• question_answer Match the following. 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)

(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