Question

The solution of
$217x+131y=913$
$131x+217y=827$ is

• A. $x=2,y=3$
• B. $x=3,y=2$
• C. $x=2,y=2$
• D. $x=3,y=3$

Answer 1

Answer: (B)

$x=3,y=2$

The given equations are

$217x+131y=913$ and $131x+217y=827$

According to the given passage, we have

$a=217,b=131,c=913$ and $d=827$

$\therefore x=\frac{1}{2}(\frac{c+d}{a+b}+\frac{c-d}{a-b})=\frac{1}{2}(\frac{913+827}{217+131}+\frac{913-827}{217-131})$

$=\frac{1}{2}(\frac{1740}{348}+\frac{86}{86})=\frac{1}{2}(5+1)=3$

and $y=\frac{1}{2}(\frac{c+d}{a+b}-\frac{c-d}{a-b})=\frac{1}{2}(\frac{913+827}{217+131}-\frac{913-827}{217-131})$

$=\frac{1}{2}(\frac{1740}{348}-\frac{86}{86})=\frac{1}{2}(5-1)=2$

$\therefore (x,y)=(3,2)$