Question

Which of the following is the solution set for the conjunction –6 < 4x + 2 < 6? x>−32 and x<32 x >–2 and x < 1 x>−32 or x<32 x <–2 or x > 1

Answer 1

$-6<4x+2<6\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-2<x<1\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-2,\:1\right)\end{bmatrix}$

Solution:

Given conjunction is:

$-6<4x+2<6$

We have to find the solution set

Let us solve the given conjuction

$\mathrm{If}\:a<u<b\:\mathrm{then}\:a<u\ \mathrm{and}\ \:u<b$

Therefore,

$-6<4x+2\quad \mathrm{and}\quad \:4x+2<6$

Let us first solve -6 < 4x + 2

$-6 < 4x + 2$

Switch the sides by flipping the inequality symbol

$4x+2>-6$

$\mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}\\\\4x+2-2>-6-2\\\\\mathrm{Simplify}\\\\4x>-8\\\\\mathrm{Divide\:both\:sides\:by\:}4\\\\\frac{4x}{4}>\frac{-8}{4}$

$x>-2$

Now let us solve 4x + 2 < 6

$4x+2<6\\\\\mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}\\\\4x+2-2<6-2\\\\\mathrm{Simplify}\\\\4x<4\\\\\mathrm{Simplify}\\\\x<1$

$\mathrm{Combine\:the\:intervals}$

$x>-2\quad \mathrm{and}\quad \:x<1$

$Merge\:Overlapping\:Intervals$

$-2<x<1$

Thus the solution set for  –6 < 4x + 2 < 6 is:

$-6<4x+2<6\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-2<x<1\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-2,\:1\right)\end{bmatrix}$