Question

Solve the following pair of simultaneous equations:

$\frac{x-1}{2}+\frac{y+1}{5}=4\frac{1}{5};\frac{x+y}{3}=y-1$

• A. $x=7,y=5$
• B. $x=2,y=4$
• C. $x=0,y=6$
• D. $x=5,y=3$

Answer 1

The correct option is A $x=7,y=5$

$\frac{x-1}{2}+\frac{y+1}{5}=4\frac{1}{5}\Rightarrow \frac{5x-5+2y+2}{10}=\frac{21}{5}\Rightarrow 5x+2y=45$ ...(i)

$\frac{x+y}{3}=y-1\Rightarrow x+y=3y-3\Rightarrow x-2y=-3$ ...(ii)

On multiplying (i) by 1 and (ii) by 5, we get
$5x+2y=45$
$\underset{–}{\underset{-}{5}x\underset{+}{-}10y=\underset{+}{-}15}$
$12y=60$

$\therefore y=5$

On putting $y=5$ in (ii), we get
$x-2\left(5\right)=-3\Rightarrow x=7$

$\therefore x=7,y=5$