Question

Solve the following system of equations by the elimination method:

$$\begin{gathered} \left(xix\right)1.25x+1.1y=1895 \\ 1.1x+1.25y=1770 \end{gathered}$$

Answer 1

Solution :-

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

$$\begin{gathered} 1.25x+1.1y=1895----\left(1\right) \\ 1.1x+1.25y=1770----\left(2\right) \\ \left(1\right)+\left(2\right)\Rightarrow 2.35x+2.35y=3666----\left(3\right) \end{gathered}$$

Dividing equation (3) with the number $2.35$

$\left(3\right)÷2.35\Rightarrow x+y=1560----\left(4\right)$

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

$$\begin{gathered} 1.25x+1.1y=1896----\left(1\right) \\ 1.1x+1.25y=1770----\left(2\right) \\ \left(1\right)-\left(2\right)\Rightarrow 0.15x-0.15y=126----\left(5\right) \end{gathered}$$

Dividing equation (5) with 0.15, we get ,

$x-y=840----\left(6\right)$

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

$$\begin{gathered} x+y=1560----\left(4\right) \\ x-y=840-----\left(6\right) \\ \left(4\right)+\left(6\right)\Rightarrow 2x=2400 \\ x=1200 \end{gathered}$$

Step 4: substitute the value of in $x=1200$ equation (4) .

$$\begin{gathered} x+y=1560 \\ 1200+y=1560 \\ y=360 \end{gathered}$$

Hence, the value of $x=1200$ and $y=360$.