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2.1: Equations with Variables on One Side

Difficulty Level: At Grade Created by: CK-12

Erin, Jillian, Stephanie and Jacob went to the movies. The total bill for the tickets and snacks came to 72.00. What is an equation that represents this situation? How much should each teen pay to split the bill evenly? Watch This Guidance When solving any equation, your job is to find the value for the letter that makes the equation true. Solving equations with variables on one side can be done with the help of models such as a balance or algebra tiles. When solving equations with variables on one side of the equation there is one main rule to follow: whatever you do to one side of the equals sign you must do the same to the other side of the equals sign. For example, if you add a number to the left side of an equals sign, you must add the same number to the right side of the equals sign. Example A 5a+2=17\begin{align*}5a + 2 = 17\end{align*} The problem can be solved if we think about the problem in terms of a balance. We know that the two sides are equal so the balance has to stay horizontal. We can place each side of the equation on each side of the balance. In order to solve the equation, we have to get the variable a\begin{align*}a\end{align*} all by itself. Always remember that we need to keep the balance horizontal. This means that whatever we do to one side of the equation, we have to do to the other side. Let’s first subtract 2 from both sides to get rid of the 2 on the left. Since 5 is multiplied by a\begin{align*}a\end{align*}, we can get a by itself (or isolate it) by dividing by 5. Remember that whatever we do to one side, we have to do to the other. If we simplify this expression, we get: Therefore a=3\begin{align*}a = 3\end{align*}. We can check our answer to see if we are correct by substituting our answer back into the original equation. 5a+25(3)+215+217=17=17=17=17 \begin{align*}5a + 2 &= 17\\ 5({\color{red}3}) + 2 &= 17\\ 15 + 2 &= 17\\ 17 &= 17 \ \ \end{align*} Example B 7b7=42\begin{align*}7b - 7 = 42\end{align*} Again, we can solve the problem if we think about the problem in terms of a balance (or a seesaw). We know that the two sides are equal so the balance has to stay horizontal. We can place each side of the equation on each side of the balance. In order to solve the equation, we have to get the variable b\begin{align*}b\end{align*} all by itself. Always remember that we need to keep the balance horizontal. This means that whatever we do to one side of the equation, we have to do to the other side. Let’s first add 7 from both sides to get rid of the 7 on the left. Since 7 is multiplied by b\begin{align*}b\end{align*}, we can get a by itself (or isolate it) by dividing by 7. Remember that whatever we do to one side, we have to do to the other. If we simplify this expression, we get: Therefore b=7\begin{align*}b = 7\end{align*}. We can check our answer to see if we are correct. 7b77(7)749742=42=42=42=42 \begin{align*}7b - 7 &= 42\\ 7({\color{red}7}) - 7 &= 42\\ 49 - 7 &= 42\\ 42 &= 42 \ \ \end{align*} Example C This same method can be extended by using algebra tiles. If we let rectangular tiles represent the variable, square tiles represent one unit, green tiles represent positives numbers, and white tiles represent the negative numbers, we can solve the equations using an alternate method. The green algebra x\begin{align*}x-\end{align*}tiles represent variables; therefore, there are 3 c\begin{align*}c\end{align*} blocks for the equation. The other green blocks represent the numbers or constants. There is a 2 on the left side of the equation so there are 2 square green blocks. There is an 11 on the right side of the equation so there are 11 square green blocks on the right side of the equation. To solve, add two negative tiles to the right and left hand sides. The same rule applies to this problem as to all of the previous problems. Whatever we do to one side we have to do to the other. This leaves us with the following: We can reorganize these to look like the following: Organizing the remaining algebra tiles allows us to realize the answer to be x=3\begin{align*}x = 3\end{align*} or for our example c=3\begin{align*}c = 3\end{align*}. Let’s do our check as with the previous two problems. 3c+23(3)+29+211=11=11=11=11 Y\begin{align*}3c + 2 &= 11\\ 3({\color{red}3}) + 2 &= 11\\ 9 + 2 &= 11\\ 11 &= 11 \ \ Y\end{align*} Concept Problem Revisited There are four teens going to the movies (Erin, Jillian, Stephanie, and Jacob). The total bill was72.00. Therefore our equation is 4x=72\begin{align*}4x = 72\end{align*}. We divide by 4 to find our answer.

xx=724=18\begin{align*}x &= \frac{72}{4}\\ x &= 18\end{align*}

Therefore each teen will have to pay \$18.00 for their movie ticket and snack.

