Free Math Worksheets

- Move all variable terms to one side. - Move constant terms to the other side. - Use inverse operations on both sides. - Divide by the final coefficient of the variable.

1. For a), subtract \(5x\): \(3x + 13 = 31\). Subtract \(13\): \(3x = 18\). Divide by \(3\): \(x = 6\). 2. For b), add \(6y\): \(14 = 8y - 18\). Add \(18\): \(32 = 8y\). Divide by \(8\): \(y = 4\). 3. For c), subtract \(3z\): \(8z - 7 = 25\). Add \(7\): \(8z = 32\). Divide by \(8\): \(z = 4\).

a) \(x = 6\) b) \(y = 4\) c) \(z = 4\)

- Combine variable terms and constant terms separately on each side. - Move all variable terms to one side and constants to the other. - Divide by the final coefficient of the variable.

1. For a), combine the left side: \(17x - 5 = 2x + 40\). Subtract \(2x\): \(15x - 5 = 40\). Add \(5\), then divide by \(15\): \(x = 3\). 2. For b), combine both sides: \(30 - 4y = 8y - 12\). Add \(4y\): \(30 = 12y - 12\). Add \(12\): \(42 = 12y\). Divide by \(12\): \(y = 3.5 = \frac{7}{2}\).

a) \(x = 3\) b) \(y = 3.5 = \frac{7}{2}\)

- Combine the variable terms on the left first. - Move all variable terms to one side and constants to the other. - Divide by the final coefficient of \(x\).

1. Combine like terms on the left: \(11x - 15 = 3x + 25\). 2. Subtract \(3x\): \(8x - 15 = 25\). 3. Add \(15\): \(8x = 40\). 4. Divide by \(8\): \(x = 5\). 5. Therefore, \(S = \{5\}\).

\(S = \{5\}\)

- Move all variable terms to one side. - Move constants to the other side. - Divide by the final decimal coefficient.

1. Subtract \(0.6x\): \(4 = 0.5x - 3.5\). 2. Add \(3.5\): \(7.5 = 0.5x\). 3. Divide by \(0.5\): \(x = 15\).

\(x = 15\)

- Move variable terms to one side. - Move constants to the other side. - Divide by the final coefficient of the variable.

1. Add \(1.2x\): \(4x - 4.2 = 7.8\). 2. Add \(4.2\): \(4x = 12\). 3. Divide by \(4\): \(x = 3\).

\(x = 3\)

- Move all variable terms to one side. - Move constants to the other side. - Divide by the remaining coefficient of \(x\).

1. Subtract \(0.6x\): \(1.2x - 4.2 = 3\). 2. Add \(4.2\): \(1.2x = 7.2\). 3. Divide by \(1.2\): \(x = 6\).

\(x = 6\)

- Distribute before moving terms. - Keep decimal coefficients aligned carefully. - Use division to undo the final multiplication.

1. Distribute on the left: \(6x - 15 = 4.5x + 6\). 2. Subtract \(4.5x\): \(1.5x - 15 = 6\). 3. Add \(15\): \(1.5x = 21\). 4. Divide by \(1.5\): \(x = 14\).

\(x = 14\)

- Distribute the decimal factor first. - Move variable terms to one side. - After solving for \(x\), substitute it into the formula for \(A\).

1. Distribute on the right: \(1.5x - 2 = 0.5x + 3\). 2. Subtract \(0.5x\): \(x - 2 = 3\). 3. Add \(2\): \(x = 5\). 4. Substitute into the second expression: \(A = 3 \cdot 5 + 5 = 20\).

\(x = 5\) and \(A = 20\).

- Move variable terms to one side. - Distribute or remove parentheses before combining like terms. - Simplify each side before isolating the variable.

1. For a), subtract \(2x\): \(5x - 12 = 13\). Add \(12\), then divide by \(5\): \(x = 5\). 2. For b), distribute: \(3x - 12 = 2x - 3\). Subtract \(2x\), then add \(12\): \(x = 9\). 3. For c), remove the parentheses: \(10y - 2y - 4 = 6y\). Then \(8y - 4 = 6y\), so \(2y = 4\) and \(y = 2\).

a) \(x = 5\) b) \(x = 9\) c) \(y = 2\)

- Combine variable terms and constants on each side first. - Move variable terms to one side and constants to the other. - Substitute your result into the original equation to check it.

1. Combine like terms on the left: \(8x - 6 = 4x + 18\). 2. Subtract \(4x\): \(4x - 6 = 18\). 3. Add \(6\): \(4x = 24\). 4. Divide by \(4\): \(x = 6\). 5. Check: The left side is \(13 \cdot 6 - 8 - 5 \cdot 6 + 2 = 42\), and the right side is \(4 \cdot 6 + 18 = 42\).

\(x = 6\)

- Move all variable terms to one side. - Move constants to the other side. - Divide by the final coefficient of \(x\).

1. Subtract \(3x\): \(4x + 15 = -5\). 2. Subtract \(15\): \(4x = -20\). 3. Divide by \(4\): \(x = -5\).

\(x = -5\)

- Handle the negative sign before parentheses carefully. - Simplify both sides before moving terms. - Move variable terms to one side and constants to the other.

1. For a), remove the parentheses: \(18 - 3x - 6 = 2x - 3\). Then \(12 - 3x = 2x - 3\), so \(15 = 5x\) and \(x = 3\). 2. For b), distribute: \(5x - 5 = 2x + 7\). Then \(3x = 12\), so \(x = 4\).

a) \(x = 3\) b) \(x = 4\)

- Simplify each side first. - Move all variable terms to one side. - Move all constant terms to the other side. - Track the signs when combining terms.

1. Combine like terms on the left: \(-5z + 15 = 3 - 2z\). 2. Add \(5z\) to both sides: \(15 = 3 + 3z\). 3. Subtract \(3\): \(12 = 3z\). 4. Divide by \(3\): \(z = 4\).

\(z = 4\)

- Combine like terms on each side first. - Move all variable terms to one side. - Move constants to the other side. - Divide by the variable coefficient.

1. Combine like terms on the left: \(8x - 14 = 3x + 11\). 2. Subtract \(3x\): \(5x - 14 = 11\). 3. Add \(14\): \(5x = 25\). 4. Divide by \(5\): \(x = 5\).

\(x = 5\)

- Distribute the decimal factor to both terms. - Combine like terms before moving terms. - Move variable terms to one side and constants to the other.

1. Distribute \(0.4\): \(2x - 4 + 1.2x = 4.2x - 9\). 2. Combine like terms on the left: \(3.2x - 4 = 4.2x - 9\). 3. Subtract \(3.2x\): \(-4 = x - 9\). 4. Add \(9\): \(x = 5\).

\(x = 5\)

- Translate each phrase into an algebraic expression before writing the equation. - How can you represent “seven times a number” and “2 greater than \(x\)”? - If one quantity is 10 greater than another, where should the 10 appear in the equation? - Distribute before collecting the variable terms. - How can you write “the number that is 2 greater than \(x\)” algebraically?

1. Translate the statement into an equation: \(7x = 3(x + 2) + 10\). 2. Distribute on the right side: \(7x = 3x + 6 + 10\). 3. Combine like terms: \(7x = 3x + 16\). 4. Subtract \(3x\) from both sides: \(4x = 16\). 5. Divide both sides by \(4\): \(x = 4\).

\(x = 4\)

- Describe how each person’s savings changes each month. - Write one expression for Lucas’s savings and another for Sarah’s savings after \(x\) months. - Set the two expressions equal to find when the amounts match.

1. Let \(x\) be the number of months. Write an equation for equal savings: \(85 + 12x = 205 - 8x\). 2. Add \(8x\) to both sides: \(85 + 20x = 205\). 3. Subtract \(85\) from both sides: \(20x = 120\). 4. Divide both sides by \(20\): \(x = 6\). 5. Substitute \(6\) into either expression: \(85 + 12 \cdot 6 = 157\). 6. After 6 months, each friend will have \(\$157\).

After 6 months, each friend will have \(\$157\).

- Notice that Sarah saves more each week but starts with less. - Determine how much Sarah closes the gap each week. - Set the two expressions equal to find when the savings match.

