Assertion (A): The pair of linear equations $px + 3y + 50 = 0$ and $2x + 6y + 118 = 0$ will have infinitely many solutions if $p = 1$.

Reason (R): If the pair of linear equations $px + 3y + 19 = 0$ a

Question

Assertion (A): The pair of linear equations $px + 3y + 50 = 0$ and $2x + 6y + 118 = 0$ will have infinitely many solutions if $p = 1$.

Reason (R): If the pair of linear equations $px + 3y + 19 = 0$ and $2x + 6y + 157 = 0$ has a unique solution, then $p 
eq 1$.

Accepted Answer

Step 1: For infinitely many solutions: $\frac{p}{2} = \frac{3}{6} = \frac{50}{118}$. $\frac{3}{6} = \frac{1}{2}$ but $\frac{50}{118} 
eq \frac{1}{2}$. So no value of $p$ gives infinitely many solutions. Assertion (A) is false.
For Reason (R): For unique solution, $\frac{p}{2} 
eq \frac{3}{6} = \frac{1}{2}$, so $p 
eq 1$. Reason (R) is true.
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

Question

Solution