# Solving Inequalities

## Solving Inequalities Revision

**Solving Inequalities**

Inequalities are not always presented to us in a straight forward way. More often than not, they’re all jumbled up – like equations often are – and therefore they need to be **rearranged and solved**.

Make sure you are happy with the following topics before continuing.

**Type 1: ****Listing values**

x is an integer such that -1\leq x \lt 4. List all numbers that satisfy this inequality.

For such questions you need consider if the inequalities are inclusive or strict, in this case we have,

x takes any value greater then or equal to -1 **and **x takes any value less than 4

Hence, the integers that satisfy the inequality are: -1,0,1,2,3

**Type 2:** **Solving Inequalities Basic**

Solve the inequality 5a - 4 > 2a + 8

Firstly, **add** 4 to both sides of the inequality to get,

\begin{aligned}(\textcolor{maroon}{+4})\,\,\,\,\,\,\,\,\, 5a -4 &\gt 2a+8 \\ 5a &\gt 2a+12 \end{aligned}

Then, **subtract** 2a from both sides to get,

\begin{aligned}(\textcolor{maroon}{-2a})\,\,\,\,\,\,\,\,\, 5a &\gt 2a+12 \\ 3a &\gt 12 \end{aligned}

Finally, **divide** both sides by 3 to get,

\begin{aligned}(\textcolor{maroon}{\div 3})\,\,\,\,\,\,\,\,\, 3a &\gt 12 \\ a &\gt 4 \end{aligned}

**Type 3: ****Solving Inequalities 2 signs**

Solve the inequality 5 \lt 2x-3 \lt 13

Firstly, **add** 3 to each side of the inequality,

(Remember what you do to one side you do to all sides, even if there are 3 sides), to get

\begin{aligned}(\textcolor{maroon}{+3})\,\,\,\,\,\,\,\,\, 5& \lt 2x-3 \lt 13 \\ 8 &\lt 2x \lt 16 \end{aligned}

Finally, **divide** both sides by 2 to get,

\begin{aligned}(\textcolor{maroon}{\div 2})\,\,\,\,\,\,\,\,\, 8 \lt 2x& \lt 16 \\ 4 \lt x& \lt 8 \end{aligned}

**Type 4: Multiplying and Dividing by a Negative Number **

When **rearranging an inequality**, you are performing the same operation to both sides of the inequality without changing it (just like as you would with an equation) but with one exception:

**If you multiply or divide by a negative number, then the inequality sign changes direction**

For example, if we have to solve the inequality -2x \gt 4, we have to divide both sides by -2,

\begin{aligned}(\textcolor{maroon}{\div -2})\,\,\,\,\,\,\,\,\, -2x &\gt 4 \\ x &\lt -2 \end{aligned}

**Example**

Solve the inequality \dfrac{4x+4}{2} > x

**[3 marks]**

We need to get rid of the fraction first by **multiplying** by 2

{4x+4} > 2x

Then **subtract** 4x

4 > -2x

Then **divide** by -2

-2 < x

**Remember** the sign changes direction when multiplying or dividing by a negative number.

## Solving Inequalities Example Questions

**Question 1:** Solve the inequality 7 - 3k > -5k + 12

**[2 marks]**

We solve this inequality by simply rearranging it to make k the subject,

\begin{aligned}7 - 3k &> -5k + 12 \\ 7 +2k&> 12 \\ 2k&>5 \\ k&>\dfrac{5}{2}\end{aligned}

Hence k can take any value greater than \dfrac{5}{2}

**Question 2:** Solve the inequality \dfrac{5x-1}{4} > 3x

**[3 marks]**

We solve this inequality by simply rearranging it to make x the subject,

\begin{aligned}\dfrac{5x-1}{4} &> 3x \\ \\ 5x-1&> 12x \\ -1&>7x \\ x&<-\dfrac{1}{7}\end{aligned}

Hence k can take any value less than -\dfrac{1}{7}

**Question 3:** Solve the inequality 2x+5 > 3x-2

**[2 marks]**

We solve this inequality by simply rearranging it to make x the subject,

\begin{aligned}2x+5 &> 3x-2 \\ 7& > x \\ x&<7\end{aligned}

Hence x can take any value less than 7

**Question 4:** Find the range of values that satisfy the inequality, 4-3x\leq19

**[3 marks]**

We solve this inequality by simply rearranging it to make x the subject in the center of the inequality,

\begin{aligned}4-3x&\leq19 \\ -3x&\leq 15 \\ 3x&\geq -15\\ x&\geq-5\end{aligned}

Hence x can take any value greater or equal to -5

**Question 5:** Find the range of values that satisfy the inequality -5<2x-3<10

**[2 marks]**

We solve this inequality by simply rearranging it to make x the subject in the center of the inequality,

\begin{aligned}-5<2x&-3<10 \\ -2<2x&<13 \\ \\ -1<x&<\frac{13}{2}\end{aligned}

Hence x can take any value greater than -1 and less than \dfrac{13}{2}