Question

How do you solve $ |2x+3|=5 $ and find any extraneous solutions?

Answer 1

Hint: In this question, we have to find the value of x. The equation given in the problem contains the absolute term. For the absolute terms, we will apply basic mathematical rules to solve the same. We first remove the absolute sign and then we get two equations. So, we solve both the equations separately and make the necessary calculations, to get two values of x. Now, for extraneous solutions, we will put the values of x in the LHS of the equation and if it is equal to the RHS, then we have no extraneous solution, which is the required solution to the problem.

Complete step by step answer:
According to the question, we have to find the value of x.
So, we will apply basic mathematical rules to solve this problem.
The equation given to us is $ |2x+3|=5 $ ----------- (1)
Now, first, we will remove the absolute sign of the equation (1), we get
 $ 2x+3=\pm 5 $
Therefore, we get
 $ 2x+3=5 $ ------------ (2) and
 $ 2x+3=-5 $ ----------- (3)
So, we will first solve equation (2), which is
 $ \Rightarrow 2x+3=5 $
Now, we will subtract 3 on both sides in the above equation, we get
 $ \Rightarrow 2x+3-3=5-3 $
As we know, the same terms with opposite signs cancel out each other, we get
 $ \Rightarrow 2x=2 $
Now, we will divide 2 on both sides of the equation, we get
 $ \Rightarrow \dfrac{2}{2}x=\dfrac{2}{2} $
On further solving, we get
 $ \Rightarrow x=1 $
Now, we will solve equation (3), which is
 $ \Rightarrow 2x+3=-5 $
Now, we will subtract 3 on both sides in the above equation, we get
 $ \Rightarrow 2x+3-3=-5-3 $
As we know, the same terms with opposite signs cancel out each other, we get
 $ \Rightarrow 2x=-8 $
Now, we will divide 2 on both sides of the equation, we get
 $ \Rightarrow \dfrac{2}{2}x=\dfrac{-8}{2} $
On further solving, we get
 $ \Rightarrow x=-4 $
Therefore, for the equation $ |2x+3|=5 $ , the two values of x are $ 1 $ and $ -4 $ .
Now, for checking the extraneous solutions, we will put the value of x in the LHS of the equation (1), that is
We will put the value $ x=1 $ in the LHS of the equation (1), we get
 $ \Rightarrow |2.(1)+3| $
 $ \Rightarrow |2+3| $
Therefore, we get
 $ \Rightarrow 5=RHS $
Therefore, $ x=1 $ is the solution of the equation $ |2x+3|=5 $ .
Now, we will put the value $ x=-4 $ in the LHS of the equation (1), we get
 $ \Rightarrow |2.(-4)+3| $
 $ \Rightarrow |-8+3| $
Therefore, we get
 $ \Rightarrow 5=RHS $
Therefore, $ x=-4 $ is the solution to the equation $ |2x+3|=5 $ .
Hence, there are no extraneous solutions in the equation $ |2x+3|=5 $.

Note:
While solving this problem, when you remove the absolute signs, keep in mind that the RHS will have the $ \pm $ sign, and not only the $ + $ sign or the $ - $ sign. Do all the calculations properly to avoid mathematical errors. For extraneous solutions, after putting the value of x in the equation, the answer must be equal to RHS, it means no extraneous solution is there.