Question

How do you solve $\left| {3x + 4} \right| - 5 = 3$ ?

Answer 1

Hint: The given expression has an absolute value bracket within it. The absolute value bracket makes any negative value positive and any positive value positive itself. So here we consider two cases where we take the negative value of the expression inside the modulus and solve and then the positive value of the expression in the modulus and then solve.

Complete step by step solution:
The given expression is, $\left| {3x + 4} \right| - 5 = 3$
So, let’s consider two cases,
The one where we take the negative value of the expression inside the modulus and solve and again take another case where the positive value of the expression in the modulus is considered and then solve it.
We get two solutions of $x$ for these types of equations.
First, let’s take the positive value of the expression and solve.
$ \Rightarrow ( + )(3x + 4) - 5 = 3$
Now let’s solve this linear equation.
$ \Rightarrow 3x + 4 - 5 = 3$
$ \Rightarrow 3x - 1 = 3$
Add $1$ on both sides of the equation.
$ \Rightarrow 3x - 1 + 1 = 3 + 1$
Now evaluate.
$ \Rightarrow 3x = 4$
Now divide the entire equation with the constant $3$
$ \Rightarrow \dfrac{{3x}}{3} = \dfrac{4}{3}$
$ \Rightarrow x = \dfrac{4}{3}$
This is one value of $x$ .
The other one will be,
let’s take the negative value of the expression and solve.
$ \Rightarrow ( - )(3x + 4) - 5 = 3$
Now let’s solve this linear equation.
$ \Rightarrow - 3x - 4 - 5 = 3$
$ \Rightarrow - 3x - 9 = 3$
Add $9$ on both sides of the equation.
$ \Rightarrow -3x - 9 + 9 = 3 + 9$
Now evaluate.
$ \Rightarrow - 3x = 12$
Now divide the entire equation with the constant $3$ and negate the equation.
$ \Rightarrow x = - \dfrac{{12}}{3}$
$ \Rightarrow x = - 4$
This is the other value of $x$ .

Hence the two values of $x$ are $\dfrac{4}{3}, - 4$.

Note: A mathematical function, absolute value converts any value to non-negative value. Absolute value is also known as the number’s distance from zero. Always keep in mind that anything outside the absolute value brackets does not affect the values inside the absolute value brackets.
We can cross-check our solution by substituting it back in the expression.
For $x = \dfrac{4}{3}$ ,
$ \Rightarrow \left| {3\left( {\dfrac{4}{3}} \right) + 4} \right| - 5 = 3$
On evaluating we get,
$ \Rightarrow \left| {4 + 4} \right| - 5 = 3$
$ \Rightarrow \left| 8 \right| - 5 = 3$
Any positive value will remain positive inside the modulus.
$ \Rightarrow 8 - 5 = 3$
$ \Rightarrow 3 = 3$
LHS=RHS.
Now let’s check for $x = - 4$
$ \Rightarrow \left| {3\left( { - 4} \right) + 4} \right| - 5 = 3$
On evaluating we get,
$ \Rightarrow \left| { - 12 + 4} \right| - 5 = 3$
$ \Rightarrow \left| { - 8} \right| - 5 = 3$
Any negative value will become positive inside the modulus.
$ \Rightarrow 8 - 5 = 3$
$ \Rightarrow 3 = 3$
LHS=RHS.
Hence our both solutions are correct.