# How do you solve 5abs(3+7m)+1=51?

Aug 3, 2016

$m = - \frac{13}{7} \text{ }$ or $\text{ } m = 1$

#### Explanation:

The first thing to do here is isolate the modules on one side of the equation. To do that, subtract $1$ from both sides of the equation

$5 \cdot | 3 + 7 m | + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} - 1 = 51 - 1$

$5 \cdot | 3 + 7 m | = 50$

Now divide both sides of the equation by $5$ to get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} \cdot | 3 + 7 m |}{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}} = \frac{50}{5}$

$| 3 + 7 m | = 10$

At this point, you have two possibilities to cover

$\textcolor{w h i t e}{a}$

• $3 + 7 m \ge 0 \implies | 3 + 7 m | = 3 + 7 m$

In this case, the above equation will get you

$3 + 7 m = 10$

$7 m = 7 \implies m = \frac{7}{7} = 1$

$\textcolor{w h i t e}{a}$

• $3 + 7 m < 0 \implies | 3 + 7 m | = - \left(3 + 7 m\right)$

In this case, the equation becomes

$- \left(3 + 7 m\right) = 10$

$- 3 - 7 m = 10$

$- 7 m = 13 \implies m = - \frac{13}{7}$

You can thus say that the original equation has two possibile solutions

$m = - \frac{13}{7} \text{ }$ or $\text{ } m = 1$

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

$5 \cdot | 3 + 7 \cdot 1 | + 1 = 51$

$5 \cdot 10 + 1 = 51 \text{ } \textcolor{g r e e n}{\sqrt{}}$

$5 \cdot | 3 + 7 \cdot \left(- \frac{13}{7}\right) | + 1 = 51$

$5 \cdot | 3 - 13 | + 1 = 51$

$5 \cdot 10 + 1 = 51 \text{ } \textcolor{g r e e n}{\sqrt{}}$