| (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-T
B) i-T ii-T iii-F
C) i-F ii-T iii-T
D) i-F ii-F iii-T
Correct Answer: D
Solution :
(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.
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