Solve equations with variables on both sides

Learning goals

Represents the equation structure correctly.
Uses inverse operations and equivalent transformations to isolate the variable.
Maintains equality through each transformation.
Checks the solution by substitution.
Recognizes when simplification is needed before solving.

Difficulty: 1 – 5

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Solve equations with variables on both sides 1

In $5x + 2 = 3x + 10$, which operation correctly moves all variable terms to the left side in one step? A) Add $3x$ to both sides. B) Subtract $3x$ from both sides. C) Subtract $5x$ from the left side only. D) Add $10$ to both sides.

Hints

- Look for an operation that removes the variable term from the right side. - Any valid equation transformation must apply the same operation to both sides.

Solution

1. To collect the variable terms on the left, apply the same operation to both sides. 2. Subtracting $3x$ from both sides leaves $2x + 2 = 10$. 3. The correct choice is B.

Answer

B) Subtract $3x$ from both sides.

Solve equations with variables on both sides 2

For $2x + 7 = x + 11$, subtract $x$ from both sides. What equation results?

Hints

- Apply the stated subtraction to both sides of the equation. - Simplify only the variable terms that are affected by subtracting $x$.

Solution

1. Subtract $x$ from each side: $2x - x + 7 = x - x + 11$. 2. Simplify to get $x + 7 = 11$.

Answer

$x + 7 = 11$

Solve equations with variables on both sides 3

Jordan is solving $-2x+7=3x-8$. Jordan says, “I can add $2x$ to the left side only to remove the $-2x$.” What is wrong with Jordan's step? Then write the correct equation that results from adding $2x$ to both sides.

Hints

- Think about what must stay true when one side of an equation is changed. - The goal is to remove the negative variable term without changing the solution set. - Apply the chosen operation to both sides before simplifying.

Solution

1. An operation performed on only one side generally changes the solution set, so Jordan's proposed step would not keep the equation equivalent. 2. Add $2x$ to both sides: $-2x+7+2x=3x-8+2x$. 3. Simplify to $7=5x-8$.

Answer

Jordan must perform the same operation on both sides to keep the equation equivalent. The correct resulting equation is $7=5x-8$.

Solve equations with variables on both sides 4

Start with $x+7=3x-1$. What equation do you get by subtracting $x$ from both sides?

Hints

- Both sides contain $x$. How does subtracting the same term affect equality? - Combine the remaining variable terms on the right.

Solution

1. Subtract $x$ from each side: $x+7-x=3x-1-x$. 2. Combine terms to get $7=2x-1$.

Answer

$7=2x-1$

Solve equations with variables on both sides 5

In $9-2x=3+x$, add $2x$ to both sides. Write the resulting equation.

Hints

- What happens to the negative variable term on the left? - Which terms can be combined on the right?

Solution

1. Add $2x$ to both sides: $9-2x+2x=3+x+2x$. 2. Simplify to $9=3+3x$.

Answer

$9=3+3x$

Solve equations with variables on both sides 6

Solve each linear equation. a) $8x + 13 = 5x + 31$ b) $14 - 6y = 2y - 18$ c) $11z - 7 = 3z + 25$

Hints

- 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.

Solution

1. For the first equation, subtract $5x$: $3x+13=31$. Then $3x=18$, so $x=6$. 2. For the second equation, add $6y$: $14=8y-18$. Then $32=8y$, so $y=4$. 3. For the third equation, subtract $3z$: $8z-7=25$. Then $8z=32$, so $z=4$.

Answer

a) $x = 6$ b) $y = 4$ c) $z = 4$

Solve equations with variables on both sides 7

Simplify each side as much as possible, then solve. a) $14x - 5 + 3x = 2x + 40$ b) $12 - 4y + 18 = 10y - 12 - 2y$

Hints

- 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.

Solution

1. For the first equation, combine the left side: $17x-5=2x+40$. Subtract $2x$: $15x-5=40$. Add $5$ and divide by $15$: $x=3$. 2. For the second equation, combine both sides: $30-4y=8y-12$. Add $4y$ and $12$: $42=12y$. Divide by $12$: $y=3.5=\frac{7}{2}$.

Answer

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

Solve equations with variables on both sides 8

Simplify both sides and solve $18x-15-7x=3x+25$.

Hints

- Combine like terms before moving variable terms across the equation. - Which operation will leave all the x-terms on one side? - Check the value in the unsimplified original equation.

Solution

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 $8$: $x=5$.

Answer

$11x-15=3x+25$, so $x=5$.

Solve equations with variables on both sides 9

Solve $0.6x + 4 = 1.1x - 3.5$.

Hints

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

Solution

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

Answer

$x = 15$

Solve equations with variables on both sides 10

Solve $1.5x - 2 = 0.5(x + 6)$.

Hints

- Distribute the decimal factor before moving any variable terms. - Collect the $x$-terms on one side of the equation. - Use the remaining constant equation to isolate $x$.

Solution

1. Distribute on the right: $1.5x-2=0.5x+3$. 2. Subtract $0.5x$: $x-2=3$. 3. Add $2$: $x=5$.

Answer

$x=5$

Solve equations with variables on both sides 11

Let $T_1=\frac{1}{2}(k+12)$ and $T_2=2k-18$. Find $k$ when $T_1=T_2$.

Hints

- Equality of $T_1$ and $T_2$ means their two expressions should be set equal. - Distribute the one-half factor before collecting the $k$-terms. - Keep the variable terms on one side and constants on the other.

Solution

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 $1.5$: $k=16$.

Answer

$k=16$

Solve equations with variables on both sides 12

Solve each equation. a) $7x - 12 = 2x + 13$ b) $3(x - 4) = 2x - 3$ c) $10y - (2y + 4) = 6y$

Hints

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

Solution

1. For the first equation, subtract $2x$: $5x-12=13$. Add $12$ and divide by $5$: $x=5$. 2. For the second equation, distribute: $3x-12=2x-3$. Subtract $2x$ and add $12$: $x=9$. 3. For the third equation, remove the parentheses: $10y-2y-4=6y$, so $8y-4=6y$. Subtract $6y$: $2y-4=0$, so $y=2$.

Answer

a) $x = 5$ b) $x = 9$ c) $y = 2$

Solve equations with variables on both sides 13

Solve $13x - 8 - 5x + 2 = 4x + 18$, then check your solution.

Hints

- 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.

Solution

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 $4 \cdot 6 + 18 = 42$.

Answer

$x = 6$

Solve equations with variables on both sides 14

Solve each equation. a) $18-(3x+6)=2x-3$ b) $5(x-1)=2x+7$

Hints

- Remove parentheses carefully before combining terms. - Bring the variable terms together on one side. - Check each result in its original equation.

Solution

1. For a), remove parentheses: $18-3x-6=2x-3$, so $12-3x=2x-3$. Add $3x$ and $3$: $15=5x$, hence $x=3$. 2. For b), distribute: $5x-5=2x+7$. Subtract $2x$ and add $5$: $3x=12$, hence $x=4$.

