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. |
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