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# Equations with Variables on Both Sides

## Move all variables to one side and solve.

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Practice Equations with Variables on Both Sides
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Solve Equations with a Variable on Both Sides

The band is selling popcorn for a fundraiser. For the past few weeks, the students have been out there taking orders with the hope of raising enough money for new uniforms.

“I really hope that we make enough to get the blue ones,” Josie said to Jake and Karen at lunch.

“Me too,” Karen said.

“I was sick a lot of the time, so I didn’t sell as many boxes as I’d hoped to,” Jake sighed.

“That’s okay Jake. Those things happen,” Karen said smiling.

That afternoon after practice, Karen went to sort through the orders that had come in. She began counting all of the sales that the students had made. She discovered that she and Josie had sold the same amount of boxes. Josie sold thirty-six more boxes than Jake after all, Jake had been sick. She sold three times as many as Jake did. Karen began to figure out how many boxes Jake had sold.

Can you figure this out? You will need to understand how to work with variables in a new way to write an equation and solve it. You will learn how to do this in the following Concept.

### Guidance

Do you remember how to solve a basic equation?

Consider the problem, $12 + t = 30$ .

The strategy for solving this equation is to use inverse operations to isolate the variable, $t$ , on one side of the equation. Since 12 is added to $t$ , you would subtract 12 from both sides of the equation to get $t$ by itself.

$12 + t & = 30\\12 - 12 + t &= 30 - 12\\0 + t &= 18\\t &= 18$

What if you needed to solve an equation like this?

$12 + t = 30 + 3t$

How do we solve an equation with variables on both sides of the equation?

To solve an equation that has the same variable on both sides of it, you will use the same basic strategy you already know. You will use inverse operations to isolate the variables on one side of the equation. You will do this by using inverse operations to get all the terms that include variables on one side of the equation and using inverse operations to get all the numerical terms on the other side. Once you do this, you will be able to solve for the variable.

Think about it logically and it makes perfect sense. You get the variables together on one side of the equation, and then you get the numbers together on the other side of the equation. Once you have done this, you can combine like terms and solve for the value of the variable.

Solve for $t$ : $12 + t = 30 + 3t$ .

The variable, $t$ , is on both sides of the equation. We can treat terms with variables the same as we treat numbers. That is, we can use inverse operations to get all of the terms with the variable, $t$ , on one side of the equation. So, just as we could subtract 12 from both sides of the equation to get all of the numerical terms on the right side of the equation, we could subtract $t$ from both sides of the equation to get all of the terms with variables on the right side of the equation.

Alternatively, we could subtract $3t$ from both sides of the equation to get all of the terms with variables on the left side of the equation. It does not matter which of these steps we take. Either will result in the correct answer. However, since it is easier to subtract $3t - t$ than it is to subtract $t - 3t$ , let's subtract $t$ from both sides of the equation. Remember, $t = 1t$ .

$12 + t &= 30 + 3t\\12 + t - t &= 30 + 3t - t\\12 + 0 &= 30 + 2t\\12 &= 30 + 2t$

Now, the only variable is on the right side of the equation. So, let's get all the numerical terms on the left side of the equation. Since 30 is added to $2t$ , we can get $2t$ by itself on the right side of the equation by subtracting 30 from both sides of the equation. Remember, subtracting 30 from 12 is the same as adding -30 to 12.

$12 &= 30 + 2t\\12 - 30 &= 30 -30 + 2t\\12 + (-30) &= 0 + 2t\\-18 &= 2t$

Now, we can use inverse operations to get the $t$ by itself on one side of the equation. Let's divide both sides by 2 to do that. Doing so involves dividing a negative integer, -18, by a positive integer, 2.

$-18 &= 2t\\\frac{-18}{2} &= \frac{2t}{2}\\-9 &= 1t\\-9 &= t$

The value of $t$ is -9.

Sometimes, an equation will have a set of parentheses and variables on both sides of the equation. The distributive property is very helpful in solving these equations.

