Question

Solve the following system of equations by the elimination method:

$$\begin{gathered} \left(xxii\right)126x+84y=420 \\ 84x+126y=420 \end{gathered}$$

Answer 1

Solution :-

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

$$\begin{gathered} 126x+84y=420----\left(1\right) \\ 84x+126y=420----\left(2\right) \\ \left(1\right)+\left(2\right)\Rightarrow 210x+210y=840----\left(3\right) \end{gathered}$$

Dividing equation (3) with the number 210.

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

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

$$\begin{gathered} 126x+84y=420----\left(1\right) \\ 84x+126y=420----\left(2\right) \\ \left(1\right)-\left(2\right)\Rightarrow 42x-42y=-----\left(5\right) \end{gathered}$$

Dividing equation (5) with the number 42.

$x-y=0----\left(6\right)$

Step 3: Adding equation (4) and equation (6) we get,

$$\begin{gathered} x+y=4----\left(4\right) \\ x-y=0----\left(6\right) \\ \left(4\right)+\left(6\right)\Rightarrow 2x=4 \\ x=2 \end{gathered}$$

Step 4: substitute the value of in $x=2$ equation (4) we get,

$$\begin{gathered} x+y=4----\left(4\right) \\ 2+y=4 \\ y=2 \end{gathered}$$

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