6.3: Inequalities with Multiplication and Division
What if the distance in miles of a bowling alley from your house is \begin{align*}\frac {1}{5}\end{align*}
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Multimedia Link: For help with solving inequalities involving multiplication and division, visit Khan Academy's website: http://khanexercises.appspot.com/video?v=PNXozoJWsWc.
Guidance
Equations are mathematical sentences in which the two sides have the same “weight.” By adding, subtracting, multiplying, or dividing the same value to both sides of the equation, the balance stays in check. However, inequalities begin offbalance. When you perform inverse operations, the inequality will remain offbalance. This is true with inequalities involving both multiplication and division.
Before we can begin to solve inequalities involving multiplication or division, you need to know two properties: the Multiplication Property of Inequality and the Division Property of Inequality.
Multiplication Property of Inequality: For all real positive numbers \begin{align*}a, \ b,\end{align*}
If \begin{align*}x < a\end{align*}
If \begin{align*}x > a\end{align*}
Division Property of Inequality: For all real positive numbers \begin{align*}a, \ b,\end{align*}
If \begin{align*}x < a\end{align*}
If \begin{align*}x > a\end{align*}
Example A
Consider the inequality \begin{align*}2x \ge 12\end{align*}
\begin{align*}2x & \ge 12\\
\frac{2x}{2} & \ge \frac{12}{2}\\
x & \ge 6\end{align*}
This solution can be expressed in four ways. One way is already written: \begin{align*}x \ge 6\end{align*}

\begin{align*}\left \{ x  x \ge 6 \right \}\end{align*}
{xx≥6} 
\begin{align*}[6, \infty)\end{align*}
[6,∞)  Using a number line:
Example B
Solve for \begin{align*}y: \frac{y}{5} \le 3\end{align*}
Solution: The inequality above is read, “\begin{align*}y\end{align*}
\begin{align*}\frac{y}{5} \cdot \frac{5}{1} & \le 3 \cdot \frac{5}{1}\\
y & \le 15\end{align*}
One method of writing the solution is \begin{align*}y \le 15.\end{align*}
The other three are:

\begin{align*}(\infty ,15]\end{align*}
(−∞,15] 
\begin{align*}\left \{ y  y \le 15 \right \}\end{align*}
{yy≤15}
Multiplying and Dividing an Inequality by a Negative Number
Notice that the two properties in this Concept focused on \begin{align*}c\end{align*}
Think of it this way. When you multiply a value by –1, the number you get is the negative of the original.
\begin{align*}6(1)=6\end{align*}
Multiplying each side of a sentence by –1 results in the opposite of both values.
\begin{align*}5x(1)&=4(1)\\
5x&=4\end{align*}
When multiplying by a negative, you are doing the “opposite” of everything in the sentence, including the verb.
\begin{align*}x&>4\\
x(1)&>4(1)\\
x&<4\end{align*}
This concept is summarized below.
Multiplication/Division Rule of Inequality: For any real number \begin{align*}a\end{align*}
If \begin{align*}x<a\end{align*}
If \begin{align*}x<a\end{align*}
As with the other properties of inequalities, these also hold true for \begin{align*}\le\end{align*}
Example C
Solve for \begin{align*}r: 3r < 9\end{align*}
Solution: To isolate the variable \begin{align*}r\end{align*}
\begin{align*}\frac{3r}{3} < \frac{9}{3}\end{align*}
Since you are dividing by –3, everything becomes opposite, including the inequality sign.
\begin{align*}r > 3\end{align*}
Example D
Solve for \begin{align*}p: 12p < 30\end{align*}
Solution: To isolate the variable \begin{align*}p\end{align*}
\begin{align*}\frac{12p}{12} < \frac{30}{12}\end{align*}
Because 12 is not negative, you do not switch the inequality sign.
\begin{align*}p < \frac{5}{2}\end{align*}
In set notation, the solution would be: \begin{align*}\left ( \infty, \frac{5}{2} \right )\end{align*}.
Guided Practice
Solve for \begin{align*}m: \frac{m}{3} < 2.4\end{align*}.
Solution:
To isolate the variable \begin{align*}m\end{align*}, we must cancel “divide by 3” using its inverse operation, multiplying by 3. We must also cancel out the negative, so we would multiply by 1. Multiplying by 3 and 1 means multiplying by 3.
\begin{align*}3 \cdot \frac{m}{3} < 3 \cdot 2.4\end{align*}
Because 3 is negative, you need to switch the inequality sign.
\begin{align*}m > 7.2\end{align*}.
In set notation, the solution would be: \begin{align*}\left ( 7.2, \infty \right )\end{align*}.
This means that \begin{align*}m\end{align*} must be greater than 7.2, but not equal to it.
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 Multiplication and Division (10:27)
 In which cases do you change the inequality sign?
Solve each inequality. Give the solution using inequality notation and with a solution graph.
 \begin{align*}3x \le 6\end{align*}
 \begin{align*}\frac{x}{5} >  \frac{3}{10}\end{align*}
 \begin{align*}10x > 250\end{align*}
 \begin{align*}\frac{x} {7} \ge 5\end{align*}
 \begin{align*}9x >  \frac{3}{4}\end{align*}
 \begin{align*}\frac{x}{15} \le 5\end{align*}
 \begin{align*}620 x > 2400\end{align*}
 \begin{align*}\frac{x}{20} \ge  \frac{7}{40}\end{align*}
 \begin{align*}0.5x \le 7.5\end{align*}
 \begin{align*}75x \ge 125\end{align*}
 \begin{align*}\frac{x}{3} >  \frac{10}{9}\end{align*}
 \begin{align*}\frac{k}{14} \le 1\end{align*}
 \begin{align*}\frac{x}{15} < 8\end{align*}
 \begin{align*}\frac{x}{2} > 40\end{align*}
 \begin{align*}\frac{x}{3} \le 12\end{align*}
 \begin{align*}\frac{x}{25} < \frac{3}{2}\end{align*}
 \begin{align*}\frac{x}{7} \ge 9\end{align*}
 \begin{align*}4x < 24\end{align*}
 \begin{align*}238 < 14d\end{align*}
 \begin{align*}19m \le 285\end{align*}
 \begin{align*}9x \ge  \frac{3}{5}\end{align*}
 \begin{align*}5x \le 21\end{align*}
Mixed Review
 After 3 dozen cookies, Anna has fewer than 24 to make.
 Write an inequality to represent this situation. Let \begin{align*}c=\end{align*} the number of cookies Anna had to make.
 Write the solutions in set notation.
 Tracey’s checking account balance is $31.85. He needs to deposit enough money to pay his satellite T.V. bill, which is $97.12.
 Write an inequality to represent this situation.
 Write the solutions as an algebraic sentence.
 Solve for \begin{align*}v: v=2(19)+6.\end{align*}
 A piggy bank is filled with dimes and quarters. The total amount of money is $26.00.
 Graph all the combinations that make this statement true.
 If $13.50 is in quarters, how many dimes must be in the piggy bank?
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Here you'll learn how to use multiplication and division to find the solutions to onestep inequalities.