Question

How do you solve \[-5-\left( 15y-1 \right)=2\left( 7y-16 \right)-y\]?

Answer 1

Hint: This is a linear equation in one variable as there is only one variable in an equation. In the given question, the variable is the letter ‘y’, to solve this question we need to get ‘y’ on one side of the “equals” sign, and all the other numbers on the other side. To solve this equation for a given variable ‘y’, we have to undo the mathematical operations such as addition, subtraction, multiplication and division that have been done to the variables. In this way we will get our required answer.

Complete step by step solution:
We have the given equation;
\[\Rightarrow -5-\left( 15y-1 \right)=2\left( 7y-16 \right)-y\]
Open all the brackets and simplify the terms in the above equation, we get
\[\Rightarrow -5-15y+1=14y-32-y\]
Combining the like terms in the above equation, we get
\[\Rightarrow -4-15y=13y-32\]
Adding ‘4’ to both the side of the equation, we get
\[\Rightarrow -4-15y+4=13y-32+4\]
Simplifying the above, we get
\[\Rightarrow -15y=13y-28\]
Subtracting 13y from both the sides of the equation, we get
\[\Rightarrow -15y-13y=13y-28-13y\]
Combining the like terms, we get
\[\Rightarrow -28y=-28\]
Dividing both the side of the equation by -28, we get
\[\Rightarrow \dfrac{-28y}{-28}=\dfrac{-28}{-28}\]
Simplifying the above, we get
\[\Rightarrow y=1\]
Therefore, the possible value of \[y\] is \[ 1\].
It is the required solution.

Note: Use addition or subtraction properties of equality to gather variable terms on one side of the equation and constant on the other side of the equation. The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. Use the multiplication or division properties of equality to form the coefficient of the variable term equivalent to 1. This is the type of question where only mathematical operations such as addition, subtraction, multiplication and division is used.