Vocabulary

Constant
A constant is also a numerical coefficient but does not contain a variable. For example in the equation 4x+72=0\begin{align*}4x + 72 = 0\end{align*}, the 72 is a constant.
Equation
An equation is a mathematical equation with expressions separated by an equals sign.
Numerical Coefficient
In mathematical equations, the numerical coefficients are the numbers associated with the variable. For example, with the expression 4x\begin{align*}4x\end{align*}, 4 is the numerical coefficient and x\begin{align*}x\end{align*} is the variable.
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 a model to solve for the variable in the problem x5=12\begin{align*}x-5=12\end{align*}.

2. Use a different model than used in question (1) to solve for the variable in the problem 3y+9=12\begin{align*}3y+9=12\end{align*}.

3. Using one of the models from the concept, solve for x\begin{align*}x\end{align*} in the equation 3x2x+16=3\begin{align*}3x-2x+16=-3\end{align*}.

1. x5=12\begin{align*}x-5=12\end{align*}

Therefore, x=17\begin{align*}x=17\end{align*}. Let's do a check to make sure.

x5(17)517512=12=12=12=12  \begin{align*}x - 5 &= 12\\ ({\color{red}17}) - 5 &= 12\\ 17 - 5 &= 12\\ 12 &= 12 \ \ \end{align*}

2. 3y+9=12\begin{align*}3y+9=12\end{align*}

First you have to subtract 9 from both sides of the equation in order to start to isolate the variable.

Now, in order to get y\begin{align*}y\end{align*} all by itself, you have to divide both sides by 3. This will isolate the variable y\begin{align*}y\end{align*}.

Therefore, y=1\begin{align*}y=1\end{align*}. Let's do a check to make sure.

3y+93(1)+93+912=12=12=12=12  \begin{align*}3y + 9 &= 12\\ 3({\color{red}1}) +9 &= 12\\ 3 + 9 &= 12\\ 12 &= 12 \ \ \end{align*}

3. 3x2x+16=3\begin{align*}3x-2x+16=-3\end{align*}

You can use any method to solve this equation. Remember to isolate the x\begin{align*}x\end{align*} variable. You will notice here that there are two x\begin{align*}x\end{align*} values on the left. First let’s combine these terms.

3x2x+16x+16=3=3\begin{align*}3x-2x+16 &= -3\\ x+16 &= -3\end{align*}

Now you can use any method to solve the equation. You now should just have to subtract 16 from both sides to isolate the x\begin{align*}x\end{align*} variable.

x+1616x=316=19\begin{align*}x+16 {\color{red}-16} &= -3 {\color{red}-16}\\ x &= -19\end{align*}

Let's do a check to make sure.

3x2x+16x+1619+163=3(original problem)=3(simplified problem)=3=3  \begin{align*}3x-2x+16 &=-3 \quad \text{(original problem)}\\ x+16 &= -3 \quad \text{(simplified problem)}\\ {\color{red}-19}+16 &= -3\\ -3 &= -3 \ \ \end{align*}

Practice

Use the model of the balance to solve for each of the following variables.

1. a+3=5\begin{align*}a+3=-5\end{align*}
2. 2b1=5\begin{align*}2b-1=5\end{align*}
3. 4c3=9\begin{align*}4c-3=9\end{align*}
4. 2d=3\begin{align*}2-d=3\end{align*}
5. 4=3e=2\begin{align*}4=3e=-2\end{align*}

Use algebra tiles to solve for each of the following variables.

1. x+3=14\begin{align*}x+3=14\end{align*}
2. 2y7=5\begin{align*}2y-7=5\end{align*}
3. 3z+6=9\begin{align*}3z+6=9\end{align*}
4. 5+3x=3\begin{align*}5+3x=-3\end{align*}
5. 2x+2=4\begin{align*}2x+2=-4\end{align*}

Use the models that you have learned to solve for the variables in the following problems.

1. 4x+13=5\begin{align*}-4x+13=5\end{align*}
2. \begin{align*}3x-5=22\end{align*}
3. \begin{align*}11-2x=5\end{align*}
4. \begin{align*}2x-4=4\end{align*}
5. \begin{align*}5x+3=28\end{align*}

For each of the following models, write a problem involving an equation with a variable on one side of the equation expressed by the model and then solve for the variable.

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Dec 19, 2012
Aug 11, 2015

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