https://gateoverflow.in/321213/isi2015-dcg-11

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Let two systems of linear equations be defined as follows:

$\begin{array}{lll} & x+y & =1 \\ P: & 3x+3y & =3 \\ & 5x+5y & =5 \end{array}$ and $\begin{array}{lll} & x+y & =3 \\ Q: & 2x+2y & =4 \\ & 5x+5y & =12 \end{array}$. Then,

- $P$ and $Q$ are inconsistent
- $P$ and $Q$ are consistent
- $P$ is consistent but $Q$ is inconsistent
- None of the above

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P

$x+y = 1$ ....$eq(i)$

$3x+3y = 3$....$eq(ii)$

$5x+5y = 5$...$eq(iii)$

As we can see

$2*eq(i) + eq(ii) = eq(iii)$ this rule is satisfied by both the LHS and RHS parts of the equations.

$\implies$ they can have infinite solution

$\implies$ They are having consistent solutions.

Q

$x+y = 3$ ....$eq(i)$

$2x+2y = 4$....$eq(ii)$

$5x+5y = 12$...$eq(iii)$

As we can see

$eq(i) + 2*eq(ii) = eq(iii)$ this rule is satisfied by the LHS part and not by the RHS parts of the equations.(i.e. $3+4*2\neq 12$)

$\implies$ They have no solution

$\implies$ They are having inconsistent solution.

$\therefore$ Option $C.$ is the correct answer

$x+y = 1$ ....$eq(i)$

$3x+3y = 3$....$eq(ii)$

$5x+5y = 5$...$eq(iii)$

As we can see

$2*eq(i) + eq(ii) = eq(iii)$ this rule is satisfied by both the LHS and RHS parts of the equations.

$\implies$ they can have infinite solution

$\implies$ They are having consistent solutions.

Q

$x+y = 3$ ....$eq(i)$

$2x+2y = 4$....$eq(ii)$

$5x+5y = 12$...$eq(iii)$

As we can see

$eq(i) + 2*eq(ii) = eq(iii)$ this rule is satisfied by the LHS part and not by the RHS parts of the equations.(i.e. $3+4*2\neq 12$)

$\implies$ They have no solution

$\implies$ They are having inconsistent solution.

$\therefore$ Option $C.$ is the correct answer

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