Question

Solve the following system of equations by the elimination method:

$$\begin{gathered} \left(xv\right)148x+231y=610 \\ 231x+148y=527 \end{gathered}$$

Answer 1

Solution :-

Step 1:Adding equation (1) and equation(2),we get

$$\begin{gathered} 148x+231y=610----\left(1\right) \\ 231x+148y=527----\left(2\right) \end{gathered}$$

$\left(1\right)+\left(2\right)\Rightarrow 379x+379y=1137----\left(3\right)$

Step 2: Dividing equation (3) with 379,we get

$\left(3\right)÷379\Rightarrow x+y=3$

$x+y=3-----\left(4\right)$

Step 3: Subtracting equation (2) from (1) we get,

$$\begin{gathered} 231x+148y=527-----\left(2\right) \\ 148x+231y=610-----\left(1\right) \end{gathered}$$

$\left(2\right)-\left(1\right)\Rightarrow 83x-83y=-83-----\left(5\right)$

Step 4: Dividing equation (5) with 83.we get

$x-y=-1-----\left(6\right)$

Step 5: Adding equation (4) and equation (6),we get

$$\begin{gathered} x+y=3-----\left(4\right) \\ x-y=-1----\left(6\right) \\ \left(4\right)+\left(6\right)\Rightarrow 2x=2 \\ x=\frac{2}{2}=1 \end{gathered}$$

Step 6: Substitute the value of the variable $x=1$ in the given equation(4) we get the value of the other variable ‘y’

$$\begin{gathered} x+y=3 \\ 1+y=3 \\ y=3-1 \\ y=2 \end{gathered}$$

Hence, the value of $x=1$ and $y=2$.