Question

Solve the following system of equations by the method of substitution:
$2x+3y=5$, $3x+4y=6$

Answer 1

Given equations are

$2x+3y=5$ ....(1)

and $3x+4y=6$ ....(2)

Therefore, $3x=6-4y\Rightarrow x=\frac{6-4y}{3}$ ....(3)

Substitute this value in equation (1), we get

$2\left(\frac{6-4y}{3}\right)+3y=5$

$\Rightarrow 12-8y+9y=15$

$\Rightarrow y=3$

Put this value in equation (3),

$x=\frac{6-12}{3}$

$\Rightarrow x=-2$

Therefore, the solution is $x=-2,y=3$.