Question

solve the compound inequality.
2x + 2 < 0 and 4x + 3 > -13
graph the solution on the number line.

Explanation:

Step1: Solve \(2x + 2<0\)

Subtract \(2\) from both sides: \(2x+2 - 2<0 - 2\), so \(2x<-2\).
Divide both sides by \(2\): \(x < - 1\).

Step2: Solve \(4x + 3>-13\)

Subtract \(3\) from both sides: \(4x+3 - 3>-13 - 3\), so \(4x>-16\).
Divide both sides by \(4\): \(x>-4\).

Answer:

The solution of the compound inequality \(2x + 2<0\) and \(4x + 3>-13\) is \(-4 < x<-1\). On the number - line, we draw an open circle at \(x = - 4\) (because \(x
eq-4\) from \(x>-4\)) and an open circle at \(x=-1\) (because \(x
eq - 1\) from \(x < - 1\)), and then shade the region between them.