Solve for $a$ : $4a + 16 =13a - (2a + 3a)$

Our first step should be to simplify the expression on the right side of the equation. According the order of operations, we should combine the like terms inside the parentheses first. Then we can simplify the rest of that expression, like this:

$4a + 16 &= 13a - (2a + 3a)\\4a + 16 &= 13a - 5a\\4a + 16 &= 8a$

Now, we notice that the variable, $a$ , is on both sides of the equation. We can use inverse operations to get all of the terms with the variable, $a$ , on one side of the equation. Since there is a number on the left side of the equation and there is no number on the right side of the equation, it is easier to try to get all of the variable terms on the right side of the equation. We can get all of the variable terms on the right side of the equation by subtracting $4a$ from both sides.

$4a + 16 &= 8a\\4a - 4a + 16 &= 8a -4a\\0 + 16 &= 4a\\16 &= 4a$

Now, the only term with a variable, $4a$ , is on the right side of the equation. The only numerical term, 16, is on the left side of the equation. To solve for $a$ , we can divide both sides of the equation by 4.

$16 &= 4a\\\frac{16}{4} &= \frac{4a}{4}\\4 &= 1a\\4 &= a$

The value of $a$ is 4.

#### Example A

$6x+3=9x+6$

Solution:  $x = -1$

#### Example B

$4x+x+2=10x-13$

Solution:  $x=3$

#### Example C

$8y+2y=20y+10$

Solution:  $y=-1$

Now let's go back to the dilemma from the beginning of the Concept.

First, we write an equation.

$x =$ the number of boxes Jake sold-this is our unknown quantity.

$x + 36 =$ the number of boxes Josie sold.

$3x =$ the number of boxes Karen sold

$x + 36 = 3x$

Now we solve it for $x$ .

$x - x + 36 &= 3x - x\\36 &= 2x\\18 &= x$

Jake sold 18 boxes of popcorn.

Karen and Josie sold the same amount. We can use Karen’s information that she sold three times as many boxes as Jake did.

$&3x\\&3(18) = 54$

Josie and Karen each sold 54 boxes of popcorn.

### Guided Practice

Here is one for you to try on your own.

$6x + 1 = 8x + 3$

Let’s break down working on this problem. First, we need to move the terms with variables to the same side of the equation. Let’s move the $6x$ . We can do this by using an inverse operation. We subtract $6x$ from both sides of the equation.

$6x + 6x + 1 &= 8x - 6x + 3\\1 &= 2x + 3$

Here we performed the inverse operation and then simplified the equation. Now we can solve this just as we would any other two step equation. Take a look and be sure to watch out if you end up working with negative numbers. Don’t mix up the signs!

$1 &= 2x + 3\\1 - 3 &= 2x + 3 - 3\\-2 &= 2x\\-1 &= x$

The value of $x$ is -1.

### Explore More

Directions: Solve each equation with variables on both sides.

1. $6x=2x+16$
2. $5y=3y+12$
3. $4y=y-18$
4. $8x=10x+20$
5. $7x=4x+24$
6. $9y=2y-21$
7. $-6x+22=5x$
8. $15y=9y+36$
9. $14x=10x-40$
10. $19y=4y-30$
11. $18x=2x-32$
12. $4x+1=2x+5$
13. $6x+4=4x+10$
14. $8x+3=5x+9$
15. $10y-4=6y-12$
16. $8x-5=10x-13$
17. $12y-8=14y+14$
18. $18x-5=20x+19$
19. $-20y+8=-8y-4$

Directions: Solve each equation with variables on both sides, by simplifying each equation first by using the distributive property.

1. $2(x+3)=8x$
2. $3(x+5)=-2x$
3. $9y=4(y-5)$

### Vocabulary Language: English

distributive property

distributive property

The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
Inverse Operation

Inverse Operation

Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.
Variable

Variable

A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.