1. Lucas's savings are \(L(w)=60+4w\). Sarah's savings are \(S(w)=10+9w\). 2. After \(5\) weeks, \(L(5)=60+4\cdot 5=80\), and \(S(5)=10+9\cdot 5=55\). 3. After \(12\) weeks, \(L(12)=60+4\cdot 12=108\), and \(S(12)=10+9\cdot 12=118\). 4. Set the expressions equal: \(60+4w=10+9w\). Subtract \(4w\) and \(10\): \(50=5w\), so \(w=10\). Each has \(\$100\).

a) Lucas: \(60+4w\); Sarah: \(10+9w\) b) After \(5\) weeks: Lucas \(\$80\), Sarah \(\$55\). After \(12\) weeks: Lucas \(\$108\), Sarah \(\$118\). c) After \(10\) weeks; each has \(\$100\).

- Distribute each factor to every term inside its parentheses. - A minus sign before parentheses changes both signs inside. - Combine like terms on each side before moving variable terms.

1. For a), distribute: \(4x - 12 = 2x + 10\). Subtract \(2x\), add \(12\), and divide by \(2\): \(x = 11\). 2. For b), remove the parentheses: \(15 - 3a - 4 = 5a - 21\). Then \(11 - 3a = 5a - 21\), so \(32 = 8a\) and \(a = 4\). 3. For c), distribute: \(6t + 8 = 5t - 10\). Subtract \(5t\) and \(8\): \(t = -18\).

a) \(x = 11\) b) \(a = 4\) c) \(t = -18\)

- Use a common denominator when combining fractional coefficients. - Distribute before combining variable terms. - Multiplying both sides by a denominator can clear a fraction. - Keep decimal points and signs aligned carefully.

1. For a), subtract \(\frac{1}{2}x\): \(\frac{1}{4}x - 5 = 2\). Add \(5\): \(\frac{1}{4}x = 7\). Multiply by \(4\): \(x = 28\). 2. For b), distribute: \(2x - 4 = 1.5x + 7\). Subtract \(1.5x\): \(0.5x - 4 = 7\). Add \(4\), then divide by \(0.5\): \(x = 22\). 3. For c), multiply by \(3\): \(2x - 4 = 3x - 15\). Subtract \(2x\), then add \(15\): \(x = 11\).

a) \(x = 28\) b) \(x = 22\) c) \(x = 11\)

- Treat decimal and fractional factors like any other factors when distributing. - A minus sign before parentheses acts like a factor of \(-1\). - Simplify both sides fully before moving variable terms. - Verify each result in the original equation.

1. For a), distribute: \(2x - 4 + 3x = 6 - x + 2\). Then \(5x - 4 = 8 - x\), so \(6x = 12\) and \(x = 2\). 2. For b), distribute: \(2y + 3 - 5 = 2y - 2 - y\). Then \(2y - 2 = y - 2\), so \(y = 0\). 3. For c), distribute: \(6m + 2 - 4m + 8 = 5m - 2\). Then \(2m + 10 = 5m - 2\), so \(12 = 3m\) and \(m = 4\).

a) \(x = 2\) b) \(y = 0\) c) \(m = 4\)

- Multiply by a common multiple of the denominators to clear the fractions. - The least common multiple of \(3\) and \(6\) is \(6\). - Move variable terms to one side and constants to the other.

1. Multiply both sides by the least common denominator, \(6\): \(4y - 24 = y + 6\). 2. Subtract \(y\): \(3y - 24 = 6\). 3. Add \(24\): \(3y = 30\). 4. Divide by \(3\): \(y = 10\).

\(y = 10\)

- Distribute the factor to each term in the parentheses. - A minus sign before parentheses changes both signs inside. - Simplify both sides before moving variable terms.

1. Distribute on the left: \(6a - 10 + 4\). 2. Remove the parentheses on the right: \(10 - a - 2\). 3. Simplify: \(6a - 6 = 8 - a\). 4. Add \(a\): \(7a - 6 = 8\). 5. Add \(6\): \(7a = 14\). Divide by \(7\): \(a = 2\).

\(a = 2\)

- Distribute carefully with decimal factors. - Combine like terms on each side before moving terms. - Keep decimal points aligned during subtraction and division.

1. Distribute on the left: \(6y - 3\). 2. Distribute and combine on the right: \(3y + 6 + 0.5y = 3.5y + 6\). 3. The equation is \(6y - 3 = 3.5y + 6\). 4. Subtract \(3.5y\): \(2.5y - 3 = 6\). 5. Add \(3\): \(2.5y = 9\). Divide by \(2.5\): \(y = 3.6\).

\(y = 3.6\)

- Distribute each factor to every term inside its parentheses. - A minus sign before parentheses changes both signs. - Simplify each side before moving variable terms.

1. For a), distribute: \(4x - 20 = 2x + 6\). Subtract \(2x\): \(2x - 20 = 6\). Add \(20\), then divide by \(2\): \(x = 13\). 2. For b), remove the parentheses and distribute: \(25 - 3z - 4 = 2z - 4\). Then \(21 - 3z = 2z - 4\). Add \(3z\) and \(4\): \(25 = 5z\), so \(z = 5\).

a) \(x = 13\) b) \(z = 5\)

- Check whether the outside factor was applied to every term inside the parentheses. - Redo the distribution before comparing later steps. - Solve the corrected equation and verify the result.

1. Max did not distribute \(2\) to both terms inside the parentheses. The correct expansion is \(2x + 8\), not \(2x + 4\). 2. Correct the equation: \(2x + 8 - 3 = 15 - x\). 3. Combine like terms: \(2x + 5 = 15 - x\). 4. Add \(x\): \(3x + 5 = 15\). 5. Subtract \(5\): \(3x = 10\). Divide by \(3\): \(x = \frac{10}{3}\).

Max failed to multiply \(4\) by \(2\) when distributing. The correct solution is \(x = \frac{10}{3}\).

- Solve each equation independently. - Distribute or divide to remove the parentheses. - Compare the final solution sets, not just the appearance of the equations.

1. For Alex’s equation, distribute: \(3a + 12 = 2a + 18\). Subtract \(2a\), then subtract \(12\): \(a = 6\). Its solution set is \(S = \{6\}\). 2. For Sofia’s equation, divide by \(2\): \(a - 3 = 3\). Add \(3\): \(a = 6\). Its solution set is \(S = \{6\}\). 3. Since both solution sets are \(\{6\}\), the equations have the same solution set.

Yes. Both equations have solution set \(S = \{6\}\).

- The phrase “the same result” indicates an equation. - How can you represent twice a number and three times a number? - If one quantity is \(4\) less than another, how can you relate the two expressions?

1. For part a, write \(2x + 5 = 3x - 3\). Subtract \(2x\): \(5 = x - 3\). Add \(3\): \(x = 8\). 2. For part b, write \(x = 5x - 4\). Subtract \(x\): \(0 = 4x - 4\). Add \(4\): \(4 = 4x\). Divide by \(4\): \(x = 1\). Because the equation has a solution, such a number exists.

a) The number is \(8\). b) Yes. The number is \(1\).

- In part b, the word “sum” signals that parentheses are needed. - You may clear the fraction in part a before collecting the variable terms. - Pay attention to which expression is \(2\) less than the other in part b.

1. For part a, write \(\frac{1}{4}x + 0.5 = 2x - 3\). Multiply both sides by \(4\): \(x + 2 = 8x - 12\). Subtract \(x\): \(2 = 7x - 12\). Add \(12\): \(14 = 7x\). Divide by \(7\): \(x = 2\). 2. For part b, write \(3(x + 4) = 5x - 2\). Distribute: \(3x + 12 = 5x - 2\). Subtract \(3x\): \(12 = 2x - 2\). Add \(2\): \(14 = 2x\). Divide by \(2\): \(x = 7\).

a) \(x = 2\) b) \(x = 7\)

- Distribute each factor to every term inside its parentheses. - A minus sign before parentheses changes both signs. - Simplify both sides before moving variable terms.

