Thomas has $50 and Jack has $100. Thomas is saving $10 per week for his new bike. Jack is saving $5 a week for his new bike. How long will it be before the two boys have the same amount of money? Can you represent this situation with an equation?

### Equations with Variable on Both Sides

The methods used for solving equations with variables on both sides of the equation are the same as the methods used to solve equations with variables on one side of the equation. What differs is that first you must add or subtract a term from both sides in order to have the variable on only one side of the equals sign.

Remember that your goal for solving any equation is to get the variables on one side and the constants on the other side. You do this by adding and subtracting terms from both sides of the equals sign. Then you isolate the variables by multiplying or dividing. You must remember in these problems, as with any equation, whatever operation (addition, subtraction, multiplication, or division) you do to one side of the equals sign, you must do to the other side. This is a big rule to remember in order for equations to remain equal or to remain in balance.

#### Let's practice solving for a variable with the following problems:

- \begin{align*}x+4=2x-6\end{align*}
x+4=2x−6

You can solve this problem using the balance method.

You could first try to get the variables all on one side of the equation. You do this by subtracting \begin{align*}x\end{align*}

Next, isolate the \begin{align*}x\end{align*}

Therefore \begin{align*}x = 10\end{align*}

\begin{align*}\text{Check:}&\\
x+4 &= 2x-6\\
({\color{red}10})+4 &= 2({\color{red}10})-6\\
14 &= 20-6\\
14 &= 14 \ \ \end{align*}

- \begin{align*}14-3y=4y\end{align*}
14−3y=4y

You can solve this equation using algebra tiles.

You first have to combine our variables \begin{align*}(x)\end{align*}

By isolating the variable \begin{align*}(y)\end{align*}

Rearranging you will get the following.

\begin{align*}\text{Check:} &\\
14 - 3y &= 4y\\
14-3({\color{red}2}) &= 4({\color{red}2})\\
14-6 &= 8\\
8 &= 8 \ \ \end{align*}

Therefore \begin{align*}y = 2\end{align*}

- \begin{align*}53a-99=42a\end{align*}
53a−99=42a

To solve this problem, you would need to have a large number of algebra tiles! It might be more efficient to use the balance method to solve this problem.

\begin{align*}\text{Check:} &\\
53a-99 &= 42a\\
53({\color{red}9}) - 99 &= 42({\color{red}9})\\
477-99 &= 378\\
378 &= 378 \ \ \end{align*}

Therefore, \begin{align*}a = 9\end{align*}

### Examples

#### Example 1

Earlier, you were told that Thomas has $50 and Jack has $100 and that Thomas is saving $10 per week for his new bike while Jack is saving $5 a week for his new bike.

How long will it be before the two boys have the same amount of money? Can can represent the situation with an equation?

If you let \begin{align*}x\end{align*}

\begin{align*}\underbrace{ 10x+50 }_{\text{Thomas's money:} \ \$10 \ \text{per week} + \$50}= \underbrace{ 5x+100 }_{\text{Jack's money:} \ \$5 \ \text{per week} + \$100}\end{align*}

The equations are equal because the boys are saving for the same bike. You can solve the equation now by first combining like terms.

\begin{align*}10x+50 &= 5x+100\\
10x {\color{red}-5x}+50 &= 5x {\color{red}-5x}+100 && \text{-moving the} \ x \ \text{variables to left side of the equation}\\
5x+50 {\color{red}-50} &= 100 {\color{red}-50} && \text{-moving the constants to right side of the equation}\\
5x &= 50\end{align*}

You can now solve for \begin{align*}x\end{align*}

\begin{align*}5x &= 50\\
\frac{5x}{5} &= \frac{50}{5}\\
x &= 10\end{align*}

Therefore, in 10 weeks Jack and Thomas will each have the same amount of money.

#### Example 2

Solve for the variable in the equation \begin{align*}6x+4=5x-5\end{align*}

\begin{align*}6x+4=5x-5\end{align*}

Therefore \begin{align*}x = -9\end{align*}

#### Example 3

Solve for the variable in the equation \begin{align*}7r-4=3+8r\end{align*}

\begin{align*}7r-4=3+8r\end{align*}

You can begin by combining the \begin{align*}r\end{align*}

You next have to isolate the variable. To do this, add 4 to both sides of the equation.

But there is still a negative sign with the \begin{align*}r\end{align*}

Therefore \begin{align*}r = -7\end{align*}

#### Example 4

Determine the most efficient method to solve for the variable in the problem \begin{align*}10b-22=29-7b\end{align*}

You could choose either method but there are larger numbers in this equation. With larger numbers, the use of algebra tiles is not an efficient manipulative. You should solve the problem using the balance method. Work through the steps to see if you can follow them.

Therefore \begin{align*}b = 3\end{align*}

### Review

Use the balance method to find the solution for the variable in each of the following problems.

- \begin{align*}5p+3=-3p-5\end{align*}
5p+3=−3p−5 - \begin{align*}6b-13=2b+3\end{align*}
6b−13=2b+3 - \begin{align*}2x-5=x+6\end{align*}
2x−5=x+6 - \begin{align*}3x-2x=-4x+4\end{align*}
3x−2x=−4x+4 - \begin{align*}4t-5t+9=5t-9\end{align*}
4t−5t+9=5t−9

Use algebra tiles to find the solution for the variable in each of the following problems.

- \begin{align*}6-2d=15-d\end{align*}
6−2d=15−d - \begin{align*}8-s=s-6\end{align*}
8−s=s−6 - \begin{align*}5x+5=2x-7\end{align*}
5x+5=2x−7 - \begin{align*}3x-2x=-4x+4\end{align*}
3x−2x=−4x+4 - \begin{align*}8+t=2t+2\end{align*}
8+t=2t+2

Use the methods that you have learned for solving equations with variables on both sides to solve for the variables in each of the following problems. Remember to choose an efficient method to solve for the variable.

- \begin{align*}4p-7=21-3p\end{align*}
4p−7=21−3p - \begin{align*}75-6x=4x-15\end{align*}
75−6x=4x−15 - \begin{align*}3t+7=15-t\end{align*}
3t+7=15−t - \begin{align*}5+h=11-2h\end{align*}
5+h=11−2h - \begin{align*}9-2e=3-e\end{align*}
9−2e=3−e

For each of the following models, write a problem to represent the model and solve for the variable for the problem.

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### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.2.