Answer

a) $x=3$ b) $x=4$

Solve equations with variables on both sides 15

Lucas has four times as much money in his piggy bank as his younger sister Mia. Lucas has $\$42$ more than Mia. How much money does Mia have? Write an equation.

Hints

- Use one variable for Mia's money. - Can Lucas's amount be described in two different ways? - Which equation sets those descriptions equal?

Solution

1. Let $x$ be Mia's amount in dollars. 2. Lucas has both $4x$ and $x+42$ dollars, so $4x=x+42$. 3. Subtract $x$: $3x=42$. Divide by $3$: $x=14$. 4. Mia has $\$14$, and Lucas has $\$56$.

Answer

Mia has $\$14$. One equation is $4x=x+42$.

Solve equations with variables on both sides 16

A crate contains apples and pears. The apples weigh three times as much as the pears. If $2\,\text{lb}$ of apples are removed and $5\,\text{lb}$ of pears are added, the pears still weigh $3\,\text{lb}$ less than the apples. How many pounds of pears were there at first? Write an equation.

Hints

- Express the apple weight using the original pear weight. - Track both changes before writing the comparison. - Which amount must be $3$ pounds larger afterward?

Solution

1. Let $b$ be the original pear weight in pounds. The apples weigh $3b$ pounds. 2. After the changes, the pears weigh $b+5$ and the apples weigh $3b-2$. 3. The pear weight is $3$ pounds less: $b+5=3b-2-3$, or $b+5=3b-5$. 4. Subtract $b$ and add $5$: $10=2b$, so $b=5$ pounds.

Answer

$5\,\text{lb}$ of pears originally. An equation is $b+5=3b-5$.

Solve equations with variables on both sides 17

Let $A=3(x+4)$ and $B=10-x$. Find the value of $x$ for which $A=B$.

Hints

- Equality of $A$ and $B$ means their two expressions should be set equal. - Distribute before collecting the variable terms. - Move all $x$-terms to one side before isolating $x$.

Solution

1. Set the expressions equal: $3(x+4)=10-x$. 2. Distribute: $3x+12=10-x$. 3. Add $x$ and subtract $12$: $4x=-2$. 4. Divide by $4$: $x=-\frac{1}{2}$.

Answer

$x=-\frac{1}{2}$

Solve equations with variables on both sides 18

One water tank contains three times as much as a second tank. After $15.5\,\text{gal}$ is removed from the first tank and $28.5\,\text{gal}$ is added to the second, the amounts become equal. How much water was originally in each tank? Write an equation.

Hints

- Use one variable for the smaller starting amount. - Describe both amounts after the removal and addition. - What must be true about those amounts afterward?

Solution

1. Let $x$ be the original amount in the second tank; the first contains $3x$ gallons. 2. Equal final amounts give $3x-15.5=x+28.5$. 3. Subtract $x$ and add $15.5$: $2x=44$, so $x=22$. 4. The first tank originally contains $3\cdot22=66$ gallons. After the changes, both have $50.5$ gallons.

Answer

First tank: $66\,\text{gal}$; second tank: $22\,\text{gal}$. One equation is $3x-15.5=x+28.5$.

Solve equations with variables on both sides 19

Taylor changes $5x+2=3x+10$ into $2x+2=3x+10$. Is the new equation equivalent to the original? Correct the step and solve for $x$.

Hints

- Compare the changes on the two sides of the equals sign. - An allowed equation step must keep the same solutions. - Try your answer in the starting equation.

Solution

1. The step is invalid because subtracting $3x$ must happen on both sides. 2. The correct equation is $2x+2=10$. 3. Subtract $2$, then divide by $2$ to get $x=4$.

Answer

No. The correct next equation is $2x+2=10$, so $x=4$.

Solve equations with variables on both sides 20

Seven times a number $x$ is 10 greater than three times the number that is 2 greater than $x$. Find $x$.

Hints

- 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? - After distributing, decide how to bring the variable terms together. - Check whether your solution makes the two verbal quantities differ by exactly $10$.

Solution

1. Translate the statement into an equation: $7x=3(x+2)+10$. 2. Distribute and combine like terms: $7x=3x+16$. 3. Subtract $3x$ from both sides: $4x=16$. 4. Divide by $4$: $x=4$.

Answer

$x = 4$

Solve equations with variables on both sides 21

A bookshelf is $\frac13$ full. After $12$ more books are added, it is $\frac12$ full. Let $x$ be the number of books the shelf holds when full. a) Write an equation with $x$ on both sides that models the situation. b) Solve your equation algebraically. Your work must show equivalent equations that remove the fractional coefficients and then isolate $x$. c) Interpret the solution in the context.

Hints

- Express the initial and final numbers of books as fractions of the same unknown capacity. - Use the least common multiple of the denominators to clear the fractions. - After clearing fractions, collect the variable terms on one side. - Check that adding $12$ to one-third of your answer gives one-half of it.

Solution

1. Initially the shelf holds $\frac13x$ books. After adding $12$ books, it holds $\frac12x$, so $ \frac13x+12=\frac12x. $ 2. Multiply both sides by $6$: $2x+72=3x$. 3. Subtract $2x$ from both sides: $72=x$. 4. Therefore, the shelf holds $72$ books when completely full.

Answer

a) $\frac13x+12=\frac12x$ b) $x=72$ c) The shelf's full capacity is $72$ books.

Solve equations with variables on both sides 22

Lucas has $\$85$ in savings and adds $\$12$ each month. Sarah has $\$205$ but spends $\$8$ from her savings each month. After how many months will they have the same amount? How much will each have? Write an equation.

Hints

- Describe each balance after the same number of months. - What does it mean for the two balances to match? - Check the resulting amount against both savings rules.

Solution

1. Let $x$ be the number of months. Equal savings means $85+12x=205-8x$. 2. Add $8x$ to both sides: $85+20x=205$. 3. Subtract $85$: $20x=120$, so $x=6$. 4. Lucas has $85+12\cdot6=157$ dollars, and Sarah has $205-8\cdot6=157$ dollars.

Answer

After $6$ months, each has $\$157$. An equation is $85+12x=205-8x$.

Solve equations with variables on both sides 23

The formula $C=\frac{F-32}{1.8}$ converts a temperature from degrees Fahrenheit to degrees Celsius. Is there a temperature at which the Celsius and Fahrenheit readings have the same numerical value? Write and solve an equation to find it.

Hints

- Use one variable for the numerical value that appears on both temperature scales. - Substitute that same variable for both $C$ and $F$ in the formula. - Clear the decimal denominator before collecting the variable terms.

Solution

1. Let $x$ be the common numerical value, so $C=x$ and $F=x$. 2. Substitute into the conversion formula: $x=\frac{x-32}{1.8}$. 3. Multiply both sides by $1.8$: $1.8x=x-32$. 4. Subtract $x$: $0.8x=-32$. Divide by $0.8$: $x=-40$. 5. Therefore, the two scales have the same numerical reading at $-40$ degrees.

