Question

Assertion :$kx+2y=5$ and $3x+y=1$ have a unique solution if $k=1$ Reason: $x+2y=3$ and $5x+ky+7=0$ have a unique solution if $k\ne 1$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

The pair of linear equations represented by $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ have a unique solution, if

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

The given linear equations $kx+2y=5$ and $3x+y=1$ can be written as $kx+2y-5=0$ and $3x+y-1=0$

Substitute $k=1$ we have $a_1=1,b_1=2,a_2=3,b_2=1$

Now, $\frac{a_1}{a_2}=\frac{1}{3}\ne \frac{2}{1}=\frac{b_1}{b_2}$

Thus, the lines given in assertion have a unique solution.

For the second pair,

The given linear equations $x+2y=3$ and $5x+ky+7=1$ can be written as $x+2y-3=0$ and $5x+ky+7=0$

We have $a_1=1,b_1=2,a_2=5,b_2=k$

Now, $\frac{a_1}{a_2}=\frac{1}{5}$

$k\ne 1$, so it may or may not take the value so that $\frac{b_1}{b_2}=\frac{1}{5}=\frac{a_1}{a_2}$

Thus, we are not sure about the second pair of linear equations that whether it has unique solution or not.

Hence, assertion is correct but reason is incorrect.