What is the solution set for the equation |4a + 6| − 4a = 10 ?

Oct 13, 2015

$a = - 2$

Explanation:

The first thing to do here is isolate the modulus on onse side of the equation by adding $4 a$ to both sides

$| 4 a + 6 | - \textcolor{red}{\cancel{\textcolor{b l a c k}{4 a}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{4 a}}} = 10 + 4 a$

$| 4 a + 6 | = 10 + 4 a$

Now, by definition, the absolute value of a real number will only return positive values, regardless of the sign of said number.

This means that the first condition that any value of $a$ must satisfy in order to be a valid solution will be

$10 + 4 a \ge 0$

$4 a \ge - 10 \implies a \ge - \frac{5}{2}$

Keep this in mind. Now, since the absolute value of a number returns a positive value, you can have two possibilities

• $4 a + 6 < 0 \implies | 4 a + 6 | = - \left(4 a + 6\right)$

In this case, the equation becomes

$- \left(4 a + 6\right) = 10 + 4 a$

$- 4 a - 6 = 10 + 4 a$

$8 a = - 16 \implies a = \frac{\left(- 16\right)}{8} = - 2$

• $\left(4 a + 6\right) \ge 0 \implies | 4 a + 6 | = 4 a + 6$

This time, the equation becomes

$\textcolor{red}{\cancel{\textcolor{b l a c k}{4 a}}} + 6 = 10 + \textcolor{red}{\cancel{\textcolor{b l a c k}{4 a}}}$

$6 \ne 10 \implies a \in \emptyset$

Therefore, the only valid solution will be $a = - 2$. Notice that it satisfies the initial condition $a \ge - \frac{5}{2}$.

Do a quick check to make sure that the calculations are correct

$| 4 \cdot \left(- 2\right) + 6 | - 4 \cdot \left(- 2\right) = 10$

$| - 2 | + 8 = 10$

$2 + 8 = 10 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$