Answer

$-40^\circ\text{C}=-40^\circ\text{F}$

Solve equations with variables on both sides 24

Solve each equation by distributing and combining like terms. a) $4(x - 3) = 2x + 10$ b) $15 - (3a + 4) = 5a - 21$ c) $2(3t + 4) = 5(t - 2)$

Hints

- 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.

Solution

1. For the first equation, distribute: $4x-12=2x+10$. Then $2x=22$, so $x=11$. 2. For the second equation, remove the parentheses: $15-3a-4=5a-21$, so $11-3a=5a-21$. Then $32=8a$, so $a=4$. 3. For the third equation, distribute: $6t+8=5t-10$. Subtract $5t$ and $8$: $t=-18$.

Answer

a) $x = 11$ b) $a = 4$ c) $t = -18$

Solve equations with variables on both sides 25

Solve each equation. a) $\frac{3}{4}x - 5 = \frac{1}{2}x + 2$ b) $0.4(5x - 10) = 1.5x + 7$ c) $\frac{2x - 4}{3} = x - 5$

Hints

- 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.

Solution

1. For the first equation, subtract $\frac{1}{2}x$: $\frac{1}{4}x-5=2$. Then $\frac{1}{4}x=7$, so $x=28$. 2. For the second equation, distribute: $2x-4=1.5x+7$. Then $0.5x=11$, so $x=22$. 3. For the third equation, multiply both sides by $3$: $2x-4=3x-15$. Add $15$ and subtract $2x$: $11=x$, so $x=11$.

Answer

a) $x = 28$ b) $x = 22$ c) $x = 11$

Solve equations with variables on both sides 26

Solve each equation for the indicated variable. a) $3(x + 4) - 2x = 5(x - 2)$ b) $8 - (2a + 5) = 3(a + 1)$ c) $4(k - 1) + 2(3 - k) = 10$

Hints

- Distribute every factor, including a negative sign before parentheses. - Combine like terms on each side. - Move variable terms to one side before isolating the variable. - Check the zero solution in part b carefully.

Solution

1. For the first equation, distribute and combine: $3x+12-2x=5x-10$, so $x+12=5x-10$. Subtract $x$ and add $10$: $22=4x$, so $x=5.5$. 2. For the second equation, remove the parentheses: $8-2a-5=3a+3$, so $3-2a=3a+3$. Subtract $3$ and add $2a$: $0=5a$, so $a=0$. 3. For the third equation, distribute and combine: $4k-4+6-2k=10$, so $2k+2=10$. Subtract $2$ and divide by $2$: $k=4$.

Answer

a) $x = 5.5$ b) $a = 0$ c) $k = 4$

Solve equations with variables on both sides 27

Solve each equation using equivalent operations. a) $0.5(4x - 8) + 3x = 6 - (x - 2)$ b) $\frac{1}{4}(8y + 12) - 5 = 2(y - 1) - y$ c) $2(3m + 1) - 4(m - 2) = 5m - 2$

Hints

- 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.

Solution

1. For the first equation, distribute: $2x-4+3x=6-x+2$, so $5x-4=8-x$. Then $6x=12$, so $x=2$. 2. For the second equation, distribute: $2y+3-5=2y-2-y$, so $2y-2=y-2$. Then $y=0$. 3. For the third equation, distribute: $6m+2-4m+8=5m-2$, so $2m+10=5m-2$. Then $12=3m$, so $m=4$.

Answer

a) $x = 2$ b) $y = 0$ c) $m = 4$

Solve equations with variables on both sides 28

Solve $\frac{2}{3}y - 4 = \frac{1}{6}y + 1$.

Hints

- Multiply by a common multiple of the denominators to clear the fractions. - Look for the least positive number that both denominators divide evenly into. - Move variable terms to one side and constants to the other.

Solution

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 $3$: $y = 10$.

Answer

$y = 10$

Solve equations with variables on both sides 29

Solve $2(3a - 5) + 4 = 10 - (a + 2)$ for $a$.

Hints

- 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.

Solution

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 $7$: $a = 2$.

Answer

$a = 2$

Solve equations with variables on both sides 30

Solve $1.5(4y - 2) = 3(y + 2) + 0.5y$.

Hints

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

Solution

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 $2.5$: $y = 3.6$.

Answer

$y = 3.6$

Solve equations with variables on both sides 31

Solve each equation. Pay close attention when removing parentheses. a) $4(x - 5) = 2(x + 3)$ b) $25 - (3z + 4) = 2(z - 2)$

Hints

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

Solution

1. For the first equation, distribute: $4x-20=2x+6$. Subtract $2x$, add $20$, and divide by $2$: $x=13$. 2. For the second equation, remove the parentheses and distribute: $25-3z-4=2z-4$, so $21-3z=2z-4$. Add $3z$ and $4$: $25=5z$, so $z=5$.

Answer

a) $x = 13$ b) $z = 5$

Solve equations with variables on both sides 32

Max tried to solve $2(x + 4) - 3 = 15 - x$, but made an error. Step 1: $2x + 4 - 3 = 15 - x$ Step 2: $2x + 1 = 15 - x$ Step 3: $3x = 14$ Step 4: $x = \frac{14}{3}$ Explain Max’s error in Step 1, then find the correct solution.

Hints

- 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.

Solution

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 $3$: $x = \frac{10}{3}$.

Answer

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

Solve equations with variables on both sides 33

Two students solve different equations. Alex solves $3(a + 4) = 2a + 18$. Sofia solves $2(a - 3) = 6$. Do the equations have the same solution set? Show your reasoning by solving both equations.

Hints

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

Solution

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.

Answer

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

Solve equations with variables on both sides 34

Analyze each number relationship by writing and solving an equation. a) Adding $5$ to twice a number gives the same result as subtracting $3$ from three times the number. What is the number? b) Is there a number that is exactly $4$ less than five times itself? Justify your answer by solving an equation.

Hints

- 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?

Solution

1. For the first relationship, write $2x+5=3x-3$. Subtract $2x$: $5=x-3$. Add $3$: $x=8$. 2. For the second relationship, write $x=5x-4$. Subtract $x$: $0=4x-4$. Add $4$ and divide by $4$: $x=1$. 3. Because the second equation has the solution $x=1$, such a number exists.

Answer

a) The number is $8$. b) Yes. The number is $1$.

Solve equations with variables on both sides 35

Solve each number riddle by writing an equation. a) One-fourth of a number plus $0.5$ equals twice the number minus $3$. b) Three times the sum of a number and $4$ is $2$ less than five times the number. What is the number?

Hints

- 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.

Solution

1. For the first riddle, write $\frac{1}{4}x+0.5=2x-3$. Multiply both sides by $4$: $x+2=8x-12$. Then $14=7x$, so $x=2$. 2. For the second riddle, write $3(x+4)=5x-2$. Distribute: $3x+12=5x-2$. Then $14=2x$, so $x=7$.

Answer

a) $x = 2$ b) $x = 7$

Solve equations with variables on both sides 36

Solve $4(3y-5)-(2y+7)=3(y+1)$.

