Question

How do you solve $\left| x+4 \right|=10$?

Answer 1

The modulus gives the absolute value of its argument. To solve problems of the type $\left| x \right|=a$, we have to take two cases and solve the two cases separately. The first case is $x=a$. And the second case is $x=-a$. We will take the same cases for the given modulus problem.

Complete step by step answer:
We are asked to solve $\left| x+4 \right|=10$. As we know to solve problems of the type $\left| x \right|=a$, we have to take two cases and solve the two cases separately. The first case is $x=a$. And the second case is $x=-a$. Here, we have $x+4$ at the place of $x$, and $a=10$. We will take the same cases for the given modulus problem.
The first case is, $x+4=10$
Subtracting 4 from both sides of the above equation, we get

& \Rightarrow x+4-4=10-4 \\\ & \therefore x=6 \\\ \end{aligned}$$ One solution value of $$x$$ is 6. The second case is, $$x+4=-10$$ Subtracting 4 from both sides of the above equation, we get $$\begin{aligned} & \Rightarrow x+4-4=-10-4 \\\ & \therefore x=-14 \\\ \end{aligned}$$ The other solution value of $$x$$ is $$-14$$. **Thus, the two values satisfying the given modulus are 6, and -14.** **Note:** It is necessary to know the properties of the modulus to solve its questions. The property of the modulus, we used to solve this problem is $$\left| x \right|=a\Rightarrow x=\pm a$$. There are many other properties of the modulus that we should know some of them are as follows, If $$\left| x \right| < a$$, then $$-a < x < a$$. Given that a is a positive and real. If $$\left| x \right|>a$$, then $$x < -a$$ or $$x>a$$. These should be remembered.