canceled.

1. Reverse the replacement by adding twice equation (I) to \(-5x+9y=-17\). 2. This gives \((-5x+9y)+2(3x-y)=-17+2\cdot8\), so the original equation (II) is \(x+7y=-1\). 3. Multiply equation (I) by \(-7\): \(-21x+7y=-56\). Subtract equation (II) to get \(-22x=-55\), so \(x=\frac{5}{2}\). 4. Substitute into equation (I): \(3\cdot\frac{5}{2}-y=8\), so \(y=-\frac{1}{2}\).

a) The original equation (II) is \(x+7y=-1\). b) The solution set is \(\left\{\left(\frac{5}{2}, -\frac{1}{2}\right)\right\}\).