1. Distribute and remove parentheses: \(12y - 20 - 2y - 7 = 3y + 3\). 2. Combine like terms: \(10y - 27 = 3y + 3\). 3. Subtract \(3y\): \(7y - 27 = 3\). 4. Add \(27\): \(7y = 30\). 5. Divide by \(7\): \(y = \frac{30}{7}\). 6. Therefore, \(S = \{\frac{30}{7}\}\).

\(S = \{\frac{30}{7}\}\)

- Both brothers get one year older each year. - Use a variable for the number of years from now. - Write an expression for each brother’s future age. - How can you represent “twice Leo’s age” algebraically?

1. Let \(x\) be the number of years from now. 2. Tim’s age will be \(8 + x\), and Leo’s age will be \(2 + x\). 3. Write the equation \(8 + x = 2(2 + x)\). 4. Distribute: \(8 + x = 4 + 2x\). 5. Subtract \(x\) and then subtract \(4\): \(x = 4\).

In 4 years, Tim will be twice Leo’s age.

- The difference between their ages stays constant, but does the ratio stay constant? - Test Jonas’s claim by calculating both ages 4 years from now. - Use a variable for the number of years in part c. - Remember that both people age by the same number of years.

1. For part a, divide the mother’s age by the son’s age: \(36 \div 6 = 6\), so the ratio of Ms. Meyer’s age to Jonas’s age is \(6\) to \(1\). 2. In 4 years, Ms. Meyer will be 40 and Jonas will be 10. Since \(40 \div 10 = 4\), Jonas will be one-fourth of her age, not one-half. 3. For part c, let \(x\) be the number of years from now. Write \(36 + x = 3(6 + x)\). 4. Distribute: \(36 + x = 18 + 3x\). 5. Subtract \(x\) and then subtract \(18\): \(18 = 2x\), so \(x = 9\).

a) The current ratio of Ms. Meyer’s age to Jonas’s age is \(6\) to \(1\). b) No. In 4 years Jonas will be 10 and Ms. Meyer will be 40, so he will be one-fourth of her age. c) Ms. Meyer will be three times Jonas’s age in 9 years.

- Let the variable represent the total trail length. - Combine the fractional parts of the first two sections. - What fraction of the whole trail remains for the last section? - The remaining fraction of the total equals \(7\) miles.

1. Let \(x\) be the trail’s total length in miles. 2. Write the equation \(\frac{1}{3}x + \frac{1}{5}x + 7 = x\). 3. Use a common denominator: \(\frac{5}{15}x + \frac{3}{15}x + 7 = x\). 4. Combine like terms: \(\frac{8}{15}x + 7 = x\). 5. Subtract \(\frac{8}{15}x\) from both sides: \(7 = \frac{7}{15}x\). 6. Multiply by \(\frac{15}{7}\): \(x = 7 \cdot \frac{15}{7} = 15\).

The hiking trail is \(15\) miles long.

- A full lap occurs when the distance difference reaches \(400\) meters. - Write a distance expression for each runner. - Divide Marie's distance by \(400\) to identify her lap. - Consider how changing speeds would affect the distance rules.

1. Sophie has lapped Marie once when her distance is \(400\) meters greater: \(3.8t = 3t + 400\). 2. Subtract \(3t\): \(0.8t = 400\), so \(t = 500\) seconds. 3. Sophie runs \(3.8 \cdot 500 = 1900\) meters. 4. Marie runs \(3 \cdot 500 = 1500\) meters. Since \(1500 \div 400 = 3.75\), she has completed three laps and is on lap \(4\). 5. Constant speeds make each distance a linear function of time. Without that assumption, these distance expressions would not remain valid.

a) Sophie laps Marie after \(500\) seconds. b) Sophie has run \(1900\) meters. c) Marie is on lap \(4\). Constant speeds are needed so distance remains proportional to time.

- Convert the delayed start to hours. - Julia's travel time is shorter than Luke's by \(\frac{1}{6}\) hour. - At the catch, both riders have traveled the same distance. - Calculate each full-trip travel time separately for part c).

1. Convert the delayed start: \(10\) minutes is \(\frac{1}{6}\) hour. 2. Luke travels \(12t\) miles. Julia travels \(18\left(t - \frac{1}{6}\right)\) miles. 3. At the catch, their distances are equal: \(12t = 18\left(t - \frac{1}{6}\right)\). 4. Distribute and solve: \(12t = 18t - 3\), so \(6t = 3\) and \(t = 0.5\) hour. 5. The catch occurs at \(3{:}30\) p.m., \(12 \cdot 0.5 = 6\) miles from the start. 6. Luke takes \(12 \div 12 = 1\) hour and arrives at \(4{:}00\) p.m. Julia takes \(12 \div 18 = \frac{2}{3}\) hour, or \(40\) minutes, and arrives at \(3{:}50\) p.m.

a) \(12t = 18\left(t - \frac{1}{6}\right)\) b) Julia catches Luke at \(3{:}30\) p.m., \(6\) miles from the start. c) Yes. Julia arrives at \(3{:}50\) p.m., and Luke arrives at \(4{:}00\) p.m.

- Solve each equation independently. - Isolate \(y\) using inverse operations. - Compare the complete solution sets.

1. Solve Equation I: \(4y + 8 = 2y - 6\). Then \(2y = -14\), so \(y = -7\) and \(S_1 = \{-7\}\). 2. Solve Equation II: \(0.5y + 10 = 3\). Then \(0.5y = -7\), so \(y = -14\) and \(S_2 = \{-14\}\). 3. Because \(S_1 \ne S_2\), the equations are not equivalent.

No. Equation I has solution \(y = -7\), and Equation II has solution \(y = -14\).

- Multiply by a common denominator to clear both fractions. - Distribute to every term inside the parentheses. - Simplify each side before moving variable terms.

1. Multiply the entire equation by \(4\) to clear the denominators: \(8(y - 1.5) - y = 2(3y + 2)\). 2. Distribute: \(8y - 12 - y = 6y + 4\). 3. Combine like terms: \(7y - 12 = 6y + 4\). 4. Subtract \(6y\) and add \(12\): \(y = 16\).

\(y = 16\)

- Distribute the decimal factor to both terms. - Move variable terms to one side and constants to the other. - Check the result in the original equation, not only the simplified one.

1. Distribute on the left: \(2x - 4 = 3x + 1\). 2. Subtract \(2x\): \(-4 = x + 1\). 3. Subtract \(1\): \(x = -5\). 4. Check: The left side is \(0.5(4(-5) - 8) = 0.5(-28) = -14\). The right side is \(3(-5) + 1 = -14\). The two sides are equal.

\(x = -5\). Substitution gives \(-14 = -14\), so the solution is correct.

- Distribute each factor and handle negative signs before parentheses carefully. - Simplify both sides before moving terms. - Move variable terms to one side and constants to the other.

1. Distribute and remove parentheses: \(5x - 15 - 2x - 2 = 4 - x + 3\). 2. Combine like terms: \(3x - 17 = 7 - x\). 3. Add \(x\): \(4x - 17 = 7\). 4. Add \(17\): \(4x = 24\). 5. Divide by \(4\): \(x = 6\).

\(x = 6\)

- Write a perimeter expression for each shape. - Use the fact that all sides of a square are equal and all sides of an equilateral triangle are equal. - Set the two perimeter expressions equal. - Solve the equation with the variable on both sides.

1. The square's perimeter is \(4(z + 2)\). 2. The equilateral triangle's perimeter is \(3(2z - 1)\). 3. Set the equal perimeters equal: \(4(z + 2) = 3(2z - 1)\). 4. Distribute: \(4z + 8 = 6z - 3\). 5. Add \(3\) and subtract \(4z\): \(11 = 2z\). 6. Divide by \(2\): \(z = 5.5\).

The value of \(z\) is \(5.5\).

- A minus sign before parentheses changes both signs inside. - Simplify both sides before moving terms. - Move variable terms to one side and constants to the other.

1. Distribute and remove parentheses: \(10x - 20 - 3x = 16 - x - 8\). 2. Combine like terms: \(7x - 20 = 8 - x\). 3. Add \(x\): \(8x - 20 = 8\). 4. Add \(20\): \(8x = 28\). 5. Divide by \(8\): \(x = \frac{7}{2} = 3.5\).

