• # question_answer Read the statements carefully and state 'T' for true and 'F' for false.            (i) The pair of linear equations $x+2y=5$and $7x+3y=13$ has unique solution$x=2.\text{ }y=1$.                       (ii) $\sqrt{2}x+\sqrt{3}y=0,$ $\sqrt{3}x-\sqrt{8}y=0$ has no solution. (iii) The values of p and q for which the following system of equations $2x-y=5,$ $(p+q)x+(2p-q)y=15$has infinite number of solutions, is $p=1$and$q=5$. A)  i-T                   ii-F        iii-TB)  i-T                   ii-T       iii-FC)  i-F                   ii-T       iii-TD)  i-F                   ii-F       iii-T

(i) Given equations are $x+2y=5$  ...(1) and  $7x+3y=13$        ...(2) Multiplying (1) by 7 and then subtracting from (2), we get $7x+3y-7x-14y=13-35$ $x=1$and $y=2$ Here, $\frac{1}{7}\ne \frac{2}{3}\ne \frac{5}{13},$ a unique solution exist. (ii) Given equations are $\sqrt{2}x+\sqrt{3}y=0$  and $\sqrt{3}x-\sqrt{8}y=0$ $\frac{\sqrt{2}}{\sqrt{3}}\ne \frac{\sqrt{3}}{-2\sqrt{2}}$ $\therefore$ Given equations have a unique solution (iii) Given equations are $2x-y=5$      .....(1) and $(p+q)x+(2p-q)y=15$            ?..(2) Putting $p=1$and $q=5$in (2), we get $6x-3y=15$  or  $2x-y=5$              .....(3) From (1) and (3), we have $\frac{2}{2}=\frac{-1}{-1}=\frac{5}{5}$  Hence, infinitely many solutions exist.