Question

Solve the following simultaneous equations by the method of equating coefficients.$\frac{x}{3}+\frac{y}{4}=4;\frac{x}{2}-\frac{y}{4}=1$

• A. $x=2,y=8$
• B. $x=6,y=8$
• C. $x=7,y=8$
• D. $x=1,y=8$

Answer 1

Answer: (B)

$x=6,y=8$

Given,

$\frac{x}{3}+\frac{y}{4}=4$ ....(i)

$\frac{x}{2}-\frac{y}{4}=1$ .....(ii)

Multiplying equation (i) by $12$ and equation (ii) by $4$. we get,

$4x+3y=48$ .......(iii)

$2x-y=4$ .....(iv)

Multiplying equation (iv) by $3$. we get,

$6x-3y=12$ .......(v)

Now, Adding (iii) and (v). we get,

$10x=60\Rightarrow x=6$

Substitute $x=6$ in equation (iii). we get,

$4\left(6\right)+3y=48$

$\Rightarrow 3y=24\Rightarrow y=8$

$\therefore x=6,y=8$