\(x = \frac{7}{2} = 3.5\)

- Equal expression values can be represented by an equation. - Solve the resulting equation for \(x\). - Substitute the solution into either expression to find the common value.

1. Set the expressions equal: \(0.8x + 12 = 1.4x - 3\). 2. Subtract \(0.8x\): \(12 = 0.6x - 3\). 3. Add \(3\): \(15 = 0.6x\). 4. Divide by \(0.6\): \(x = 25\). 5. Substitute into either expression: \(T_1(25) = 0.8 \cdot 25 + 12 = 32\). Also, \(T_2(25) = 32\).

The expressions are equal when \(x = 25\), and their common value is \(32\).

- Start with the required solution and choose numbers that make both sides equal. - Substitute \(-6\) to test each equation. - For the final equation, include a nonzero \(x\)-term on each side.

1. For the first form, choose \(a = 10\). Since \(-6 + 10 = 4\), one equation is \(x + 10 = 4\). 2. For the second form, choose \(c = 5\). Since \(5(-6) = -30\), one equation is \(5x = -30\). 3. For an equation with \(x\) on both sides, one example is \(2x = x - 6\). Subtracting \(x\) gives \(x = -6\). 4. Substituting \(-6\) makes both sides equal in all three equations.

1. One answer is \(x + 10 = 4\). 2. One answer is \(5x = -30\). 3. One answer is \(2x = x - 6\).

- Distribute the fractional and whole-number factors. - Simplify both sides before moving variable terms. - Check the solution in the original equation.

1. Distribute: \(6x - 9 - 2x = 5x - 20 + 1\). 2. Combine like terms: \(4x - 9 = 5x - 19\). 3. Subtract \(4x\): \(-9 = x - 19\). 4. Add \(19\): \(x = 10\). 5. Check: The left side is \(\frac{3}{4}(8 \cdot 10 - 12) - 2 \cdot 10 = 31\). The right side is \(5(10 - 4) + 1 = 31\).

\(x = 10\)

- Distribute on both sides first. - Combine like terms before moving terms. - Move variable terms to one side and constants to the other.

1. Distribute: \(8x - 6 = 5x + 10 + 2\). 2. Combine like terms on the right: \(8x - 6 = 5x + 12\). 3. Subtract \(5x\): \(3x - 6 = 12\). 4. Add \(6\): \(3x = 18\). 5. Divide by \(3\): \(x = 6\).

\(x = 6\)

- Clear the fractions using a common multiple of all denominators. - Multiply every term by the same number. - Move variable terms to one side after clearing the fractions.

1. Multiply every term by the least common denominator, \(12\): \(9x + 6 = 4x - 12\). 2. Subtract \(4x\): \(5x + 6 = -12\). 3. Subtract \(6\): \(5x = -18\). 4. Divide by \(5\): \(x = -\frac{18}{5} = -3.6\).

\(x = -\frac{18}{5} = -3.6\)

- Clear the fractions with the least common denominator. - Distribute after the denominators are removed. - Move variable terms to one side.

1. Multiply every term by \(10\): \(4(x + 3) = 5x - 1\). 2. Distribute: \(4x + 12 = 5x - 1\). 3. Subtract \(4x\): \(12 = x - 1\). 4. Add \(1\): \(x = 13\).

\(x = 13\)

- Choose either fractions or decimals and use that form consistently. - Distribute before moving variable terms. - Move variable terms to one side and constants to the other.

1. Distribute: \(\frac{3}{4}x - \frac{3}{2} = \frac{1}{2}x + 1.5\). 2. Rewrite \(\frac{3}{2}\) as \(1.5\): \(0.75x - 1.5 = 0.5x + 1.5\). 3. Subtract \(0.5x\): \(0.25x - 1.5 = 1.5\). 4. Add \(1.5\): \(0.25x = 3\). 5. Divide by \(0.25\): \(x = 12\).

\(x = 12\)

- Express the longer side in terms of the shorter side. - Write the two changed side lengths. - A square has equal side lengths, so set those expressions equal. - Use the original dimensions to calculate the perimeter.

1. Let \(s\) inches be the shorter side. The longer side is \(3s\) inches. 2. After the changes, the side lengths are \(s + 5\) and \(3s - 5\). 3. Because the new figure is a square, set the side lengths equal: \(3s - 5 = s + 5\). 4. Subtract \(s\) and add \(5\): \(2s = 10\), so \(s = 5\). 5. The original longer side is \(3 \cdot 5\,\text{in.} = 15\,\text{in.}\). 6. The original perimeter is \(2(5 + 15) = 40\,\text{in.}\).

The original side lengths are \(5\,\text{in.}\) and \(15\,\text{in.}\), and the perimeter is \(40\,\text{in.}\).

- Identify the expression that appears as a factor on both sides. - When can two different multiples of the same number be equal? - What value must the common factor have? - Solve the resulting one-step equation.

1. Both sides contain the common factor \(x - 7\), so the equation has the form \(10A = 5A\), where \(A = x - 7\). 2. Because \(10\) and \(5\) are different, the products can be equal only when the common factor is \(0\). 3. Set \(x - 7 = 0\). 4. Add \(7\): \(x = 7\). 5. Check: \(10(7 - 7) = 0\) and \(5(7 - 7) = 0\).

\(x = 7\). The unequal coefficients produce equal products only when the common factor \(x - 7\) is \(0\).

- Decide whether dividing first or distributing first is more efficient. - Move variable terms to one side when they appear on both sides. - Apply a factor to every term inside parentheses. - Use inverse operations to isolate the variable.

1. Divide by \(5\): \(x - 3 = 5\). Add \(3\): \(x = 8\). 2. Subtract \(4y\): \(8y - 7 = 17\). Add \(7\): \(8y = 24\). Divide by \(8\): \(y = 3\). 3. Distribute \(3\): \(6a + 12 - 4a = 20\). Combine like terms: \(2a + 12 = 20\). Subtract \(12\): \(2a = 8\). Divide by \(2\): \(a = 4\).

1) \(x = 8\) 2) \(y = 3\) 3) \(a = 4\)

- Write a perimeter expression for each shape. - Express the triangle's side in terms of the square's side. - Set the two perimeters equal. - Solve the equation with the variable on both sides.

1. Let \(s\) inches be the side length of the square. The triangle's side length is \(s + 2.5\) inches. 2. Set the perimeters equal: \(4s = 3(s + 2.5)\). 3. Distribute: \(4s = 3s + 7.5\). 4. Subtract \(3s\): \(s = 7.5\). 5. The triangle's side length is \(7.5\,\text{in.} + 2.5\,\text{in.} = 10\,\text{in.}\).

The square''s side length is \(7.5\,\text{in.}\), and the triangle''s side length is \(10\,\text{in.}\).

- Distribute each negative sign through its parentheses. - Combine like terms on each side. - Move variable terms to one side and constants to the other. - Substitute your result into both sides to check it.

1. Distribute the subtraction on both sides: \(45 - 3x + 12 = 15 - x + 4\). 2. Combine like terms: \(57 - 3x = 19 - x\). 3. Add \(3x\): \(57 = 19 + 2x\). 4. Subtract \(19\): \(38 = 2x\). 5. Divide by \(2\): \(x = 19\). 6. Check: the left side is \(45 - (3 \cdot 19 - 12) = 0\), and the right side is \(15 - (19 - 4) = 0\).

\(x = 19\)

- Distribute before moving terms. - Move all variable terms to one side. - Move constants to the other side. - Simplify each side before solving.

1. Distribute on both sides: \(3x + 12 = 4x - 2\). Subtract \(3x\) and add \(2\): \(14 = x\), so \(x = 14\). 2. Distribute: \(5x - 6 = 3x + 6\). Subtract \(3x\): \(2x - 6 = 6\). Add \(6\): \(2x = 12\). Divide by \(2\): \(x = 6\). 3. Distribute \(0.5\): \(3x - 5 = 2x + 7\). Subtract \(2x\): \(x - 5 = 7\). Add \(5\): \(x = 12\).

1) \(x = 14\) 2) \(x = 6\) 3) \(x = 12\)

- Distribute on both sides first. - Simplify each side before moving terms. - Move variable terms to one side and constants to the other. - Check by substituting your result into the original equation.

