<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts - Honors Go to the latest version.

# 2.2: Equations with Variables on Both Sides

Difficulty Level: Advanced Created by: CK-12
Estimated23 minsto complete
%
Progress
Practice Equations with Variables on Both Sides
Progress
Estimated23 minsto complete
%

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

### Guidance

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 there is the added step of combining like terms with the variables before isolating the variable to find the solution.

Combining like terms means that you are putting all of the variables that match on the same side of the equation. A like term is one in which the degrees match and the variables match. So, for example, 3x\begin{align*}3x\end{align*} and 4x\begin{align*}4x\end{align*} are like terms, 3x\begin{align*}3x\end{align*} and 4z\begin{align*}4z\end{align*} are not. Three apples and four apples are like terms, three apples and four oranges are not.

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.

#### Example A

x+4=2x6\begin{align*}x+4=2x-6\end{align*}

We will 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 x\begin{align*}x\end{align*} from both sides of the equation.

Next, isolate the x\begin{align*}x\end{align*} variable by adding 6 to both sides.

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

Check:x+4(10)+41414=2x6=2(10)6=206=14

#### Example B

143y=4y\begin{align*}14-3y=4y\end{align*}

We will solve this problem using algebra tiles.

We first have to combine our variables (x)\begin{align*}(x)\end{align*} tiles onto the same side of the equation. We do this by adding 3x\begin{align*}3 x\end{align*} tiles to both sides of the equals sign. In this way the 3y\begin{align*}-3y\end{align*} will be eliminated from the left hand side of the equation.

By isolating the variable (y)\begin{align*}(y)\end{align*} we are left with these algebra tiles.

Rearranging we will get the following. {Note: Remember that rearranging is not necessary, it simply allows you to quickly see what the value for the variable is.}

Check:143y143(2)1468=4y=4(2)=8=8

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

#### Example C

We can use these same methods for any of the equations involving variables. Sometimes, however, numbers are so large that one method is more valuable than the other. Let’s look at the following problem.

53a99=42a\begin{align*}53a-99=42a\end{align*}

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

Check:53a9953(9)47799378=42a=42(9)=378=378

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

#### Concept Problem Revisited

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.

If we let w\begin{align*}w\end{align*} be the number of weeks, we can write the following equation.

10x+50Thomas's money: $10 per week+$50=5x+100Jack's money: $5 per week+$100

We can solve the equation now by first combining like terms.

10x+5010x5x+505x+50505x=5x+100=5x5x+100=10050=50-combining the x variables to left side of the equation-combining the constants to left side of the equation

We can now solve for x\begin{align*}x\end{align*} to find the number of weeks until the boys have the same amount of money.

5x5x5x=50=505=10

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

### Vocabulary

Degree
The degree is the exponent on the variable in a term. For example, in the term 4x\begin{align*}4x\end{align*}, the exponent is 1 so the degree is 1.
Like Terms
Like terms refer to terms in which the degrees match and the variables match. For example 3x\begin{align*}3x\end{align*} and 4x\begin{align*}4x\end{align*} are like terms.
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is sometimes referred to as the literal coefficient.

### Guided Practice

1. Use algebra tiles to solve for the variable in the problem 6x+4=5x5\begin{align*}6x+4=5x-5\end{align*}.

2. Use the balance (seesaw) method to solve for the variable in the problem 7r4=3+8r\begin{align*}7r-4=3+8r\end{align*}.

3. Determine the most efficient method to solve for the variable in the problem 10b22=297b\begin{align*}10b-22=29-7b\end{align*}. Explain your choice of method for solving this problem.

1. 6x+4=5x5\begin{align*}6x+4=5x-5\end{align*}

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

Check:6x+46(9)+454+450=5x5=5(9)5=455=50

2. 7r4=3+8r\begin{align*}7r-4=3+8r\end{align*}

You can begin by combining the r\begin{align*}r\end{align*} terms. Subtract 8r\begin{align*}8r\end{align*} from both sides of the equation.

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 r\begin{align*}r\end{align*} term. You now have to divide both sides by -1 to finally isolate the variable.

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

Check:7r47(7)449453=3+8r=3+8(7)=356=53

3. 10b22=297b\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 (or seesaw) method. Work through the steps to see if you can follow them.

Therefore b=17\begin{align*}b = 17\end{align*}.

Check:10b2210(3)2230228=297b=297(3)=2921=8

### Practice

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

1. 5p+3=3p5\begin{align*}5p+3=-3p-5\end{align*}
2. 6b13=2b+3\begin{align*}6b-13=2b+3\end{align*}
3. 2x5=x+6\begin{align*}2x-5=x+6\end{align*}
4. 3x2x=4x+4\begin{align*}3x-2x=-4x+4\end{align*}
5. 4t5t+9=5t9\begin{align*}4t-5t+9=5t-9\end{align*}

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

1. 62d=15d\begin{align*}6-2d=15-d\end{align*}
2. 8s=s6\begin{align*}8-s=s-6\end{align*}
3. 5x+5=2x7\begin{align*}5x+5=2x-7\end{align*}
4. 3x2x=4x+4\begin{align*}3x-2x=-4x+4\end{align*}
5. 8+t=2t+2\begin{align*}8+t=2t+2\end{align*}

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.

1. 4p7=213p\begin{align*}4p-7=21-3p\end{align*}
2. 756x=4x15\begin{align*}75-6x=4x-15\end{align*}
3. 3t+7=15t\begin{align*}3t+7=15-t\end{align*}
4. 5+h=112x\begin{align*}5+h=11-2x\end{align*}
5. 92e=3e\begin{align*}9-2e=3-e\end{align*}

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

### Vocabulary Language: English

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.

Jan 16, 2013

Aug 11, 2015