Question

Use the elimination method to solve the below equations:
$x-\dfrac{y}{2}=\dfrac{5}{3}$
$\dfrac{3}{2}x-\dfrac{3}{8}y=1$

Answer 1

Hint: Here we have to simplify the given linear equation by using the elimination method. Firstly we will make the coefficient of one unknown variable the same in both the equations by multiplying them with a number. Then we will subtract the first equation by the second equation and get the value of one variable. Finally we will substitute the value in the equation and get the value of the second variable and our desired answer.

Complete step by step solution:
We have to solve the given equations by elimination method:
$x-\dfrac{y}{2}=\dfrac{5}{3}$……$\left( 1 \right)$
$\dfrac{3}{2}x-\dfrac{3}{8}y=1$…..$\left( 2 \right)$
Now as we have to make the coefficient of one variable same in both the equation we will multiply equation (1) by $\dfrac{3}{2}$ as follows:
$\Rightarrow \dfrac{3}{2}\left( x-\dfrac{y}{2} \right)=\dfrac{5}{3}\times \dfrac{3}{2}$
$\Rightarrow \dfrac{3}{2}x-\dfrac{3y}{4}=\dfrac{5}{2}$….$\left( 3 \right)$
Subtract equation (3) from equation (2) as follows:
$\Rightarrow \left( \dfrac{3}{2}x-\dfrac{3y}{4} \right)-\left( \dfrac{3}{2}x-\dfrac{3y}{8} \right)=\dfrac{5}{2}-1$
$\Rightarrow \dfrac{3}{2}x-\dfrac{3y}{4}-\dfrac{3}{2}x+\dfrac{3}{8}y=\dfrac{5-2}{2}$
Simplifying further we get,
$\Rightarrow \left( \dfrac{3}{2}x-\dfrac{3}{2}x \right)+\left( \dfrac{3}{8}y-\dfrac{3}{4}y \right)=\dfrac{3}{2}$
$\Rightarrow 0+\left( \dfrac{3y-3\times 2y}{8} \right)=\dfrac{3}{2}$
So we get,
$\Rightarrow \dfrac{3y-6y}{8}=\dfrac{3}{2}$
$\Rightarrow \dfrac{-3y}{8}=\dfrac{3}{2}$
Further simplifying,
$\Rightarrow y=\dfrac{3}{2}\times \dfrac{-8}{3}$
$\Rightarrow y=-4$
Substitute $y=-4$ in equation (3) we get,
$\Rightarrow \dfrac{3}{2}x-\dfrac{3\times -4}{4}=\dfrac{5}{2}$
$\Rightarrow \dfrac{3}{2}x+3=\dfrac{5}{2}$
Simplify further we get,
$\Rightarrow \dfrac{3}{2}x=\dfrac{5}{2}-3$
$\Rightarrow \dfrac{3}{2}x=\dfrac{5-6}{2}$
So we get,
$\Rightarrow \dfrac{3}{2}x=\dfrac{-1}{2}$
$\Rightarrow x=\dfrac{-1}{2}\times \dfrac{2}{3}$
So,
$\Rightarrow x=\dfrac{-1}{3}$
So we got $x=\dfrac{-1}{3},y=-4$ .
Hence on solving equation $x-\dfrac{y}{2}=\dfrac{5}{3}$ and $\dfrac{3}{2}x-\dfrac{3}{8}y=1$ we get the answer as $x=\dfrac{-1}{3},y=-4$.

Note:
By using the elimination method we eliminate an unknown variable and get an equation in one variable form which can be solved easily. As we have eliminated the variable $x$ in this question we can also eliminate $y$. But always try to remove the variable which has a lower coefficient or which requires less calculation. We can cross check our answer by substituting the obtained values in the equation given and see whether they satisfy the equation or not as follows:
Put $x=\dfrac{-1}{3}$ and $y=-4$ in equation (1) as follows:
$\Rightarrow \dfrac{-1}{3}-\dfrac{-4}{2}=\dfrac{5}{3}$
$\Rightarrow \dfrac{-1}{3}+2=\dfrac{5}{3}$
So,
$\Rightarrow \dfrac{-1+6}{3}=\dfrac{5}{3}$
$\Rightarrow \dfrac{5}{3}=\dfrac{5}{3}$
Our equation is satisfied as LHS is equal to RHS.
Next put $x=\dfrac{-1}{3}$ and $y=-4$ in equation (1) as follows:
$\Rightarrow \dfrac{3}{2}\times \dfrac{-1}{3}-\dfrac{3}{8}\times -4=1$
$\Rightarrow \dfrac{-1}{2}+\dfrac{3}{2}=1$
So,
$\Rightarrow \dfrac{-1+3}{2}=1$
$\Rightarrow \dfrac{2}{2}=1$
This is,
$\Rightarrow 1=1$
So our equation is satisfied as LHS is equal to RHS.