Given inequality:
[tex]2|\text{ x- 4| + 3 }\leq\text{ 15}[/tex]
Subtract 3 from both sides:
[tex]\begin{gathered} 2\text{ |x-4| + 3 -3 }\leq\text{ 15- 3} \\ 2|x\text{ -4| }\leq\text{ 12} \end{gathered}[/tex]
Divide both sides by 2:
[tex]\begin{gathered} \frac{2|x-4|}{2}\text{ }\leq\frac{12}{2} \\ |x\text{ -4| }\leq\text{ 6} \end{gathered}[/tex]
Simplifying:
Apply absolute rule:
[tex]\text{if }|u|\text{ }\leq\text{ a, }a\text{ }>\text{ 0 }then\text{ -a }\leq\text{ u }\leq\text{ a}[/tex]
Hence:
[tex]-6\text{ }\leq\text{ x-4 }\leq\text{ 6}[/tex][tex]\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}[/tex]
Merge overlapping intervals, we have the solution to be:
[tex]-2\leq\text{ x }\leq\text{ 10}[/tex]
The graph of the solution to the inequality is shown below:
