How do you solve the system using the elimination method for $2x-3y=7$ and $3x+y=5$?

Hint: Mark the equations as (1) and (2). Multiply equation (1) with 1 and equation (2) with 3 and mark the new equations as (3) and (4) respectively. Add equation (4) with equation (3) to get the value of ‘x’, Put the value of ‘x’ in any of the equations to get the value of ‘y’.

Complete step by step solution:
Elimination method for solving systems: In elimination methods we have to either add or subtract the equations if the coefficients of one variable are the same. If the coefficients of one variable are different, then we have to multiply one equation with a required constant to make the coefficients equal and then we have to add or subtract accordingly.
Now, considering our equations
$2x-3y=7$……….(1)
$3x+y=5$……….(2)
To make the coefficients of ‘y’ equal, we have to multiply the equation (1) by 1 and the equation (2) by 3
$eq(1)\times 1\Rightarrow 2x-3y=7$……….(3)
$eq(2)\times 3\Rightarrow 9x+3y=15$……….(4)
Adding equation (4) with equation (3), we get
$\begin{align}
  & eq(3)+eq(4) \\
 & \Rightarrow 2x-3y+9x+3y=7+15 \\
 & \Rightarrow 11x=22 \\
 & \Rightarrow x=\dfrac{22}{11} \\
 & \Rightarrow x=2 \\
\end{align}$
Putting the value of ‘x’ in equation (2), we get
$\begin{align}
  & 3x+y=5 \\
 & \Rightarrow 3\times 2+y=5 \\
 & \Rightarrow 6+y=5 \\
 & \Rightarrow y=5-6 \\
 & \Rightarrow y=-1 \\
\end{align}$

Hence, the solution of the system $2x-3y=7$ and $3x+y=5$ is $\left( x,y \right)=\left( 2,-1 \right)$.

Note: The given set of linear equations can also be solved by taking as simultaneous equations. From the first equation, we have to find the value of ‘x’ in terms of ‘y’. Then we have to put theta value in the second equation to find the value of ‘y’. Again putting that ‘y’ value in one of the equations, ‘x’ value can be obtained.