Question

How do you solve \[3\left( 2a+1 \right)-2=2\left( a-2 \right)+3\left( a+1 \right)\] ?

Answer 1

Hint: We will solve this question using basic linear equation concepts. First we will remove parentheses on both sides of the equation. After removing parentheses we will group like terms and then solve for the variable we have to find the value. By simplifying it we will get the solution.

Complete step by step solution:
Given equation
\[3\left( 2a+1 \right)-2=2\left( a-2 \right)+3\left( a+1 \right)\]
First we have to remove the parentheses on both sides.
Now we will remove the parentheses on the LHS side by multiplying with 3.
We will get
\[\Rightarrow 6a+3-2=2\left( a-2 \right)+3\left( a+1 \right)\]
Now we will remove the parentheses on the RHS side by multiplying with 2.
We will get
\[\Rightarrow 6a+3-2=2a-4+3\left( a+1 \right)\]
Now we will multiply with 3 to get rid of the parenthesis.
\[\Rightarrow 6a+3-2=2a-4+3a+3\]
By simplifying we will get
\[\Rightarrow 6a+1=2a+3a-1\]
By further simplifying we will get
\[\Rightarrow 6a+1=5a-1\]
Now we have to group the like terms I.e., we have to isolate terms containing variable a and remaining on the other side.
So group the terms we have to subtract 5a on both sides of the equation.
By subtracting We will get
\[\Rightarrow 6a+1-5a=5a-1-5a\]
By simplifying we get
 \[\Rightarrow a+1=-1\]
Now we will add subtract 1 on both sides of the equation.
We will get
\[\Rightarrow a+1-1=-1-1\]
By simplification
\[\Rightarrow a=-2\]
So by solving the given equation we get the value as \[a=-2\].

Note: We can verify the solution by back substituting the value. substitute the a value in the equation and check whether the LHS and RHS are equal or not. We can solve this question in many simplification ways but above discussed will contain only basic operations to perform.