Hints

- Apply each outside factor or subtraction to all terms in its parentheses. - Simplify both sides before moving variable terms. - Can you leave the final answer as an exact fraction?

Solution

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 $27$: $7y=30$, so $y=\frac{30}{7}$.

Answer

$y=\frac{30}{7}$

Solve equations with variables on both sides 37

A rectangular poster is $10\,\text{in.}$ wide. If the width is increased by $5\,\text{in.}$ and the height is decreased by $2\,\text{in.}$, the area increases by $10\,\text{in.}^2$. Let $h$ be the original height. Write an equation with $h$ on both sides that compares the new area with the original area, solve it algebraically, and report the original height.

Hints

- Write separate expressions for the original and new areas. - Translate “increases by $10\,\text{in.}^2$” as a relationship between those area expressions. - Make sure the variable appears in both area expressions before solving. - Collect the variable terms on one side after distributing.

Solution

1. The original area is $10h$. 2. The new width is $15\,\text{in.}$, and the new height is $h-2$, so the new area is $15(h-2)$. 3. The new area is $10\,\text{in.}^2$ greater: $15(h-2)=10h+10$. 4. Distribute: $15h-30=10h+10$. 5. Subtract $10h$ and add $30$: $5h=40$, so $h=8$.

Answer

Equation: $15(h-2)=10h+10$ Original height: $8\,\text{in.}$

Solve equations with variables on both sides 38

Tim is $8$ years old and his younger brother Leo is $2$. In how many years will Tim be twice Leo's age? Write and solve an equation.

Hints

- Both boys age by the same number of years. - Write each future age using the same variable. - Which future age is supposed to be twice the other?

Solution

1. Let $x$ be the number of years from now. 2. Their ages then are $8+x$ and $2+x$, so $8+x=2(2+x)$. 3. Distribute and simplify: $8+x=4+2x$. Subtract $x$ and then $4$: $x=4$. 4. In $4$ years, they will be $12$ and $6$, so Tim will be twice Leo's age.

Answer

In $4$ years. One equation is $8+x=2(2+x)$.

Solve equations with variables on both sides 39

Ms. Meyer is $36$ years old, and her son Jonas is $6$. In how many years will Ms. Meyer be three times Jonas's age? Write an equation.

Hints

- Both ages increase by the same number of years. - How can you express “three times Jonas's age” after those years? - Check that your two future ages have the correct ratio.

Solution

1. Let $x$ be the years that pass. Their future ages are $36+x$ and $6+x$. 2. Write $36+x=3(6+x)$. 3. Distribute: $36+x=18+3x$, hence $18=2x$ and $x=9$. 4. In $9$ years, they will be $45$ and $15$, which have a ratio of three to one.

Answer

In $9$ years. One equation is $36+x=3(6+x)$.

Solve equations with variables on both sides 40

Grandpa Henry is $50$ years old, and his granddaughter Marie is $10$ years old. In how many years will Grandpa Henry be exactly three times Marie’s age? Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Use one variable for the number of years in the future. - Add that same variable to both current ages. - Translate “three times” into an equation and then collect the variable terms.

Solution

1. Let $y$ be the number of years from now. 2. Their future ages are $50+y$ and $10+y$, so write $50+y=3(10+y)$. 3. Distribute: $50+y=30+3y$. 4. Subtract $y$ and then subtract $30$: $20=2y$, so $y=10$.

Answer

Original equation: $50+y=3(10+y)$. After distribution and collecting variable terms: $50=30+2y$. Grandpa Henry will be three times Marie's age in $10$ years.

Solve equations with variables on both sides 41

A hiking trail has three sections. The first is one-third of the total length, the second is one-fifth, and the last is $7$ miles long. Find the total length. Write and solve an equation.

Hints

- Let one variable represent the whole trail. - What fraction of the total is covered by the first two parts? - The remaining part has a known length.

Solution

1. Let $x$ be the total length in miles. Write $\frac{1}{3}x+\frac{1}{5}x+7=x$. 2. Combine fractions: $\frac{8}{15}x+7=x$. 3. Subtract $\frac{8}{15}x$: $7=\frac{7}{15}x$. 4. Multiply by $\frac{15}{7}$ to get $x=15$ miles. The three sections are $5$, $3$, and $7$ miles.

Answer

$15$ miles. One equation is $\frac{1}{3}x+\frac{1}{5}x+7=x$.

Solve equations with variables on both sides 42

Marie and Sophie run on a $400$-meter track. Marie runs at a steady $3\,\text{m/s}$, and Sophie runs at $3.8\,\text{m/s}$. After how many seconds has Sophie gained exactly one full lap on Marie? Write an equation.

Hints

- How far ahead is one complete lap? - Write each runner's total distance with the same time variable. - At what time is their distance difference equal to one lap?

Solution

1. Let $t$ be the running time in seconds. A full-lap lead means Sophie has covered $400$ meters more. 2. Write $3.8t=3t+400$. 3. Subtract $3t$ to get $0.8t=400$. Divide by $0.8$: $t=500$.

Answer

After $500$ seconds. One equation is $3.8t=3t+400$.

Solve equations with variables on both sides 43

Luke and Julia ride bikes to a lake $12$ miles away. Luke starts at $3{:}00$ p.m. at $12\,\text{mph}$. Julia starts on the same route $10$ minutes later at $18\,\text{mph}$. a) Write an equation for the time $t$, in hours after Luke starts, when Julia catches him. b) Find the time and distance from the start when she catches him. c) Do both riders reach the lake by $4{:}00$ p.m.? Explain.

Hints

- Convert the later start to a fraction of an hour. - At a catch-up point, what is the relationship between their distances? - Calculate full-trip times from their own departure times for part c).

Solution

1. Ten minutes is $\frac{1}{6}$ hour. Luke rides $12t$ miles and Julia rides $18(t-\frac{1}{6})$ miles. Equal distances give $12t=18(t-\frac{1}{6})$. 2. Distribute and solve: $12t=18t-3$, so $t=0.5$ hour. They meet at $3{:}30$ p.m., $12\cdot0.5=6$ miles from the start. 3. Luke takes $12\div12=1$ hour, reaching the lake at $4{:}00$ p.m. Julia takes $\frac{12}{18}=\frac{2}{3}$ hour after starting, reaching it at $3{:}50$ p.m.

Answer

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

Solve equations with variables on both sides 44

Two equations are equivalent when they have the same solution set. Determine whether these equations are equivalent. Equation I: $4(y + 2) = 2y - 6$ Equation II: $0.5y + 10 = 3$

Hints

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

Solution

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.

Answer

No. Equation I has solution $y = -7$, and Equation II has solution $y = -14$.

Solve equations with variables on both sides 45

A tractor leaves a farm at $7{:}00$ a.m. and travels at $25\,\text{mph}$. At $7{:}42$ a.m., a truck leaves the farm in the same direction at $60\,\text{mph}$. When and how far from the farm does the truck catch the tractor? Write an equation.

Hints

- Convert the time difference between departures to hours. - Which vehicle travels for longer before the catch-up point? - Their distances from the farm match at the catch.

