Question

How do you solve \[12\left( 5+2y \right)=4y-\left( 6-9y \right)\] ?

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
\[12\left( 5+2y \right)=4y-\left( 6-9y \right)\]
First we have to remove the parentheses on both sides.
Now we will remove the parentheses on LHS side by multiplying with 12.
We will get
\[\Rightarrow 60+24y=4y-\left( 6-9y \right)\]
Now we will remove the parentheses on the RHS side by multiplying with -.
We will get
\[\Rightarrow 60+24y=4y-6+9y\]
By simplifying we will get
\[\Rightarrow 60+24y=13y-6\]
Now we have to group the like terms I.e., we have to isolate terms containing variable y and remaining on the other side.
So group the terms we have to subtract 13y on both sides of the equation.
By subtracting We will get
\[\Rightarrow 60+24y-13y=13y-13y-6\]
By simplifying we get
 \[\Rightarrow 60+11y=-6\]
Now we will subtract 60 on both sides of the equation.
We will get
\[\Rightarrow 60+11y-60=-6-60\]
By simplification
\[\Rightarrow 11y=-66\]
Now we have to divide the equation with 11 on both sides .
We will get
\[\Rightarrow \dfrac{11y}{11}=\dfrac{-66}{11}\]
By simplifying we will get
\[\Rightarrow y=-6\]
So by solving the given equation we get the value of y as -6.

Note:
We can verify the solution by back substituting the value. Substitute the y 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.