6.2: Inequalities with Addition and Subtraction
Suppose your favorite baseball team had \begin{align*}w\end{align*}
Guidance
Inequalities Using Addition or Subtraction
To solve inequalities, you need some properties.
Addition Property of Inequality: For all real numbers \begin{align*}a, \ b,\end{align*}
If \begin{align*}x < a\end{align*}
If \begin{align*}x < a\end{align*}
The two properties above are also true for \begin{align*}\le\end{align*}
Because subtraction can also be thought of as “add the opposite,” these properties also work for subtraction situations.
Just like onestep equations, the goal is to isolate the variable, meaning to get the variable alone on one side of the inequality symbol. To do this, you will cancel the operations using inverses.
Example A
Solve for \begin{align*}x:\ x  3 < 10\end{align*}
Solution: To isolate the variable \begin{align*}x\end{align*}
\begin{align*}x3+3 &< 10 + 3\\
x &< 13\end{align*}
Now, check your answer. Choose a number less than 13 and substitute it into your original inequality. If you choose 0, and substitute it you get:
\begin{align*}0  3 < 10 = 3 < 10\end{align*}
What happens at 13? What happens with numbers greater than 13?
Example B
Solve for \begin{align*}x: \ x + 4 > 13\end{align*}
Solution:
\begin{align*}\text{} && x + 4 &> 13\\
\text{Subtract 4 from both sides of the inequality.} && x + 4  4 &> 13  4\\
\text{Simplify.} && x &> 9\end{align*}
The solution is shown below in a graph:
Example C
Solve for \begin{align*}x\end{align*}
Solution:
\begin{align*}\text{} && x+\frac{2}{3}&\ge \frac{1}{3}\\
\text{Subtract }\frac{2}{3}\text{ from both sides of the inequality.} && x+\frac{2}{3}\frac{2}{3}&\ge \frac{1}{3}\frac{2}{3}\\
\text{Simplify.} && x &\ge 1\end{align*}
Guided Practice
Solve for \begin{align*}y\end{align*}
Solution:
\begin{align*}\text{} && 5.6&>y3.3\\
\text{Add 3.3 to both sides of the inequality.} && 5.6+3.3&>y3.3+3.3\\
\text{Simplify.} && 8.9 &>y\end{align*}
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK12 Basic Algebra: Inequalities Using Addition and Subtraction (7:48)
Solve each inequality and graph the solution on a number line.

\begin{align*}x  1 > 10\end{align*}
x−1>−10 
\begin{align*}x  1 \le 5\end{align*}
x−1≤−5 
\begin{align*}20 + a \ge 14\end{align*}
−20+a≥14 
\begin{align*}x + 2 < 7\end{align*}
x+2<7 
\begin{align*}x + 8 \le 7\end{align*}
x+8≤−7 
\begin{align*}5 + t \ge \frac{3}{4}\end{align*}
5+t≥34 
\begin{align*}x  5 < 35\end{align*}
x−5<35 
\begin{align*}15 + g \ge 60\end{align*}
15+g≥−60 
\begin{align*}x  2 \le 1\end{align*}
x−2≤1 
\begin{align*}x  8 > 20\end{align*}
x−8>−20 
\begin{align*}11 + q > 13\end{align*}
11+q>13 
\begin{align*}x + 65 < 100\end{align*}
x+65<100 
\begin{align*}x  32 \le 0\end{align*}
x−32≤0 
\begin{align*}x + 68 \ge 75\end{align*}
x+68≥75 
\begin{align*}16 + y \le 0\end{align*}
16+y≤0
Mixed Review
 Write an equation containing (3, –6) and (–2, –2).
 Simplify: \begin{align*}211 \times 3 + 1\end{align*}
2−11×3+1 .  Graph \begin{align*}y=5\end{align*}
y=−5 on a coordinate plane. 
\begin{align*}y\end{align*}
y varies directly as \begin{align*}x\end{align*}x . When \begin{align*}x=1, \ y = \frac{4}{5}\end{align*}x=−1, y=45 . Find \begin{align*}y\end{align*}y when \begin{align*}x = \frac{16}{3}\end{align*}x=163 .  Rewrite in slopeintercept form: \begin{align*}2x + 7y = 63\end{align*}
−2x+7y=63 .
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Image Attributions
Here you'll learn how to use addition and subtraction to find the solutions to onestep inequalities.