Solution

1. The truck leaves $42$ minutes, or $0.7$ hour, later. 2. Let $t$ be the tractor's travel time in hours. Then the truck travels for $t-0.7$ hours. 3. Equate distances: $25t=60(t-0.7)$. 4. Distribute: $25t=60t-42$, giving $35t=42$ and $t=1.2$ hours. 5. The catch is at $8{:}12$ a.m. and $25\cdot1.2=30$ miles from the farm.

Answer

At $8{:}12$ a.m., $30$ miles from the farm. One equation is $25t=60(t-0.7)$.

Solve equations with variables on both sides 46

Solve $5(x - 3) - 2(x + 1) = 4 - (x - 3)$.

Hints

- 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.

Solution

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 $4$: $x = 6$.

Answer

$x = 6$

Solve equations with variables on both sides 47

Two wire frames use the same total length of wire. One frame is a square with side length $z + 2$ inches. The other is an equilateral triangle with side length $2z - 1$ inches. Find the value of $z$.

Hints

- 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.

Solution

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 $2$: $z = 5.5$.

Answer

The value of $z$ is $5.5$.

Solve equations with variables on both sides 48

Solve $5(2x - 4) - 3x = 16 - (x + 8)$.

Hints

- 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.

Solution

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 $8$: $x = \frac{7}{2} = 3.5$.

Answer

$x = \frac{7}{2} = 3.5$

Solve equations with variables on both sides 49

Write three different linear equations that each have $x = -6$ as the solution and meet the stated form. 1. An equation of the form $x + a = b$ 2. An equation of the form $cx = d$ 3. An equation with $x$ on both sides of the equal sign

Hints

- 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.

Solution

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.

Answer

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

Solve equations with variables on both sides 50

Solve $\frac{3}{4}(8x - 12) - 2x = 5(x - 4) + 1$, then check your solution.

Hints

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

Solution

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 $5(10 - 4) + 1 = 31$.

Answer

$x = 10$

Solve equations with variables on both sides 51

Solve $2(4x - 3) = 5(x + 2) + 2$.

Hints

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

Solution

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 $3$: $x = 6$.

Answer

$x = 6$

Solve equations with variables on both sides 52

Solve $\frac{3}{4}x + \frac{1}{2} = \frac{1}{3}x - 1$.

Hints

- 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.

Solution

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 $5$: $x = -\frac{18}{5} = -3.6$.

Answer

$x = -\frac{18}{5} = -3.6$

Solve equations with variables on both sides 53

Solve $\frac{2}{5}(x + 3) = \frac{1}{2}x - \frac{1}{10}$.

Hints

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

Solution

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 $1$: $x = 13$.

Answer

$x = 13$

Solve equations with variables on both sides 54

The longer side of a rectangle is three times the shorter side. If the longer side is shortened by $5\,\text{in.}$ and the shorter side is lengthened by $5\,\text{in.}$, the new figure is a square. Find the original side lengths and perimeter. Show your equation.

Hints

- Use a variable for the shorter side. - How would you express the two sides after their lengths change? - What must be true of the changed sides if they make a square?

Solution

1. Let $s$ inches be the shorter side, so the longer side is $3s$ inches. 2. After the changes, the two sides are equal: $3s-5=s+5$. 3. Subtract $s$ and add $5$: $2s=10$, hence $s=5$. 4. The original longer side is $15$ inches and the perimeter is $2(5+15)=40$ inches.

Answer

Original side lengths: $5\,\text{in.}$ and $15\,\text{in.}$. Perimeter: $40\,\text{in.}$. One equation is $3s-5=s+5$.

Solve equations with variables on both sides 55

Consider $10(x - 7) = 5(x - 7)$. Find the value of $x$ that makes the equation true. Briefly explain how you can solve it without distributing.

Hints

- 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.

Solution

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 $5(7 - 7) = 0$.

Answer

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

Solve equations with variables on both sides 56

A square and an equilateral triangle have the same perimeter. Each triangle side is $2.5\,\text{in.}$ longer than a square side. Find the side length of each shape.

Hints

- How do you express each shape's perimeter using its side length? - Use one variable and the given side-length difference. - Why must the perimeter expressions be equal?

Solution

1. Let $s$ inches be the side length of the square; the triangle's side length is $s+2.5$ inches. 2. Their perimeters are equal: $4s=3(s+2.5)$. 3. Distribute: $4s=3s+7.5$, hence $s=7.5$. 4. The triangle side is $7.5+2.5=10$ inches. Both perimeters equal $30$ inches.

Answer

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

Solve equations with variables on both sides 57

Solve the equation step by step, then check your solution. $45 - (3x - 12) = 15 - (x - 4)$

Hints

- 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.

Solution

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 $15 - (19 - 4) = 0$.

Answer

$x = 19$

Solve equations with variables on both sides 58

Solve each equation for $x$. 1) $3(x + 4) = 2(2x - 1)$ 2) $5(x - 1.2) = 3(x + 2)$ 3) $0.5(6x - 10) = 2x + 7$

Hints

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

Solution

1. For the first equation, distribute on both sides: $3x+12=4x-2$. Subtract $3x$ and add $2$: $14=x$, so $x=14$. 2. For the second equation, distribute: $5x-6=3x+6$. Subtract $3x$ and add $6$: $2x=12$, so $x=6$. 3. For the third equation, distribute $0.5$: $3x-5=2x+7$. Subtract $2x$ and add $5$: $x=12$.

Answer

1) $x = 14$ 2) $x = 6$ 3) $x = 12$

Solve equations with variables on both sides 59

Solve the equation and check your answer. $3(2x - 4) = 2(x + 5) - 2$

Hints

- 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.

Solution

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 $2(5 + 5) - 2 = 18$.

Answer

$x = 5$

Solve equations with variables on both sides 60

For what value of $y$ do expressions $A$ and $B$ have the same value? $A = 5(3y - 4) - 2(7y - 8)$ $B = 4(y + 2) - 3(2y - 1)$

Hints

- 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.

Solution

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 $3$: $y = 5$.

Answer

$y = 5$

Solve equations with variables on both sides 61

Seven times an integer is $31$ greater than five times the next integer. Find the two consecutive integers.

Hints

- 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.

Solution

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 $18 + 1 = 19$.

Answer

The consecutive integers are $18$ and $19$.

Solve equations with variables on both sides 62

Solve $20 - 3[x - (2x - 5)] = 2x + 1$.

Hints

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

Solution

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 $5$: $x = -4$.

Answer

$x = -4$

Solve equations with variables on both sides 63

Luke and Sarah hike at different speeds. Sarah hikes at $1.5$ times Luke's speed. Luke hikes for $4$ hours, while Sarah hikes for $2$ hours. Luke travels $3$ miles farther than Sarah. a) Write an equation that can be used to find Luke's speed $v$, in miles per hour. b) Find both hiking speeds. c) How far would Sarah travel if she hiked for $4$ hours?

Hints

- 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.