1. Distribute on both sides: \(6x - 12 = 2x + 10 - 2\). 2. Combine like terms on the right: \(6x - 12 = 2x + 8\). 3. Subtract \(2x\): \(4x - 12 = 8\). 4. Add \(12\): \(4x = 20\). 5. Divide by \(4\): \(x = 5\). 6. Check: the left side is \(3(2 \cdot 5 - 4) = 18\), and the right side is \(2(5 + 5) - 2 = 18\).

\(x = 5\)

- Equal expression values can be represented with an equation. - Simplify each expression separately. - Move variable terms to one side and constants to the other. - Check by evaluating both original expressions at your result.

1. Set the expressions equal: \(5(3y - 4) - 2(7y - 8) = 4(y + 2) - 3(2y - 1)\). 2. Simplify the left side: \(15y - 20 - 14y + 16 = y - 4\). 3. Simplify the right side: \(4y + 8 - 6y + 3 = -2y + 11\). 4. Solve \(y - 4 = -2y + 11\). Add \(2y\): \(3y - 4 = 11\). 5. Add \(4\): \(3y = 15\). Divide by \(3\): \(y = 5\).

\(y = 5\)

- Distribute a subtraction sign to every term inside the parentheses. - Simplify each side before moving terms. - Move variable terms to one side and constants to the other.

1. For a), distribute each subtraction: \(15 - 2x - 7 = 4x - x + 2\). 2. Combine like terms: \(8 - 2x = 3x + 2\). Subtract \(2\) and add \(2x\): \(6 = 5x\). Divide by \(5\): \(x = 1.2\). 3. For b), distribute: \(6x - 2 - 4x - 12 = 0\). 4. Combine like terms: \(2x - 14 = 0\). Add \(14\) and divide by \(2\): \(x = 7\).

a) \(x = 1.2\) b) \(x = 7\)

- Represent an integer and the next integer using one variable. - Translate “\(31\) greater than” carefully when writing the equation. - Distribute before collecting the variable terms.

1. Let \(x\) be the first integer. The next integer is \(x + 1\). 2. Write the equation \(7x = 5(x + 1) + 31\). 3. Distribute and combine constants: \(7x = 5x + 5 + 31 = 5x + 36\). 4. Subtract \(5x\): \(2x = 36\). 5. Divide by \(2\): \(x = 18\). 6. The next integer is \(18 + 1 = 19\).

The consecutive integers are \(18\) and \(19\).

- Work from the innermost parentheses outward. - Distribute each negative sign to every term inside its parentheses. - Simplify both sides before moving terms.

1. Simplify the inner parentheses: \(x - (2x - 5) = x - 2x + 5 = -x + 5\). 2. Substitute this result: \(20 - 3(-x + 5) = 2x + 1\). 3. Distribute \(-3\): \(20 + 3x - 15 = 2x + 1\). 4. Combine constants: \(3x + 5 = 2x + 1\). 5. Subtract \(2x\) and then subtract \(5\): \(x = -4\).

\(x = -4\)

- Convert \(45\) minutes to hours. - Express both distances as rate times time. - Use “twice as fast” to relate the speeds. - Translate the \(5\)-mile distance difference into an equation.

1. Let \(v\) be the runner's speed in miles per hour. The cyclist's speed is \(2v\). 2. The runner travels \(2v\) miles. 3. The cyclist rides for \(0.75\) hour and travels \(0.75(2v) = 1.5v\) miles. 4. The runner travels \(5\) miles farther, so \(2v = 1.5v + 5\). 5. Subtract \(1.5v\): \(0.5v = 5\), so \(v = 10\). 6. The cyclist's speed is \(2 \cdot 10 = 20\,\text{mph}\).

The runner runs at \(10\,\text{mph}\), and the cyclist rides at \(20\,\text{mph}\).

- Write each distance as speed times time. - Express Sarah's speed in terms of Luke's speed. - Use the \(3\)-mile difference to form the equation. - For part c), use Sarah's solved speed with the new time.

1. Luke travels \(4v\) miles. Sarah's speed is \(1.5v\), so in \(2\) hours she travels \(2(1.5v) = 3v\) miles. 2. Luke travels \(3\) miles farther, so \(4v = 3v + 3\). 3. Subtract \(3v\): \(v = 3\). 4. Luke hikes at \(3\,\text{mph}\), and Sarah hikes at \(1.5 \cdot 3 = 4.5\,\text{mph}\). 5. In \(4\) hours, Sarah would travel \(4 \cdot 4.5 = 18\) miles.

a) \(4v = 3v + 3\) b) Luke hikes at \(3\,\text{mph}\), and Sarah hikes at \(4.5\,\text{mph}\). c) Sarah would travel \(18\) miles.

- Use a variable for Mia’s amount. - Write Lucas’s amount in two different ways. - Set the two expressions for Lucas’s amount equal. - Collect the variable terms on one side.

1. Let \(x\) be the amount Mia has. 2. Lucas’s amount can be written as both \(4x\) and \(x + 42\), so write \(4x = x + 42\). 3. Subtract \(x\): \(3x = 42\). 4. Divide by \(3\): \(x = 14\).

Mia has \(\$14\) in her piggy bank.

- Express the original apple weight in terms of the pear weight. - Update both expressions after fruit is removed or added. - Pay attention to which new weight is \(3\,\text{lb}\) greater. - Translate the final comparison into an equation.

1. Let \(b\) be the original weight of the pears in pounds. The apples weigh \(3b\). 2. After the changes, the apples weigh \(3b - 2\), and the pears weigh \(b + 5\). 3. Because the new pear weight is \(3\,\text{lb}\) less than the new apple weight, write \(b + 5 = (3b - 2) - 3\). 4. Simplify: \(b + 5 = 3b - 5\). 5. Add \(5\) and subtract \(b\): \(10 = 2b\). 6. Divide by \(2\): \(b = 5\).

The pears originally weighed \(5\,\text{lb}\).

- A proposed solution must make both expressions equal. - Set \(A\) equal to \(0\) for part 2). - Distribute before solving the equation in part 3). - Move variable terms to one side when they appear on both sides.

1. At \(x = 2\), \(A = 3(2 + 4) = 18\) and \(B = 10 - 2 = 8\). Since \(18 \ne 8\), \(x = 2\) is not a solution. 2. Solve \(3(x + 4) = 0\). Divide by \(3\): \(x + 4 = 0\). Subtract \(4\): \(x = -4\). 3. Set the expressions equal: \(3(x + 4) = 10 - x\). Distribute: \(3x + 12 = 10 - x\). 4. Add \(x\) and subtract \(12\): \(4x = -2\). Divide by \(4\): \(x = -\frac{1}{2} = -0.5\).

1) No. At \(x = 2\), \(A = 18\) and \(B = 8\). 2) \(x = -4\) 3) \(x = -\frac{1}{2} = -0.5\)

- Distribute the fraction to both terms inside the parentheses. - Simplify the products before moving terms. - Move variable terms to one side. - Divide by the coefficient of \(x\).

1. Distribute \(\frac{3}{4}\): \(6x - 9 = 2x + 5\). 2. Subtract \(2x\): \(4x - 9 = 5\). 3. Add \(9\): \(4x = 14\). 4. Divide by \(4\): \(x = \frac{14}{4} = \frac{7}{2} = 3.5\).

\(x = \frac{7}{2} = 3.5\)

- Use a variable for the amount in the smaller tank. - Express the first tank’s amount as a multiple of the second tank’s amount. - Update both amounts after water is removed or added. - Set the resulting amounts equal.

1. Let \(x\) be the original amount in the second tank, in gallons. The first tank contains \(3x\). 2. After the changes, the first tank contains \(3x - 15.5\), and the second contains \(x + 28.5\). 3. Set the new amounts equal: \(3x - 15.5 = x + 28.5\). 4. Subtract \(x\) and add \(15.5\): \(2x = 44\). 5. Divide by \(2\): \(x = 22\). 6. The first tank originally contained \(3 \cdot 22 = 66\,\text{gal}\).

The first tank originally contained \(66\,\text{gal}\), and the second contained \(22\,\text{gal}\).

