SSC Sample Paper Mock Test-20 SSC CGL Tear-II Paper-1

If \[x+y=a+b\] and \[ax-by={{a}^{2}}-{{b}^{2}},\] then\[x\] and \[y\] are respectively

A) \[-a,b\]

B) \[a,b\]

C) \[\frac{1}{a},b\]

D) \[-a,-b\]

Correct Answer: B

Solution :

\[x+y=a+b\] ...(i)

and \[ax-by={{a}^{2}}-{{b}^{2}}\] ...(ii)

On multiplying Eq. (i) by b and adding, we get

\[\frac{\begin{align} & bx+by=ab+{{b}^{2}} \\ & ax-by={{a}^{2}}-{{b}^{2}} \\ \end{align}}{(a+b)x=a\,\,(a+b)}\]\[\Rightarrow \]\[x=a\]

Put the value of x in Eq. (i),\[a+y=a+b\]

\[\Rightarrow \] \[y=b\]

So, x = a and y = b is solution.