Question

The solution of
$37x+41y=70$
$41x+37y=86$ is

• A. $x=3,y=1$
• B. $x=3,y=-1$
• C. $x=-3,y=1$
• D. $x=1,y=3$

Answer 1

The correct option is B $x=3,y=-1$

The given equations are

$37x+41y=70$ and $41x+37y=86$

According to the given passage

$a=37,b=41,c=70$ and $d=86$

$\therefore x=\frac{1}{2}(\frac{c+d}{a+b}+\frac{c-d}{a-b})=\frac{1}{2}(\frac{70+86}{37+41}+\frac{70-86}{37-41})$

$=\frac{1}{2}(\frac{156}{78}+\frac{16}{4})=\frac{1}{2}(2+4)=3$

and $y=\frac{1}{2}(\frac{c+d}{a+b}-\frac{c-d}{a-b})=\frac{1}{2}(\frac{70+86}{37+41}-\frac{70-86}{37-41})$

$=\frac{1}{2}(\frac{156}{78}-\frac{16}{4})=\frac{1}{2}(2-4)=-1$

$\therefore (x,y)=(3,-1)$