- Express Lucas’s original card count in terms of Marie’s. - Apply the quadrupling and tripling to the entire original amounts. - Translate “Marie has \(10\) more” into an equation. - Use parentheses around Lucas’s original expression.

1. Let \(x\) be Marie’s original number of cards. 2. Lucas originally has \(x + 12\) cards. 3. After the changes, Marie has \(4x\), and Lucas has \(3(x + 12)\). 4. Since Marie then has \(10\) more cards, write \(4x = 3(x + 12) + 10\). 5. Distribute and combine constants: \(4x = 3x + 36 + 10 = 3x + 46\). 6. Subtract \(3x\): \(x = 46\). 7. Lucas originally had \(46 + 12 = 58\) cards.

Marie originally had \(46\) cards, and Lucas originally had \(58\) cards.

- Write an expression for the money each class has left. - Class B’s spending amount is a compound expression, so use parentheses when subtracting it. - The Class A remainder is three times the Class B remainder.

1. Let \(x\) be the amount Class A spends. 2. Class B spends \(x + 20\). 3. The remaining amounts are \(120 - x\) for Class A and \(100 - (x + 20) = 80 - x\) for Class B. 4. Write the relationship \(120 - x = 3(80 - x)\). 5. Distribute: \(120 - x = 240 - 3x\). 6. Add \(3x\) and subtract \(120\): \(2x = 120\), so \(x = 60\). 7. Class B spends \(60 + 20 = 80\) dollars.

Class A spends \(\$60\), and Class B spends \(\$80\).

- Write a linear expression for each person’s savings after \(x\) weeks. - Multiply Lucas’s entire savings expression by \(1.2\). - Set Sarah’s expression equal to the scaled Lucas expression.

1. After \(x\) weeks, Lucas has \(50 + 5x\), and Sarah has \(20 + 8x\). 2. Write the equation \(20 + 8x = 1.2(50 + 5x)\). 3. Distribute: \(20 + 8x = 60 + 6x\). 4. Subtract \(6x\) and then subtract \(20\): \(2x = 40\). 5. Divide by \(2\): \(x = 20\).

Sarah will have \(1.2\) times Lucas’s savings after \(20\) weeks.

- Use a variable for Mia’s current age. - Express both ages \(6\) years from now. - Translate “twice Mia’s age” into an equation for the future ages.

1. Let \(x\) be Mia’s current age. Lucas is \(3x\) years old. 2. In \(6\) years, write \(3x + 6 = 2(x + 6)\). 3. Distribute: \(3x + 6 = 2x + 12\). 4. Subtract \(2x\) and then subtract \(6\): \(x = 6\). 5. Lucas is \(3 \cdot 6 = 18\) years old.

Mia is \(6\) years old, and Lucas is \(18\) years old.

- Use one variable to describe both removal amounts. - Write expressions for the water remaining in each barrel. - Translate “\(15\,\text{gal}\) more” into an equation comparing the remainders.

1. Let \(x\) be the amount removed from Barrel A. Then \(3x\) is removed from Barrel B. 2. The remaining amounts are \(85 - x\) and \(110 - 3x\). 3. Since Barrel A has \(15\,\text{gal}\) more afterward, write \(85 - x = (110 - 3x) + 15\). 4. Simplify: \(85 - x = 125 - 3x\). 5. Add \(3x\) and subtract \(85\): \(2x = 40\), so \(x = 20\). 6. Barrel B loses \(3 \cdot 20 = 60\,\text{gal}\).

\(20\,\text{gal}\) is removed from Barrel A, and \(60\,\text{gal}\) is removed from Barrel B.

- Use a variable for the smaller charge loss. - Subtract each loss from the corresponding starting charge. - Place the extra \(200\,\text{mAh}\) on the side with the smaller remaining charge. - Check that the final remaining charges differ by \(200\,\text{mAh}\).

1. Let \(x\) be the charge lost by Battery 2. Battery 1 loses \(2x\). 2. The remaining charges are \(4200 - 2x\) and \(3500 - x\). 3. Since Battery 2 has \(200\,\text{mAh}\) more remaining, write \(3500 - x = (4200 - 2x) + 200\). 4. Combine constants: \(3500 - x = 4400 - 2x\). 5. Add \(2x\) and subtract \(3500\): \(x = 900\). 6. Battery 1 loses \(2 \cdot 900 = 1800\,\text{mAh}\).

Battery 1 lost \(1800\,\text{mAh}\), and Battery 2 lost \(900\,\text{mAh}\).

- Express the planned area as daily rate times planned days. - Express the actual area using the greater daily rate and \(2\) fewer days. - Relate the actual area to the original assignment plus the additional area. - Use the planned number of days as the variable.

1. Let \(x\) be the planned number of workdays. 2. The original assigned area is \(400x\,\text{ft}^2\). 3. At the actual rate, the team works \(x - 2\) days and completes \(500(x - 2)\,\text{ft}^2\). 4. The actual area is \(200\,\text{ft}^2\) greater than the original assignment, so write \(400x + 200 = 500(x - 2)\). 5. Distribute and solve: \(400x + 200 = 500x - 1000\), so \(1200 = 100x\) and \(x = 12\). 6. The original area was \(400 \cdot 12 = 4800\,\text{ft}^2\).

The original assignment covered \(4800\,\text{ft}^2\).

- Use a variable for the second pump’s running time. - The first pump runs \(5\) minutes longer. - Original amount equals amount removed plus amount remaining. - The two original amounts were equal.

1. Let \(x\) be the running time of the second pump in minutes. The first pump runs for \(x + 5\) minutes. 2. The original amount in the first tank is \(10(x + 5) + 50\,\text{gal}\). 3. The original amount in the second tank is \(15x + 30\,\text{gal}\). 4. Set the original amounts equal: \(10(x + 5) + 50 = 15x + 30\). 5. Distribute and solve: \(10x + 100 = 15x + 30\), so \(70 = 5x\) and \(x = 14\). 6. Substitute: \(15 \cdot 14 + 30 = 240\,\text{gal}\).

Each tank originally contained \(240\,\text{gal}\) of water.

- Represent the number of people in one group with a variable. - Write each group's starting amount as ticket cost plus money left. - Set the two expressions equal. - Substitute the group size back into either expression.

1. Let \(x\) be the number of people in Group A. 2. Group A received \(15x + 10\) dollars. 3. Group B has \(x + 5\) people, so it received \(12(x + 5) + 7\) dollars. 4. Set the equal amounts equal: \(15x + 10 = 12(x + 5) + 7\). 5. Distribute and simplify: \(15x + 10 = 12x + 67\). 6. Subtract \(12x + 10\): \(3x = 57\), so \(x = 19\). 7. The amount received was \(15 \cdot 19 + 10 = 295\) dollars.

Each group received \(\$295.00\).

- Write each runner's distance from the common starting point. - Include Ben's head start in his distance expression. - Set the distances equal for part a), then convert hours to minutes. - For part b), compare the two speeds.

1. Let \(t\) be the number of hours Anna runs before catching Ben. 2. Anna's distance is \(8t\) miles. Ben's distance from the common starting point is \(6t + 0.3\) miles. 3. At the catch-up point, \(8t = 6t + 0.3\). 4. Subtract \(6t\): \(2t = 0.3\), so \(t = 0.15\) hour. 5. Convert to minutes: \(0.15 \cdot 60 = 9\). Anna catches Ben after \(9\) minutes. 6. If Ben runs at \(9\,\text{mph}\), he is faster than Anna. His head start grows rather than shrinks, so she cannot catch him.

a) Anna catches Ben after \(9\) minutes. b) At \(9\,\text{mph}\), Ben runs faster than Anna, so the distance between them continually increases.

- Write the downstream and upstream speeds in terms of the current. - Use distance equals rate times time in each direction. - The distance between the docks is the same both ways. - Distribute before collecting the variable terms.

1. Let \(c\,\text{mph}\) be the speed of the current. 2. The downstream speed is \(15 + c\), and the upstream speed is \(15 - c\). 3. The distance between the docks is the same in both directions, so \(2(15 + c) = 3(15 - c)\). 4. Distribute: \(30 + 2c = 45 - 3c\). 5. Add \(3c\) and subtract \(30\): \(5c = 15\). 6. Divide by \(5\): \(c = 3\).

