# How do you solve  3 abs( x + 5 ) < 21 ?

Apr 29, 2015

$x < 2$ or $x > - 12$

Solution

$3 \left\mid x + 5 \right\mid < 21$

$\left\mid x + 5 \right\mid < \frac{21}{3}$

$\left\mid x + 5 \right\mid < 7$

It has two solutions

Either

$x + 5 < 7$ ......(i)

Or

$x + 5 > - 7$.....(ii)

Solving (i)

$x + 5 - 5 < 7 - 5$

$x + \cancel{5} - \cancel{5} < 2$

$x < 2$

Now, on solving (ii)

$x + 5 - 5 > - 7 - 5$

$x + \cancel{5} - \cancel{5} > - 12$

$x > - 12$

Apr 29, 2015

We can divide both sides by 3 to get

$\left\mid x + 5 \right\mid < \frac{21}{3}$

$\to \left\mid x + 5 \right\mid < 7$

This tells us that $x + 5$ will lie between $- 7 \mathmr{and} 7$:

$- 7 < x + 5 < 7$

$- 7 - 5 < x + \cancel{5} - \cancel{5} < 7 - 5$

color(green)( -12 < x < 2

To verify your answer , choose appropriate values of $x$ and see if the inequality is satisfied

• Say $x = - 10$

Left hand side = $3 \cdot \left\mid - 10 + 5 \right\mid = 3 \cdot \left\mid - 5 \right\mid = 3 \cdot 5 = 15$ ($< 21$)

• Say $x = 1$

Left hand side = $3 \cdot \left\mid 1 + 5 \right\mid = 3 \cdot \left\mid 6 \right\mid = 3 \cdot 6 = 18$ ($< 21$)