Question

How do you solve $\left| 5x+5 \right|-7=18$ ?

Answer 1

Hint: In this problem, we are to find the solution of the given equation. As there is a modulus sign given in the problem, we are to choose it using the positive and the negative value. Using the positive value we are getting one value and negative one getting another value. Putting the value back in the equation we get the right solutions of the equation.

Complete step by step answer:
According to the question, we are to solve $\left| 5x+5 \right|-7=18$.
Now, to start with, when you take off the absolute value we have two separate equations.
So, for the positive one, we have, $5x+5-7=18$, for x > -1.
And again, for x < -1, we get, $-\left( 5x+5 \right)-7=18$.
Now, we will use both equations to find the solution. Both answers may not always work. We need to put the solutions back to check our solution.
For the first equation, $5x+5-7=18$
Simplifying, $5x-2=18$
Adding 2 on both sides, $5x=20$
Dividing the equation by 5 on both sides, $x=\dfrac{20}{5}=4$ .
Again, for the second equation, $-\left( 5x+5 \right)-7=18$
Simplifying, $-5x-5-7=18$
$\Rightarrow -5x-12=18$
Adding 12 on both sides, $-5x=30$.
Thus we get, $x=\dfrac{30}{-5}=-6$.
Now, we need to check if the values are satisfying the given equation.
Putting the value x = 4, we get, on the left side, $\left| 5\times 4+5 \right|-7$.
Simplifying we get, $\left| 20+5 \right|-7=25-7=18$.
This is equal to the right hand side. So, this value satisfies.
 Putting the value x = -6, we get, on the left side, $\left| 5\times \left( -6 \right)+5 \right|-7$.
Simplifying we get, $\left| -30+5 \right|-7=\left| -25 \right|-7$.
This can be written as, $25-7=18$.
This is equal to the right hand side. So, this value satisfies.
So, we have the value of x as, x = 4, -6.

Note: In the solution, we have used the properties of the absolute value of a number. Absolute value describes the distance from zero that a number is on the number line, without considering direction. The absolute value of a number is never negative.