Question

Solve the following system of equations by using the method of elimination by equating the coefficients:

$\frac{x}{10}+\frac{y}{5}+1=15$ and $\frac{x}{8}+\frac{y}{6}=15$.

Answer 1

The given system of equations is
$\frac{x}{10}+\frac{y}{5}=14$

$\frac{x}{8}+\frac{y}{6}=15$

This system of equations can be re-written as
$x+2y=140$ ..(i)
$3x+4y=360$ .(ii)

Let us eliminate y from the equations (i) and (ii). The coefficients of y in the given equations are $2$ and $4$ respectively. The l.c.m of $2$ and $4$ is $4$.
$2x+4y=280$ ..(iii)
$3x+4y=360$ (iv)

Subtracting (iv) from (iii), we get
$-x=-80\Rightarrow x=80$

Putting $x=80$ in equation (i), we get
$80+2y=140\Rightarrow 2y=60\Rightarrow y=30$

Hence, the solution of the given system of equations is $x=80,y=30$.