Question

Solve the given pair of linear equations:
$\begin{align}
  & 4x+\dfrac{x-y}{8}=17 \\
 & 2y+x-\dfrac{5y+2}{3}=2 \\
\end{align}$

Answer 1

Hint:Here, we will use elimination method to find the solution to the given pair of linear equations. We will eliminate one of the variables and then find the value of the other. After that, we will use the obtained value to find the value of the variable that was eliminated.

Complete step-by-step answer:
We know that an equation is said to be a linear equation in two variables if it is written in the form of ax+by+c=0, where a, b and c are real numbers and are not equal to zero.
The solution for such an equation is a pair of values, one for x and one for y which further makes the two sides of an equation equal.
Here, the pair of linear equations given to us are:
$\begin{align}
  & 4x+\dfrac{x-y}{8}=17.........\left( 1 \right) \\
 & 2y+x-\dfrac{5y+2}{3}=2.........\left( 2 \right) \\
\end{align}$
In equation (1), on taking the lowest common multiple of 1 and 8, we can write it as:
$\begin{align}
  & \dfrac{8\times 4x+\left( x-y \right)}{8}=17 \\
 & \Rightarrow 32x+x-y=8\times 17 \\
 & \Rightarrow 33x-y=136...........\left( 3 \right) \\
\end{align}$
Now, in the equation (2), on taking the LCM of 1 and 3, we can write it as:
$\begin{align}
  & \dfrac{3\left( 2y+x \right)-\left( 5y+2 \right)}{3}=2 \\
 & \Rightarrow 6y+3x-5y-2=2\times 3 \\
 & \Rightarrow 3x+y-2=6 \\
 & \Rightarrow 3x+y=8.............\left( 4 \right) \\
\end{align}$
Now, on applying the method of elimination, we will add equation (3) and equation (4) and find the value of x.
On adding equation (2) and (4), we get:
$\begin{align}
  & 33x-y+3x+y=136+8 \\
 & \Rightarrow 36x=14 \\
 & \Rightarrow x=\dfrac{144}{36}=4 \\
\end{align}$
So, the value of x comes out to be 4.
Now, on putting x = 4 in equation (4), we can get the value of y as:
$\begin{align}
  & 3\times 4+y=8 \\
 & \Rightarrow 12+y=8 \\
 & \Rightarrow y=-4 \\
\end{align}$
So, the value of y comes out to be -4.
Hence, the solution of the given pair of linear equations is x = 4 and y = -4.

Note: Students should note here that it is not necessary that we eliminate x using the given equations. One can also eliminate y and then find the value of x. One can verify the answer by putting the obtained values in the equations and it must satisfy i.e LHS=RHS should be obtained else the solutions is not correct.