A) \[\frac{-11}{10},\frac{9}{10}\]
B) \[\frac{-9}{10},\frac{11}{10}\]
C) \[\frac{-33}{2},6\]
D) \[-6,\frac{33}{5}\]
Correct Answer: C
Solution :
| [c] The pair of equations are \[5x-3y-9=0\] |
| and \[(a-b)\,x-(a+b-3)\,y-(a-4b)=0\] |
| Here, \[{{a}_{1}}=5,\,\,{{b}_{1}}=-3,\,{{c}_{1}}=-9,\,\,{{a}_{2}}=a-b,\] |
| \[{{b}_{2}}=-(a+b-3),\,\,{{c}_{2}}=-(a-4b)\] |
| For the equations to be coincident, we have |
| \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\frac{5}{a-b}=\frac{-3}{-(a+b-3)}=\frac{-9}{-(a-4b)}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\frac{5}{a-b}=\frac{3}{a+b-3}\] and \[\frac{3}{a+b-3}=\frac{9}{a-4b}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,5a+5b-15=3a-3b\] |
| and \[3a-12b=9a+9b-27\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,2a+8b-15=0\] ..(1) |
| and \[2a+7b-9=0\] (2) |
| Subtracting eq. (2) from eq. (1). we get |
| \[b-6=0\,\,\,\Rightarrow \,\,b=6\] |
| Substituting this value of b in eq. (1). we get |
| \[2a+8\times 6-15=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,2a+48-15=0\,\,\,\Rightarrow \,\,2a=-33\,\,\,\Rightarrow \,\,a=\frac{-33}{2}\] |
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