# How do you solve 4d + 7 = - abs(2d + 41) ?

$d = - 8$

#### Explanation:

Given that

$4 d + 7 = - | 2 d + 41 |$

Case 1: When $2 d + 41 \setminus \ge 0 \setminus \setminus \mathmr{and} \setminus \setminus d \setminus \ge - \frac{41}{2}$

$4 d + 7 = - \left(2 d + 41\right)$

$4 d + 7 = - 2 d - 41$

$6 d = - 48$

$d = - \frac{48}{6}$

$d = - 8$

Since $d = - 8 > - \frac{41}{2}$ hence $d = - 8$ is a solution

Case 2: When $2 d + 41 < 0 \setminus \setminus \mathmr{and} \setminus \setminus d < - \frac{41}{2}$

$4 d + 7 = - \left(- \left(2 d + 41\right)\right)$

$4 d + 7 = 2 d + 41$

$2 d = 34$

$d = 17$

But $d < - \frac{41}{2}$ hence $d = 17$ is not a solution

Hence, the only solution to the given equation is

$d = - 8$