Question

How do you solve \[7y-2\left( 8y+1 \right)=4\]?

Answer 1

Hint: This type of question is based on the concept of converting linear equations with one variable. We have to first use distributive property, that is, \[a\left( b+c \right)=ab+ac\]. We then have to group the y terms and constants in the left-hand side of the equation and right-hand side of the equation respectively. Then, divide the whole equation by -9 to get the value of y which is the required solution.

Complete step-by-step solution:
According to the question, we are asked to find the value of y in the equation \[7y-2\left( 8y+1 \right)=4\].
We have been given the equation is \[7y-2\left( 8y+1 \right)=4\]. ----------(1)
Here the variable is y.
First, we have to consider equation (1) and use distributive property \[a\left( b+c \right)=ab+ac\] in the left-hand side of the equation (1).
\[\Rightarrow 7y-2\times 8y-2\times 1=4\]
On further simplification, we get
\[\Rightarrow 7y-16y-2=4\]
Take y common from the first two terms of the equation.
\[\Rightarrow \left( 7-16 \right)y-2=4\]
Since 7-16=-9, we get
\[\Rightarrow \left( -9 \right)y-2=4\]
Let us now add 2 from both the sides of the equation. We get
\[\Rightarrow \left( -9 \right)y+2-2=4+2\]
Terms with the same magnitude and opposite signs cancel out.
Therefore, \[\left( -9 \right)y=4+2\]
On further simplification, we get
\[\Rightarrow -9y=6\]
Let us now divide the whole equation by -9 to convert the coefficient of y to 1.
\[\Rightarrow \dfrac{-9y}{-9}=\dfrac{6}{-9}\]
Since the numerator and denominator have -9 as common, we can cancel them out.
\[\Rightarrow y=\dfrac{6}{-9}\]
On further simplification, we get
\[\Rightarrow y=-\dfrac{2\times 3}{3\times 3}\]
Cancelling out the common term 3 from the numerator and denominator, we get
\[y=-\dfrac{2}{3}\]
Hence, the value of y in the equation \[7y-2\left( 8y+1 \right)=4\] is \[-\dfrac{2}{3}\].

Note: We should always keep the variable in the left-hand side and the constants in the right-hand side of the equation. We should not make calculation mistakes based on sign conventions.
We can check whether the obtained solution is correct or not.
Substitute the value of y in the given equation \[7y-2\left( 8y+1 \right)=4\].
Consider the LHS.
LHS=\[7y-2\left( 8y+1 \right)\]
On substituting \[y=-\dfrac{2}{3}\], we get
LHS=\[7\left( \dfrac{-2}{3} \right)-2\left( 8\left( \dfrac{-2}{3} \right)+1 \right)\]
On using distributive property, we get
LHS=\[7\left( \dfrac{-2}{3} \right)-2\times 8\left( \dfrac{-2}{3} \right)-2\]
LHS=\[7\left( \dfrac{-2}{3} \right)-16\left( \dfrac{-2}{3} \right)-2\]
Taking the common terms out, we get
LHS=\[\left( 7-16 \right)\left( \dfrac{-2}{3} \right)-2\]
LHS=\[\left( -9 \right)\left( \dfrac{-2}{3} \right)-2\]
Cancelling the common terms from the numerator and denominator, we get
LHS=\[\left( -3\times -2 \right)-2\]
LHS=\[6-2\]
LHS=4
We know that RHS=4
Therefore, LHS=RHS.
Hence, the obtained answer is correct.