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.
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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}{+2})\,\,\,\,\,\,\,\,\, 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 2x3 \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 2x3 \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.
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Solving Inequalities Online Exam
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{5x1}{4} > 3x
[3 marks]
We solve this inequality by simply rearranging it to make x the subject,
\begin{aligned}\dfrac{5x1}{4} &> 3x \\ \\ 5x1&> 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 > 3x2
[2 marks]
We solve this inequality by simply rearranging it to make x the subject,
\begin{aligned}2x+5 &> 3x2 \\ 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, 43x\leq19
[3 marks]
We solve this inequality by simply rearranging it to make x the subject in the center of the inequality,
\begin{aligned}43x&\leq19 \\ 3x&\leq 15 \\ 3x&\geq 15\\ x&\geq5\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<2x3<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}
Worksheets and Exam Questions
(NEW) Solving Inequalities Exam Style Questions
Level 45 New Official MMEDrill Questions
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