Solution

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 $3v$: $v=3$. Luke hikes at $3\,\text{mph}$, and Sarah hikes at $1.5\cdot3=4.5\,\text{mph}$. 4. In $4$ hours, Sarah would travel $4\cdot4.5=18$ miles.

Answer

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.

Solve equations with variables on both sides 64

Lucas has $12$ more trading cards than Marie. If Marie quadruples her number of cards and Lucas triples his number, Marie will have $10$ more cards than Lucas. How many cards did each person have originally?

Hints

- 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.

Solution

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.

Answer

Marie originally had $46$ cards, and Lucas originally had $58$ cards.

Solve equations with variables on both sides 65

Container 1 holds $10\,\text{gal}$ more water than Container 2. If $5\,\text{gal}$ is poured from Container 2 into Container 1, Container 1 will contain exactly three times as much water as Container 2. a) Find the original amount of water in each container. b) If $x$ is the original amount in Container 2, explain what $x - 5$ represents.

Hints

- 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.

Solution

1. Let $x$ be the original amount in Container 2. Then Container 1 initially contains $x+10$. 2. After the transfer, Container 2 contains $x-5$, and Container 1 contains $x+15$. The condition gives $x+15=3(x-5)$. 3. Distribute and solve: $x+15=3x-15$, so $30=2x$ and $x=15$. Therefore, Container 2 originally held $15\,\text{gal}$, and Container 1 originally held $25\,\text{gal}$. 4. The expression $x-5$ is the amount remaining in Container 2 after $5\,\text{gal}$ is poured out.

Answer

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.

Solve equations with variables on both sides 66

Ms. Chen's class has saved $\$120$ for a trip. Mr. Patel's class has saved $\$100$. Both classes buy snacks. Mr. Patel's class spends $\$20$ more than Ms. Chen's class. Afterward, Ms. Chen's class has three times as much money left as Mr. Patel's class. How much does each class spend? Show an equation.

Hints

- Express both spending amounts using one variable. - Subtract the spending from each class's original savings. - Which remaining balance must be three times the other?

Solution

1. Let $x$ be what Ms. Chen's class spends. Mr. Patel's class spends $x+20$ dollars. 2. They have $120-x$ and $100-(x+20)=80-x$ dollars left. 3. Set the remaining amounts in a ratio: $120-x=3(80-x)$. 4. Distribute and collect: $120-x=240-3x$, so $2x=120$ and $x=60$. 5. Mr. Patel's class spends $60+20=80$ dollars. The remaining balances are $60$ and $20$ dollars.

Answer

Ms. Chen's class spends $\$60$; Mr. Patel's class spends $\$80$. One equation is $120-x=3(80-x)$.

Solve equations with variables on both sides 67

Barrel A has $85\,\text{gal}$ of water, and Barrel B has $110\,\text{gal}$. Three times as much water is removed from Barrel B as from Barrel A. Afterward, Barrel A has $15\,\text{gal}$ more than Barrel B. How many gallons were removed from each barrel? Use an equation with variables on both sides.

Hints

- What one variable could describe both amounts removed? - Compare the water left in each barrel, not the amounts taken out. - Which barrel has more water after the removal?

Solution

1. Let $x$ be the gallons removed from Barrel A. Then $3x$ gallons are removed from Barrel B. 2. Their remaining amounts satisfy $85-x=110-3x+15$. 3. Add $3x$ to both sides and subtract $85$: $2x=40$, so $x=20$. 4. Barrel B loses $3\cdot20=60$ gallons. The remaining amounts are $65$ and $50$ gallons, a difference of $15$.

Answer

Barrel A: $20\,\text{gal}$; Barrel B: $60\,\text{gal}$. With $x$ gallons removed from Barrel A, $85-x=110-3x+15$.

Solve equations with variables on both sides 68

Two phone batteries start with charges of $4200\,\text{mAh}$ and $3500\,\text{mAh}$. Battery 1 loses twice as much charge as Battery 2. After use, Battery 2 has $200\,\text{mAh}$ more charge left than Battery 1. How much charge did each battery lose? Use an equation with variables on both sides.

Hints

- Let the smaller charge loss be $x$. - Write the charge left in each battery after use. - Which remaining charge is larger? Use that comparison to build the equation.

Solution

1. Let $x$ be the charge lost by Battery 2. Battery 1 loses $2x$. 2. The remaining charges are $4200-2x$ and $3500-x$. Thus $3500-x=4200-2x+200$. 3. Add $2x$ and subtract $3500$: $x=900$. 4. Battery 1 loses $1800\,\text{mAh}$ and Battery 2 loses $900\,\text{mAh}$. Their remaining charges are $2400\,\text{mAh}$ and $2600\,\text{mAh}$.

Answer

Battery 1: $1800\,\text{mAh}$; Battery 2: $900\,\text{mAh}$. With $x$ the loss from Battery 2, $3500-x=4200-2x+200$.

Solve equations with variables on both sides 69

A landscaping team is assigned a park area. The plan is to complete $400\,\text{ft}^2$ per day. The team actually completes $500\,\text{ft}^2$ per day. By a date $2$ days earlier than the planned completion date, the team has completed the entire original assignment plus an additional $200\,\text{ft}^2$. What was the area of the original assignment?

Hints

- 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.

Solution

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 $400 \cdot 12 = 4800\,\text{ft}^2$.

Answer

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

Solve equations with variables on both sides 70

Two tanks start with the same amount of water. Water is pumped from the first at $10\,\text{gal/min}$ and from the second at $15\,\text{gal/min}$. The first pump runs $5$ minutes longer. After both pumps stop, the first tank has $50\,\text{gal}$ left and the second has $30\,\text{gal}$ left. How much water was originally in each tank? Use an equation with variables on both sides.

Hints

- Express both pumping times using one variable. - For either tank, how do the amount removed and the amount remaining relate to the initial amount? - What do you know about the two initial amounts?

Solution

1. Let $x$ be the second pump's running time in minutes. The first runs for $x+5$ minutes. 2. Each original amount is the amount pumped out plus what remains, so $10(x+5)+50=15x+30$. 3. Distribute: $10x+100=15x+30$. Thus $70=5x$, so $x=14$. 4. The original amount in each tank was $15\cdot14+30=240\,\text{gal}$.

Answer

$10(x+5)+50=15x+30$, where $x$ is the second pump's running time in minutes. Each tank began with $240\,\text{gal}$.

Solve equations with variables on both sides 71

Two youth groups receive the same amount of money for outings. Group A buys zoo tickets for $\$15.00$ per person and has $\$10.00$ left. Group B visits an adventure park that costs $\$12.00$ per person. Group B has $5$ more people than Group A and has $\$7.00$ left. How much money did each group receive?

Hints

- 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.

Solution

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.

Answer

Each group received $\$295.00$.

Solve equations with variables on both sides 72

Ben is already $0.3$ mile along a straight running path when Anna starts. Ben continues at $6\,\text{mph}$ while Anna runs at $8\,\text{mph}$ along the same path. a) After how many minutes does Anna catch Ben? Use an equation with variables on both sides. b) Explain why Anna would never catch Ben if he ran at $9\,\text{mph}$ instead.

