we have the inequality
[tex]\lvert5x+4\rvert\leq10[/tex]step 1
Find out the first solution (positive case)
[tex]\begin{gathered} +(5x+4)\leq10 \\ 5x\leq10-4 \\ 5x\leq6 \\ x\leq\frac{6}{5} \\ x\leq1.20 \end{gathered}[/tex]The first solution is all real numbers less than or equal to 1.20
Interval (-infinite,1.20]
step 2
Find out the second solution (negative case)
[tex]-(5x+4)\leq10[/tex]Multiply by -1 both sides
[tex]\begin{gathered} (5x+4)\ge-10 \\ 5x\ge-10-4 \\ 5x\ge-14 \\ x\ge-\frac{14}{5} \\ x\ge-2.8 \end{gathered}[/tex]The second solution is all real numbers greater than or equal to -2.8
the interval [-2.8, infinite)
step 3
Find out the solution to the given inequality
The solution is
[-2.8, infinite) ∩ (-infinite,1.20]=[-2.8,1.20]
the solution is the interval [-2.8,1.20]
see the attached figure to better understand the problem
