Question

Solve and graph the equality 2|x-4|+3<15 on a number line

Answer 1

Given inequality:

`2|\text + 3 \leq\text{ 15}`

Subtract 3 from both sides:

`\begin{gathered} 2\textx-4\leq\text{ 15- 3} \\ 2|x\text \leq\text{ 12} \end{gathered}`

Divide both sides by 2:

`\begin{gathered} (2|x-4|)/(2)\text{ }\leq(12)/(2) \\ |x\text -4\leq\text{ 6} \end{gathered}`

Simplifying:

Apply absolute rule:

`\text{if }|u|\text{ }\leq\text{ a, }a\text{ }>\text{ 0 }then\text{ -a }\leq\text{ u }\leq\text{ a}`

Hence:

`-6\text{ }\leq\text{ x-4 }\leq\text{ 6}`
`\begin{gathered} x\text{ -4 }\ge\text{ -6} \\ x\text{ }\ge\text{ -6 + 4} \\ x\ge\text{ -2 } \\ \\ \text{and } \\ \\ x\text{ -4 }\leq\text{ 6} \\ x\text{ }\leq\text{ 6 + 4} \\ x\text{ }\leq\text{ 10} \end{gathered}`

Merge overlapping intervals, we have the solution to be:

`-2\leq\text{ x }\leq\text{ 10}`

The graph of the solution to the inequality is shown below: