A) 1, -1
B) -1, 1
C) 1, 0
D) 0, 2
Correct Answer: A
Solution :
| The given system can be rewritten as |
| \[ax+by=a-b\] ...(i) |
| and \[bx-ay=a+b\] ...(ii) |
| From Eq. (i), we get |
| \[by=a-b-ax\] |
| \[\Rightarrow \,\,\,\,y=\frac{a-b-ax}{b}\] (iii) |
| On substituting the value of y in Eq. (ii), we get |
| \[bx-a\left[ \frac{a-b-ax}{b} \right]=a+b\] |
| \[\Rightarrow \,\,\,{{b}^{2}}x-a\left( a-b-ax \right)=b\left( a+b \right)\] |
| [multiplying both sides by b] |
| \[\Rightarrow \,\,\,{{b}^{2}}x-{{a}^{2}}+ab+{{a}^{2}}x=ab+{{b}^{2}}\] |
| \[\Rightarrow \,\,\,\left( {{b}^{2}}+{{a}^{2}} \right)x=ab+{{b}^{2}}+{{b}^{2}}-ab\] |
| \[\Rightarrow \,\,\,x=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=1\] |
| On substituting x = 1 in Eq. (iii), we get |
| \[y=\frac{a-b-a}{b}\,\,\Rightarrow y=\frac{-b}{b}=-1\] |
| Hence, solution of the given system is x = 1 and y = - 1. |
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