The river current flows at \(3\,\text{mph}\).

- Rewrite the fraction and decimals in a convenient form. - Distribute on both sides. - Move variable terms to one side and constants to the other. - Divide by the final coefficient of \(x\).

1. Distribute on both sides: \(3x - 2 + 0.2x = 2.4x + 12\). 2. Combine like terms on the left: \(3.2x - 2 = 2.4x + 12\). 3. Subtract \(2.4x\): \(0.8x - 2 = 12\). 4. Add \(2\): \(0.8x = 14\). 5. Divide by \(0.8\): \(x = 17.5\).

\(x = 17.5\)

- Recall the sum of all angles around a point. - Express the sum of the other three angles in terms of the unknown angle. - Translate “one-fourth of” into multiplication by a fraction.

1. Let \(x\) degrees be the angle measure. The four angles around the intersection add to \(360^\circ\). 2. The sum of the other three angles is \(360 - x\). 3. Translate the relationship into an equation: \(x = \frac{1}{4}(360 - x)\). 4. Multiply by \(4\): \(4x = 360 - x\). 5. Add \(x\): \(5x = 360\). Divide by \(5\): \(x = 72\).

The angle measures \(72^\circ\).

- For part a, compare the Celsius values of two Fahrenheit temperatures that differ by \(18\). - For part b, look for a value that produces the same number when substituted for both \(C\) and \(F\).

1. In part a, an increase of \(18\) in \(F\) increases the numerator \(F - 32\) by \(18\). Dividing by \(1.8\) gives \(18 \div 1.8 = 10\), so the Celsius temperature increases by \(10^\circ\text{C}\). 2. For part b, let \(x\) be the common numerical value. Write \(x = \frac{x - 32}{1.8}\). 3. Multiply both sides by \(1.8\): \(1.8x = x - 32\). 4. Subtract \(x\) from both sides: \(0.8x = -32\). 5. Divide by \(0.8\): \(x = -40\).

a) The temperature increases by \(10^\circ\text{C}\). b) The common numerical value is \(-40\): \(-40^\circ\text{C} = -40^\circ\text{F}\).

- Let \(x\) represent Marie’s current age. How can you express Grandpa Henry’s age? - First use their total age to find their current ages. - Then write a new equation for their ages in the future. - Both people will be older by the same number of years.

1. Let \(x\) be Marie’s current age. Then Grandpa Henry’s age is \(5x\). 2. Write the equation \(x + 5x = 60\). 3. Combine like terms and solve: \(6x = 60\), so \(x = 10\). Marie is 10 and Grandpa Henry is 50. 4. Let \(y\) be the number of years from now. Write \(50 + y = 3(10 + y)\). 5. Distribute and solve: \(50 + y = 30 + 3y\), so \(20 = 2y\) and \(y = 10\).

a) Marie is 10 years old, and Grandpa Henry is 50 years old. b) He will be three times her age in 10 years.

- Convert the delayed start to hours. - Express the truck's travel time in terms of the tractor's travel time. - Set their distances equal at the catch. - Convert the solved time back to a clock time.

1. Convert the delayed start: \(42\) minutes is \(\frac{42}{60} = 0.7\) hour. 2. Let \(t\) be the tractor's travel time in hours when the catch occurs. The truck travels for \(t - 0.7\) hours. 3. Their distances are equal at the catch: \(25t = 60(t - 0.7)\). 4. Distribute and solve: \(25t = 60t - 42\), so \(35t = 42\) and \(t = 1.2\) hours. 5. One hour and \(12\) minutes after \(7{:}00\) a.m. is \(8{:}12\) a.m. 6. The distance from the farm is \(25 \cdot 1.2 = 30\) miles.

The truck catches the tractor at \(8{:}12\) a.m., \(30\) miles from the farm.

- Set \(T_1\) and \(T_2\) equal first. - Distribute the one-half factor carefully. - Check the value of \(k\) in both expressions. - Use the evaluated expression values in the formula for \(S\).

1. Set the expressions equal: \(\frac{1}{2}(k + 12) = 2k - 18\). 2. Distribute: \(0.5k + 6 = 2k - 18\). 3. Subtract \(0.5k\) and add \(18\): \(24 = 1.5k\). 4. Divide by \(1.5\): \(k = 16\). 5. Evaluate the expressions: \(T_1 = 0.5(16 + 12) = 14\) and \(T_2 = 2 \cdot 16 - 18 = 14\). 6. Then \(S = \frac{14 + 14}{16} = \frac{28}{16} = \frac{7}{4} = 1.75\).

\(k = 16\) and \(S = \frac{7}{4} = 1.75\).

- If the smaller number is \(y\), express the larger number using the known sum. - Translate “twice” and “three times” into equations. - For part b, solve the revised equation and compare the larger number with the result from part a. - Because the sum stays fixed, a decrease in one number causes an equal increase in the other.

1. Let \(y\) be the smaller number. Then the larger number is \(100 - y\). 2. For part a, write \((100 - y) - 10 = 2y\). 3. Simplify: \(90 - y = 2y\), so \(90 = 3y\) and \(y = 30\). 4. The larger number is \(100 - 30 = 70\). 5. For part b, write \(90 - y = 3y\). Then \(90 = 4y\), so \(y = 22.5\). 6. The larger number becomes \(100 - 22.5 = 77.5\), so it increases.

a) The numbers are \(70\) and \(30\). b) The larger number must increase, from \(70\) to \(77.5\). The smaller number decreases to \(22.5\), so the sum remains \(100\).

- Begin with the desired solution \(x = -4\). - Apply reversible operations to create a more complex equivalent equation. - For the second equation, add a variable term to both sides. - Verify that each equation has exactly one solution.

1. Start with an equation that becomes true at \(x = -4\) and uses parentheses. One example is \(3(x + 6) = 6\), because substituting \(x = -4\) gives \(3(2) = 6\). Solving the equation also gives \(x + 6 = 2\), so \(x = -4\). 2. For an equation with \(x\) on both sides, one example is \(2x + 10 = x + 6\). Subtracting \(x\) and then \(10\) gives \(x = -4\). 3. Many other correct equations are possible if each condition is met and the only solution is \(x = -4\).

Possible answers: 1) \(3(x + 6) = 6\) 2) \(2x + 10 = x + 6\)

- Express both original amounts using one variable. - Track how the transfer changes each container’s amount. - The factor of \(3\) applies to the amount remaining in Container 2. - A subtraction from a starting value represents what remains after removal.

1. Let \(x\) be the original amount in Container 2. Container 1 contains \(x + 10\). 2. After the transfer, Container 2 contains \(x - 5\), and Container 1 contains \(x + 15\). 3. Write the new relationship: \(x + 15 = 3(x - 5)\). 4. Distribute: \(x + 15 = 3x - 15\). 5. Add \(15\) and subtract \(x\): \(30 = 2x\), so \(x = 15\). 6. Container 1 originally contained \(15 + 10 = 25\,\text{gal}\). 7. The expression \(x - 5\) represents the amount left in Container 2 after \(5\,\text{gal}\) is poured out.

a) Container 1 originally held \(25\,\text{gal}\), and Container 2 held \(15\,\text{gal}\). b) \(x - 5\) is the number of gallons remaining in Container 2 after the transfer.

- Express both original counts using one variable. - Update each count after the transfer of \(9\) laptops. - The fraction \(\frac{4}{5}\) describes a multiplicative relationship between the new counts. - Decide whether any laptops enter or leave the school.

1. Let \(x\) be the original number of laptops in Lab Beta. Lab Alpha has \(2x\). 2. After the move, Beta has \(x + 9\), and Alpha has \(2x - 9\). 3. Write the equation \(x + 9 = \frac{4}{5}(2x - 9)\). 4. Multiply both sides by \(5\): \(5x + 45 = 4(2x - 9)\). 5. Distribute: \(5x + 45 = 8x - 36\). 6. Add \(36\) and subtract \(5x\): \(81 = 3x\), so \(x = 27\). 7. Lab Alpha originally had \(2 \cdot 27 = 54\) laptops. 8. The total stays the same because laptops are moved within the school; none are added or removed.

