| System of equations | Solutions |
| (P) \[2x-3y+15=0\] \[3x-5=0\] | (i) \[x=3,\text{ }y=5\] |
| (Q) \[2x-y=1\] \[4x+3y=27\] | (ii) \[x=1,y=4\] |
| (R) \[x+2y-3=0\] \[3x-2y+7=0\] | (iii) \[x=\frac{5}{3},y=\frac{55}{9}\] |
| (S) \[4x+\frac{y}{3}=\frac{16}{3}\] \[\frac{x}{2}+\frac{2y}{4}=\frac{5}{2}\] | (iv) \[x=-1,y=2\] |
A) (P) \[\to \] (iii); (Q) \[\to \] (iv); (R) \[\to \] (i); (S) \[\to \] (ii)
B) (P) \[\to \] (iii); (Q) \[\to \] (i); (R) \[\to \] (iv); (S) \[\to \] (ii)
C) (P) \[\to \] (iii); (Q) \[\to \] (ii); (R) \[\to \] (iv); (S) \[\to \] (i)
D) (P) \[\to \] (ii); (Q) \[\to \] (i); (R) \[\to \] (iv); (S) \[\to \] (iii)
Correct Answer: B
Solution :
(P) We have, \[2x-3y=-15\] .....(1) and \[3x-5=0\] \[\Rightarrow \] \[x=\frac{5}{3}\] From (1), \[2\left( \frac{5}{3} \right)-3y=-15\] \[\Rightarrow \] \[3y=\frac{10}{3}+15\,\,\Rightarrow y=\frac{55}{9}\] (Q) We have, \[2x-y=1\] ?.(1) and \[4x+3y=27\] ?.(2) Multiplying (1) by 2 and then subtracting from (2), we get, \[y=5\]and \[x=3\] (R) We have, \[x+2y=3\] ?.(1) and \[3x-2y=-7\] ?...(2) Multiplying (1) by 3 and then subtracting from (2), we get\[y=2\]and \[x=-1\] (S) We have,\[4x+\frac{y}{3}=\frac{16}{3}\,\,\Rightarrow \,\,12x+y=16\] ...(1) and \[\frac{x}{2}+\frac{y}{2}=\frac{5}{2}\,\,\Rightarrow \,\,x+y=5\] ...(2) Subtracting (2) from (1), we get x= 1 and y = 4
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