What if you had an algebraic equation involving addition or subtraction like \begin{align*}x  \frac{2}{5} = \frac{5}{8}\end{align*}
Watch This
CK12 Foundation: 0301S Solving Equations with Addition and Subtraction (H264)
Guidance
Nadia is buying a new mp3 player. Peter watches her pay for the player with a $100 bill. She receives $22.00 in change, and from only this information, Peter works out how much the player cost. How much was the player?
In algebra, we can solve problems like this using an equation. An equation is an algebraic expression that involves an equals sign. If we use the letter \begin{align*}x\end{align*}
Another way we could write the equation would be \begin{align*}x = 100  22\end{align*}
In this chapter, we will learn how to solve for the variable in a onevariable linear equation. Linear equations are equations in which each term is either a constant, or a constant times a single variable (raised to the first power). The term linear comes from the word line, because the graph of a linear equation is always a line.
We’ll start with simple problems like the one in the last example.
Solving Equations Using Addition and Subtraction
When we work with an algebraic equation, it’s important to remember that the two sides have to stay equal for the equation to stay true. We can change the equation around however we want, but whatever we do to one side of the equation, we have to do to the other side. In the introduction above, for example, we could get from the first equation to the second equation by subtracting 22 from both sides:
\begin{align*} x + 22 &= 100\\ x + 22  22 &= 100  22\\ x &= 100  22\end{align*}
Similarly, we can add numbers to each side of an equation to help solve for our unknown.
Example A
Solve \begin{align*}x  3 = 9\end{align*}
Solution
To solve an equation for \begin{align*}x\end{align*}
\begin{align*} x  3 &= 9\\ x  3 + 3 &= 9 + 3\\ x + 0 &= 9 + 3\\ x &= 12\end{align*}
Example B
Solve \begin{align*}z  9.7 = 1.026\end{align*}
Solution
It doesn’t matter what the variable is—the solving process is the same.
\begin{align*} z  9.7 &= 1.026\\ z  9.7 + 9.7 &= 1.026 + 9.7\\ z &= 8.674\end{align*}
Make sure you understand the addition of decimals in this example!
Example C
Solve \begin{align*}x + \frac{4}{7} = \frac{9}{5}\end{align*}
Solution
To isolate \begin{align*}x\end{align*}
\begin{align*} x + \frac{4}{7} &= \frac{9}{5}\\ x + \frac{4}{7}  \frac{4}{7} &= \frac{9}{5}  \frac{4}{7}\\ x &= \frac{9}{5}  \frac{4}{7}\end{align*}
Now we have to subtract fractions, which means we need to find the LCD. Since 5 and 7 are both prime, their lowest common multiple is just their product, 35.
\begin{align*}x &= \frac{9}{5}  \frac{4}{7}\\ x &= \frac{7 \cdot 9}{7 \cdot 5}  \frac{4 \cdot 5}{7 \cdot 5}\\ x &= \frac{63}{35}  \frac{20}{35}\\ x &= \frac{63  20}{35}\\ x &= \frac{43}{35}\end{align*}
Make sure you’re comfortable with decimals and fractions! To master algebra, you’ll need to work with them frequently.
Watch this video for help with the Examples above.
CK12 Foundation: Solving Equations with Addition and Subtraction
Vocabulary
 An equation in which each term is either a constant or the product of a constant and a single variable is a linear equation.
 We can add, subtract, multiply, or divide both sides of an equation by the same value and still have an equivalent equation.
 To solve an equation, isolate the unknown variable on one side of the equation by applying one or more arithmetic operations to both sides.
Guided Practice
Solve \begin{align*}x +10 = 17\end{align*}
Solution
To solve an equation for \begin{align*}x\end{align*}
\begin{align*} x +10 &= 17\\ x +10 10 &= 17 10\\ x + 0 &= 1710\\ x &= 7\end{align*}
Practice
For 15, solve the following equations for \begin{align*}x\end{align*}

\begin{align*}x  11 = 7\end{align*}
x−11=7 
\begin{align*}x  1.1 = 3.2\end{align*}
x−1.1=3.2 
\begin{align*}x +0.257 = 1\end{align*}
x+0.257=1 
\begin{align*}x + \frac{5}{2} = \frac{2}{3}\end{align*}
x+52=23 
\begin{align*}x  \frac{5}{6} = \frac{3}{8}\end{align*}
x−56=38
For 610, solve the following equations for the unknown variable.
 \begin{align*}q  13 = 13\end{align*}
 \begin{align*}z + 1.1 = 3.0001\end{align*}
 \begin{align*}r + 1 = \frac{2}{5}\end{align*}
 \begin{align*}t + \frac{1}{2} = \frac{1}{3} \end{align*}
 \begin{align*}\frac{3}{4} = \frac{1}{2}  y \end{align*}