Question

Find '$x+y$' for the following simultaneous equations : $4x+3y=8$ and $3x+4y=20$.

Answer 1

Let $4x+3y=8$ ...(1)
and $3x+4y=20$ ....(2)
From equation (2), find the value of $x$ in terms of $y$:
$x=\frac{20-4y}{3}$ ...(3)

Substituting $x=\frac{20-4y}{3}$ in equation (1), we get
$\Rightarrow 4\left(\frac{20-4y}{3}\right)+3y=8$
$\Rightarrow \frac{80-16y}{3}+3y=8$
$\Rightarrow \frac{80}{3}-\frac{16y}{3}+3y=8$
$\Rightarrow 3y-\frac{16y}{3}=8-\frac{80}{3}$
$\Rightarrow \frac{-7}{3}y=\frac{-56}{3}$
$\Rightarrow y=\frac{56}{7}=8$
$\Rightarrow y=8$
Substituting $y=8$ in equation (3), we get
$x=\frac{20-4\times 8}{3}$
$\Rightarrow x=\frac{20-32}{3}$
$\Rightarrow x=\frac{-12}{3}=-4$
Therefore, the value of $x+y$ is $-4+8=4$.