Hints

- Measure both runners' positions from Anna's starting point. - Include Ben's lead in his position expression. - At the moment Anna catches Ben, what is true about their positions? - For part b), compare the speeds before trying any calculations.

Solution

1. Let $t$ be the hours Anna runs. She travels $8t$ miles while Ben's position, measured from her starting point, is $6t+0.3$ miles. 2. At the meeting, $8t=6t+0.3$, so $2t=0.3$ and $t=0.15$ hour. 3. Convert: $0.15\cdot60=9$ minutes. 4. At $9\,\text{mph}$, Ben would keep increasing his lead because he would be faster than Anna.

Answer

a) $8t=6t+0.3$, where $t$ is hours after Anna starts. She catches Ben after $9$ minutes. b) Ben would run faster than Anna, so his lead would keep growing.

Solve equations with variables on both sides 73

A boat travels at $15\,\text{mph}$ in still water. It takes $2$ hours to travel downstream between two docks and $3$ hours to return along the same route. Find the river current's speed using an equation with variables on both sides.

Hints

- How does a current change the boat's downstream and upstream speeds? - What quantity is the same for both trips? - Use the two travel times to express the common distance.

Solution

1. Let $c$ be the current's speed in miles per hour. Downstream speed is $15+c$, and upstream speed is $15-c$. 2. The distance is the same both ways: $2(15+c)=3(15-c)$. 3. Distribute: $30+2c=45-3c$. Add $3c$ and subtract $30$: $5c=15$, so $c=3$. 4. Check: the boat covers $36$ miles downstream in $2$ hours and $36$ miles upstream in $3$ hours.

Answer

$2(15+c)=3(15-c)$, where $c$ is the current's speed in mph, gives $c=3\,\text{mph}$.

Solve equations with variables on both sides 74

Fence posts are installed with equal spacing. The gap from the first post to the beginning of the fence line and the gap from the last post to the end are the same as the gaps between posts. With $12\,\text{ft}$ gaps, a certain number of posts is needed. With $7.5\,\text{ft}$ gaps, $3$ more posts are needed. Let $x$ be the number of posts in the first arrangement. Write an equation with $x$ on both sides by equating the two expressions for the same fence length. Solve that equation algebraically, then find the fence length.

Hints

- Count gaps, including both end gaps, for each arrangement. - Write one length expression for each spacing choice using the same variable. - Because the physical fence length is unchanged, set those two variable expressions equal. - Solve the resulting equation by collecting the variable terms on one side.

Solution

1. With $x$ posts, the first arrangement has $x+1$ equal gaps, so its length is $12(x+1)$. 2. The second arrangement has $x+3$ posts and $x+4$ gaps, so its length is $7.5(x+4)$. 3. Equate the two length expressions: $12(x+1)=7.5(x+4)$. 4. Distribute: $12x+12=7.5x+30$. Subtract $7.5x$ and $12$: $4.5x=18$, so $x=4$. 5. The fence length is $12(4+1)=60\,\text{ft}$.

Answer

Equation: $12(x+1)=7.5(x+4)$ Solution: $x=4$ Fence length: $60\,\text{ft}$

Solve equations with variables on both sides 75

Two numbers add to $100$. Subtracting $10$ from the larger gives twice the smaller. a) Find both numbers. b) If subtracting $10$ from the larger instead gives three times the smaller, while their sum stays $100$, does the larger number increase or decrease? Explain with an equation.

Hints

- Use the fixed sum to express one number in terms of the other. - How does “twice” change to “three times” in the second equation? - Compare the larger number before and after the change.

Solution

1. Let $y$ be the smaller number, so the larger is $100-y$. 2. For a), $100-y-10=2y$, so $90=3y$ and $y=30$. The larger is $70$. 3. For b), $100-y-10=3y$, so $90=4y$ and $y=22.5$. 4. The revised larger number is $100-22.5=77.5$, which is greater than $70$.

Answer

a) Larger $70$, smaller $30$. b) $100-y-10=3y$ gives $y=22.5$. The larger number increases from $70$ to $77.5$.

Solve equations with variables on both sides 76

A chemist has $200\,\text{g}$ of a $10\%$ salt solution. The chemist adds $x$ grams of a $40\%$ salt solution to make a mixture that is exactly $28\%$ salt. a) Write an equation with $x$ on both sides that equates the amount of salt before and after mixing. b) Solve the equation algebraically. Show at least two equivalent-equation steps after writing the model. c) Interpret the solution.

Hints

- Compute the grams of salt already present. - Express both the added salt and the final total mass using $x$. - The final salt amount equals $28\%$ of the final mixture mass. - After distributing, collect the variable terms on one side.

Solution

1. The original solution contains $0.10\cdot200=20$ grams of salt. The added solution contributes $0.40x$ grams of salt. 2. The final mixture has mass $200+x$ grams and is $28\%$ salt, so $ 20+0.40x=0.28(200+x). $ 3. Distribute: $20+0.40x=56+0.28x$. 4. Subtract $0.28x$: $20+0.12x=56$. 5. Subtract $20$: $0.12x=36$. 6. Divide by $0.12$: $x=300$. 7. The chemist must add $300\,\text{g}$ of the $40\%$ solution.

Answer

a) $20+0.40x=0.28(200+x)$ b) $x=300$ c) Add $300\,\text{g}$ of the $40\%$ solution.

Solve equations with variables on both sides 77

Write two different linear equations that each have $x = -4$ as their only solution. 1. The first equation must contain parentheses and have $x$ on both sides of the equal sign. 2. The second equation must have $x$ on both sides of the equal sign and no parentheses. Verify that each equation has $x = -4$ as its only solution.

Hints

- Begin with the desired solution $x = -4$. - Apply reversible operations to create equivalent equations with variable terms on both sides. - Include a grouped expression only in the first equation. - Solve each equation you create to verify that it has exactly one solution.

Solution

1. One equation meeting the first condition is $3(x + 6) = x + 10$. Distributing gives $3x + 18 = x + 10$, so $2x = -8$ and $x = -4$. 2. One equation meeting the second condition is $2x + 10 = x + 6$. Subtracting $x$ and then $10$ gives $x = -4$. 3. Both equations use variable terms on both sides and each has exactly one solution, $x = -4$.

Answer

1. $3(x + 6) = x + 10$ 2. $2x + 10 = x + 6$ Both equations have $x = -4$ as their only solution.

Solve equations with variables on both sides 78

A school has two computer labs. Lab Alpha initially has twice as many laptops as Lab Beta. $9$ laptops are moved from Alpha to Beta. After the move, Beta has $\frac{4}{5}$ as many laptops as Alpha. Find the original number of laptops in each lab. Briefly explain why the school’s total number of laptops does not change.

Hints

- 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.

Solution

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.

Answer

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

Solve equations with variables on both sides 79

A cordless drill originally costs five times as much as a drill-bit set. During a sale, the price of each item is reduced by $\$12$. After the discounts, the drill costs eight times as much as the drill-bit set. Find the original price of each item.

