Question

Solve the following system of equations by the elimination method:

$$\begin{gathered} \left(xviii\right)5x+3y=47 \\ 3x+5y=57 \end{gathered}$$

Answer 1

Solution :-

Step 1: Adding equation (1) and equation (2) and then dividing the new equation obtained with the common factor.

$$\begin{gathered} 5x+3y=47----\left(1\right) \\ 3x+5y=57----\left(2\right) \\ \left(1\right)+\left(2\right)\Rightarrow 8x+8y=104----\left(3\right) \end{gathered}$$

Dividing equation (3) with 8, we get ,

$x+y=13---\left(4\right)$

Step 2: Subtracting equation (1) from equation (2) and then dividing it with common factor.

$\left(1\right)-\left(2\right)\Rightarrow 2x-2y=-10----\left(5\right)$

Dividing equation (5) with 2, we get ,

$x-y=-5----\left(6\right)$

Step 3: Adding equation (4) and equation (6).

$$\begin{gathered} x+y=13----\left(4\right) \\ x-y=-5----\left(6\right) \\ \left(4\right)+\left(6\right)\Rightarrow 2x=8 \\ x=\frac{8}{2}=4 \end{gathered}$$

Step 4: substitute the value of in $x=4$ equation (4) .

$$\begin{gathered} x+y=13----\left(4\right) \\ 4+y=13 \\ y=9 \end{gathered}$$

Hence, the value of $x=4$ and $y=9$.