Question

Solve the following simultaneous equations:

$\frac{x}{3}+\frac{y}{4}=4;\frac{5x}{6}-\frac{y}{8}=4$

• A. $x=5,y=8$
• B. $x=2,y=8$
• C. $x=1,y=8$
• D. $x=6,y=8$

Answer 1

Answer: (D)

$x=6,y=8$

$\frac{x}{3}+\frac{y}{4}=4\Rightarrow 4x+3y=48$ ....(i)

$\frac{5x}{6}-\frac{y}{8}=4\Rightarrow 20x-3y=96$ ....(ii)

Since, coefficients of $y$ are equal in both equations but with opposite sign.

So on adding (i) and (ii). we get,

$4x+3y+20x-3y=48+96$

$\Rightarrow 24x=144$

$\Rightarrow x=6$

On substituting $x=6$ in (i). we get,

$4\left(6\right)+3y=48\Rightarrow 3y=24\Rightarrow y=8$

$\therefore x=6,y=8$