Question

If a set of equations are given as $3x-2y=5$and $3y-2x=3$ then find the value of $\left( x+y \right)$.

Answer 1

Hint: We have a given set of equations:
\[\begin{align}
  & 3x-2y=5......(1) \\
 & 3y-2x=3......(2) \\
\end{align}\]
We need to find the value of $\left( x+y \right)$. For this, firstly we need to get the values of x and y and then substitute in the given equation.
To get the values of x and y we need to solve the given set of linear equations.

Complete step by step answer:
One method to solve a set of equations is the Substitution method. In the substitution method, we get the value of one variable in terms of other variables from one equation and substitute it in the other equation to get the value of another variable.
So, by applying substitution method for the given set of linear equations:
From equation (2), we can write:
$\begin{align}
  & 3y-2x=3 \\
 & \Rightarrow 3y=3+2x \\
 & \Rightarrow y=\dfrac{3+2x}{3}......(3) \\
\end{align}$
Now, we get the value of y in terms of x.
Put the value of y in equation (1), we get:
$\begin{align}
  & \Rightarrow 3x-2y=5 \\
 & \Rightarrow 3x-2\left( \dfrac{3+2x}{3} \right)=5 \\
 & \Rightarrow 9x-4x=15+6 \\
 & \Rightarrow 5x=21 \\
 & \Rightarrow x=\dfrac{21}{5}......(4) \\
\end{align}$
Put the value of x in equation (3) to get the value of y:
So, we have:
\[\begin{align}
  & y=\dfrac{3+2\left( \dfrac{21}{5} \right)}{3} \\
 & =\dfrac{15+42}{15} \\
 & =\dfrac{57}{15} \\
 & =\dfrac{19}{5}.....(5)
\end{align}\]
Now we need to find the value of $\left( x+y \right)$. From equation (4) and (5), get the value of x and y and add to get the answer:
$\begin{align}
  & x+y=\dfrac{21}{5}+\dfrac{19}{5} \\
 & =\dfrac{40}{5} \\
 & =8
\end{align}$

Hence $\left( x+y \right)$ equals to 8.

Note: There is another method to solve a given set of linear equations, i.e. Elimination method. In the elimination method, we multiply or divide both the equations with a number to eliminate one variable in both the equations by adding or subtracting them and get the value of another value. Then we put the value in one of the equations and get the value of the eliminated variable.
For the given set of equations:
\[\begin{align}
  & 3x-2y=5......(1) \\
 & 3y-2x=3......(2) \\
\end{align}\]
Multiply equation (1) by 2 and equation (2) by 3, we get:
$\begin{align}
  & 6x-4y=10......(3) \\
 & -6x+9y=9......(4) \\
\end{align}$
Add equation (3) and (4); we get:
$\begin{align}
  & \Rightarrow -4y+9y=10+9 \\
 & \Rightarrow 5y=19 \\
 & \Rightarrow y=\dfrac{19}{5}......(5) \\
\end{align}$
Now put the value of y from equation (5) in any of the above equations to get the value of x.
Let’s put value of y in equation (1), we get:
$\begin{align}
  & \Rightarrow 3x-2y=5 \\
 & \Rightarrow 3x-2\left( \dfrac{19}{5} \right)=5 \\
 & \Rightarrow 3x=\dfrac{63}{5} \\
 & \Rightarrow x=\dfrac{21}{5} \\
\end{align}$
Now we need to find the value of $\left( x+y \right)$. From equation (4) and (5), get the value of x and y and add to get the answer:
$\begin{align}
  & x+y=\dfrac{21}{5}+\dfrac{19}{5} \\
 & =\dfrac{40}{5} \\
 & =8
\end{align}$
Hence $\left( x+y \right)$ equals to 8.