Question

solve each compound inequality and g

1. ( 6 x - 1 < - 1 + 10 x leq x + 8 )

Explanation:

Step1: Solve the left - hand inequality \(6x - 1\lt - 1+10x\)

Subtract \(6x\) from both sides:
\(6x-6x - 1\lt - 1+10x-6x\)
\(-1\lt - 1 + 4x\)
Add \(1\) to both sides:
\(-1 + 1\lt - 1+1 + 4x\)
\(0\lt4x\)
Divide both sides by \(4\):
\(0\lt x\) or \(x\gt0\)

Step2: Solve the right - hand inequality \(-1 + 10x\leq x+8\)

Subtract \(x\) from both sides:
\(-1+10x - x\leq x - x+8\)
\(-1 + 9x\leq8\)
Add \(1\) to both sides:
\(-1+1 + 9x\leq8 + 1\)
\(9x\leq9\)
Divide both sides by \(9\):
\(x\leq1\)

Answer:

The solution of the compound inequality is \(0\lt x\leq1\). On the number line, we have an open circle at \(x = 0\) (because \(x\gt0\)) and a closed circle at \(x = 1\) (because \(x\leq1\)) and the line segment connecting them.