Hints

- 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.

Solution

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.

Answer

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

Solve equations with variables on both sides 80

Two digital photo albums contain pictures. Album A has four times as many photos as Album B. If $12$ photos are moved from Album A to Album B, Album A then has exactly twice as many photos as Album B. Starting from the original amounts, how many photos must be moved from Album A to Album B so the albums contain the same number?

Hints

- 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.

Solution

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.

Answer

A total of $27$ photos must be moved from Album A to Album B.

Solve equations with variables on both sides 81

A school choir has boys and girls in the ratio $3:7$. After $8$ boys join and $4$ girls leave, the ratio becomes $1:2$. How many boys and girls were originally in the choir?

Hints

- 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.

Solution

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.

Answer

The choir originally had $60$ boys and $140$ girls.

Solve equations with variables on both sides 82

Tim runs at $5\,\text{mph}$. Jordan starts from the same place $12$ minutes later and runs at $6\,\text{mph}$. a) How many hours after Jordan starts will Jordan catch Tim? Use an equation with variables on both sides. b) Jordan wants to catch Tim exactly at the $3$-mile mark. How fast must Jordan run if Tim stays at $5\,\text{mph}$ and Jordan still starts $12$ minutes later?

Hints

- Express the $12$-minute head start in hours. - At the catch-up point, both runners have covered the same distance. - For part b), first work out how long Tim takes to reach the marker. - How much time remains for Jordan to reach that same marker?

Solution

1. Jordan starts $0.2$ hour later. Let $t$ be the number of hours Jordan runs to catch Tim. 2. The distances are equal: $6t=5(t+0.2)$. Thus $6t=5t+1$, so $t=1$. 3. Tim reaches the $3$-mile mark after $\frac{3}{5}=0.6$ hour. 4. Jordan has $0.6-0.2=0.4$ hour to cover $3$ miles. The speed required is $3\div0.4=7.5\,\text{mph}$.

Answer

a) $6t=5(t+0.2)$, so $t=1$ hour after Jordan starts. b) $7.5\,\text{mph}$.

Solve equations with variables on both sides 83

Ava's hiking group leaves for a lodge at $8{:}30$ a.m., walking at $2\,\text{mph}$. Noah's group starts on the same trail at $10{:}00$ a.m., walking at $4\,\text{mph}$. a) When does Noah's group catch Ava's? Use an equation with variables on both sides. b) How far have the groups traveled when they meet? c) The lodge is $7$ miles from the start. Do they meet before reaching it? Explain.

Hints

- How much earlier does Ava's group start? - Express both distances using Noah's hiking time. - Convert Noah's hiking time to a clock time. - Compare the meeting distance with the lodge's distance.

Solution

1. Ava's group has a $1.5$-hour head start. Let $t$ be Noah's hiking time before catching up. 2. Equal distances give $2(t+1.5)=4t$. Thus $2t+3=4t$, so $t=1.5$ hours. 3. Noah's group starts at $10{:}00$ a.m. and catches up at $11{:}30$ a.m. 4. The groups meet $4\cdot1.5=6$ miles from the start, which is $1$ mile before the lodge.

Answer

a) $2(t+1.5)=4t$, where $t$ is Noah's hiking time in hours. Thus they meet at $11{:}30$ a.m. b) $6$ miles from the start. c) Yes, $1$ mile before the lodge.

Solve equations with variables on both sides 84

A part-time worker agrees to work for $8$ weeks in exchange for $\$420$ and a tablet. The worker must leave after $5$ weeks, so the compensation is adjusted in proportion to the time worked. The worker receives $\$60$ and keeps the tablet. What value was assigned to the tablet?

Hints

- 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.

Solution

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 $x = 540$.

Answer

The tablet was valued at $\$540$.

Solve equations with variables on both sides 85

At a community center, one-third of the students choose cooking and $25\%$ choose art. The number choosing dance is three times the difference between the cooking and art groups. The remaining $10$ students choose table tennis. a) Write an equation with variables on both sides to find the total number of students $x$. b) How many students are at the center?

Hints

- Convert the percentage to a fraction of $x$. - First express how many students choose dance. - Every student belongs to exactly one of the four groups. - Does your final total give whole numbers in all groups?

Solution

1. The cooking group has $\frac{1}{3}x$ students and the art group has $\frac{1}{4}x$. The dance group has $3(\frac{1}{3}x-\frac{1}{4}x)$. 2. The groups add to the total: $x=\frac{1}{3}x+\frac{1}{4}x+3(\frac{1}{3}x-\frac{1}{4}x)+10$. 3. The dance group has $\frac{1}{4}x$ students, so $x=\frac{5}{6}x+10$. 4. Subtract $\frac{5}{6}x$: $\frac{1}{6}x=10$, giving $x=60$. 5. Check: cooking $20$, art $15$, dance $15$, table tennis $10$; total $60$.

Answer

a) $x=\frac{1}{3}x+\frac{1}{4}x+3(\frac{1}{3}x-\frac{1}{4}x)+10$. b) $60$ students.

Solve equations with variables on both sides 86

Two classes set the same goal for a paper-recycling drive. Class A collects an average of $12\,\text{lb}$ per day, and Class B collects $15\,\text{lb}$ per day. Class B started later and has collected for $2$ fewer days. At one point, Class A is $40\,\text{lb}$ short of the goal, while Class B is $46\,\text{lb}$ short. What is the common collection goal?

Hints

- 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.

Solution

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 $12 \cdot 8 + 40 = 136\,\text{lb}$.

Answer

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

Solve equations with variables on both sides 87

A cyclist and a runner start together at point P and head toward point Q. The cyclist reaches Q, turns around immediately, and meets the runner $40$ minutes after the start, at a point $2$ miles from Q. The runner takes exactly $1$ hour to travel from P to Q. Both move at constant speeds. Find the distance from P to Q and each person's speed. Use an equation with variables on both sides.

Hints

- Use the runner's one-hour trip to connect the distance and speed. - At the meeting, how far has the runner traveled toward Q? - Write the runner's traveled distance in two ways after converting minutes to hours. - Remember that the cyclist travels beyond the meeting point before turning around.

Solution

1. Let $s$ be the distance from P to Q in miles. Because the runner takes $1$ hour, the runner's speed is $s\,\text{mph}$. 2. In $\frac{2}{3}$ hour, the runner travels $s-2$ miles, so $s-2=\frac{2}{3}s$. 3. Subtract $\frac{2}{3}s$: $\frac{1}{3}s=2$, so $s=6$. 4. The runner's speed is $6\,\text{mph}$. The cyclist travels $6+2=8$ miles in $\frac{2}{3}$ hour, so the cyclist's speed is $8\div\frac{2}{3}=12\,\text{mph}$.

Answer

P to Q: $6$ miles. Runner: $6\,\text{mph}$. Cyclist: $12\,\text{mph}$. The two-sided equation is $s-2=\frac{2}{3}s$, where $s$ is the route length in miles.

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Grade 8 math worksheets

Solve equations with variables on both sides

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