Question

Solve the following system of equations by the method of substitution:
$3x-4y=10$, $4x+3y=5$

Answer 1

Given equations are

$3x-4y=10$ ....(1)

and $4x+3y=5$ ....(2)

Therefore, $3x=10+4y\Rightarrow x=\frac{10+4y}{3}$ ....(3)

Substitute this value of $x$ in equation (2), we get

$4\left(\frac{10+4y}{3}\right)+3y=5$

$\Rightarrow 40+16y+9y=15$

$\Rightarrow 25y=-25$

$\Rightarrow y=-1$

Substitute this value of $y$ in equation (3)

$x=\frac{10-4}{3}=2$

Therefore, solution is $x=2,y=-1$.