Lab Alpha originally had \(54\) laptops, and Lab Beta had \(27\). The total remains \(81\) because the laptops only change locations.

- Express the original drill price in terms of the drill-bit set price. - Subtract the discount from both original prices. - Use the new price ratio to write an equation. - Distribute carefully on the right side.

1. Let \(b\) dollars be the original price of the drill-bit set. The drill originally costs \(5b\) dollars. 2. After the discounts, write \(5b - 12 = 8(b - 12)\). 3. Distribute: \(5b - 12 = 8b - 96\). 4. Add \(96\) and subtract \(5b\): \(84 = 3b\). 5. Divide by \(3\): \(b = 28\). 6. The drill originally costs \(5 \cdot 28 = 140\) dollars.

The drill-bit set originally cost \(\$28\), and the cordless drill originally cost \(\$140\).

- First use the \(12\)-photo transfer to find the original amounts. - Moving photos between albums does not change the total number of photos. - Equal albums must each contain half of the total. - Compare Album A’s original amount with the equal-share amount.

1. Let \(b\) be the original number of photos in Album B. Album A has \(4b\). 2. Use the first transfer condition: \(4b - 12 = 2(b + 12)\). 3. Distribute and solve: \(4b - 12 = 2b + 24\), so \(2b = 36\) and \(b = 18\). 4. The original amounts are \(18\) photos in Album B and \(4 \cdot 18 = 72\) photos in Album A. 5. There are \(72 + 18 = 90\) photos total, so equal albums would each contain \(90 \div 2 = 45\) photos. 6. Album A must lose \(72 - 45 = 27\) photos.

A total of \(27\) photos must be moved from Album A to Album B.

- Represent the original ratio with a common factor \(x\). - Update each group after students join or leave. - A \(1:2\) ratio means the second group is twice the first. - Write and solve the resulting proportion.

1. Let the original numbers be \(3x\) boys and \(7x\) girls. 2. After the changes, there are \(3x + 8\) boys and \(7x - 4\) girls. 3. Write the new ratio equation \(\frac{3x + 8}{7x - 4} = \frac{1}{2}\). 4. Cross multiply: \(2(3x + 8) = 7x - 4\). 5. Distribute and solve: \(6x + 16 = 7x - 4\), so \(x = 20\). 6. The original choir had \(3 \cdot 20 = 60\) boys and \(7 \cdot 20 = 140\) girls.

The choir originally had \(60\) boys and \(140\) girls.

- Convert the delayed start to hours. - At the catch-up point, both runners have traveled the same distance. - For part b), first find when Tim reaches the \(3\)-mile mark. - Subtract the delayed start from Tim's travel time.

1. Convert the head start: \(12\) minutes is \(0.2\) hour. 2. Let \(t\) be Jordan's running time in hours. At the catch-up point, the distances are equal, so \(6t = 5(t + 0.2)\). 3. Solve: \(6t = 5t + 1\), so \(t = 1\). 4. For part b), Tim reaches the \(3\)-mile mark in \(\frac{3}{5} = 0.6\) hour. 5. Jordan has \(0.6 - 0.2 = 0.4\) hour to run \(3\) miles. 6. Jordan's required speed is \(\frac{3}{0.4} = 7.5\,\text{mph}\).

a) Jordan catches Tim \(1\) hour after Jordan starts. b) Jordan must run at \(7.5\,\text{mph}\).

- Find Group A's head start in hours. - Set the two distance expressions equal. - Add the solved travel time to Group B's starting time. - Compare the meeting distance with the distance to the lodge.

1. Group A has a \(1.5\)-hour head start. 2. Let \(t\) be the number of hours Group B hikes before catching Group A. Equal distances give \(2(t + 1.5) = 4t\). 3. Solve: \(2t + 3 = 4t\), so \(t = 1.5\). 4. Group B catches Group A \(1.5\) hours after \(10{:}00\) a.m., which is \(11{:}30\) a.m. 5. The meeting distance is \(4 \cdot 1.5 = 6\) miles. 6. Since \(6 < 7\), Group B catches Group A before the lodge.

a) Group B catches Group A at \(11{:}30\) a.m. b) The groups meet \(6\) miles from the starting point. c) Yes. The meeting point is \(1\) mile before the lodge.

- Represent the tablet’s value with a variable. - Find an expression for the total value of the \(8\)-week agreement. - What fraction of the full work period is represented by \(5\) weeks? - Set the proportional value equal to the cash plus the tablet.

1. Let \(x\) be the value of the tablet in dollars. 2. The total value of the original \(8\)-week compensation is \(x + 420\). 3. The value earned for \(5\) weeks is \(\frac{5}{8}(x + 420)\). 4. The compensation actually received is worth \(x + 60\), so write \(\frac{5}{8}(x + 420) = x + 60\). 5. Multiply by \(8\): \(5(x + 420) = 8(x + 60)\). 6. Distribute and solve: \(5x + 2100 = 8x + 480\), so \(1620 = 3x\) and \(x = 540\).

The tablet was valued at \(\$540\).

- Rewrite \(25\%\) as a fraction. - Compare the cooking and makerspace group sizes before writing their difference. - All four activity groups must add to the total number of students. - Simplify the expression for the dance group before solving.

1. Let \(x\) be the total number of students. 2. The cooking group has \(\frac{1}{3}x\) students, and the makerspace group has \(\frac{1}{4}x\) students. 3. The dance group has \(3(\frac{1}{3}x - \frac{1}{4}x)\) students. 4. Add all four groups: \(x = \frac{1}{3}x + \frac{1}{4}x + 3(\frac{1}{3}x - \frac{1}{4}x) + 10\). 5. Simplify the dance term to \(\frac{1}{4}x\). Then \(x = \frac{1}{3}x + \frac{1}{4}x + \frac{1}{4}x + 10\). 6. Combine terms: \(x = \frac{5}{6}x + 10\), so \(\frac{1}{6}x = 10\). 7. Multiply by \(6\): \(x = 60\).

a) One suitable equation is \(x = \frac{1}{3}x + \frac{1}{4}x + 3(\frac{1}{3}x - \frac{1}{4}x) + 10\). b) There are \(60\) students at the center.

- Identify the quantity that is the same for both classes. - If Class A collects for \(x\) days, express Class B’s number of days. - For each class, write collected amount plus amount still needed. - Set the two goal expressions equal.

1. Let \(x\) be the number of days Class A has collected. Class B has collected for \(x - 2\) days. 2. Class A’s goal can be written as \(12x + 40\). 3. Class B’s goal can be written as \(15(x - 2) + 46\). 4. Set the expressions equal: \(12x + 40 = 15(x - 2) + 46\). 5. Distribute and solve: \(12x + 40 = 15x + 16\), so \(24 = 3x\) and \(x = 8\). 6. The goal is \(12 \cdot 8 + 40 = 136\,\text{lb}\).

The collection goal is \(136\,\text{lb}\) for each class.

- Use the runner's \(1\)-hour travel time to connect distance and speed. - At the meeting point, the runner is \(2\) miles short of Q. - Convert \(40\) minutes to hours. - The cyclist travels to Q and then \(2\) miles back.

1. Let \(s\) miles be the distance from P to Q. 2. Because the runner takes \(1\) hour to travel \(s\) miles, the runner's speed is \(s\,\text{mph}\). 3. The meeting occurs after \(40\) minutes, or \(\frac{2}{3}\) hour. By then, the runner has traveled \(s - 2\) miles. 4. Write \(s - 2 = \frac{2}{3}s\). 5. Subtract \(\frac{2}{3}s\): \(\frac{1}{3}s = 2\), so \(s = 6\). 6. The runner's speed is therefore \(6\,\text{mph}\). 7. The cyclist travels \(6 + 2 = 8\) miles in \(\frac{2}{3}\) hour, so the cyclist's speed is \(8 \div \frac{2}{3} = 12\,\text{mph}\).

The distance from P to Q is \(6\) miles. The runner travels at \(6\,\text{mph}\), and the cyclist travels at \(12\,\text{mph}\).