A) \[x=a\]and \[y=b\]
B) \[x=-\,\,a\]and \[y=-b\]
C) \[x=b\]and \[y=a\]
D) \[x=-\,\,b\]and \[y=-\,\,a\]
Correct Answer: A
Solution :
| Here, in Eq. (i) |
| \[{{a}_{1}}=a,\]\[{{b}_{1}}=-\,b,\]\[{{c}_{1}}=-\,\,({{a}^{2}}-{{b}^{2}})\]and |
| In Eq. (ii) \[{{a}_{2}}=1,\]\[{{b}_{2}}=1,\]\[{{c}_{2}}=-\,(a+b)\] |
| So, \[x=\frac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}=\frac{(b)(a+b)+(1)({{a}^{2}}-{{b}^{2}})}{a\,\,(1)-(1)(-b)}\] |
| \[x=\frac{ab+{{b}^{2}}+{{a}^{2}}-{{b}^{2}}}{a+b}=\frac{a\,\,(a+b)}{(a+b)}=a\] |
| and \[y=\frac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\] |
| \[y=\frac{-\,\,({{a}^{2}}-{{b}^{2}})(1)+(a+b)(a)}{a+b}\] \[=\frac{b\,\,(a+b)}{(a+b)}=b\] |
| So, \[x=a\]and \[y=b\]is solution. |
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