Question

The equations $3x-4y=5$ and $12x-16y=20$ have

• A. Infinitely many solutions
• B. Exactly one common solution
• C. Exactly two common solutions
• D. No common solution.

Answer 1

The correct option is A Infinitely many solutions

The given linear equations are $3x-4y=5$ and $12x-16y=20$.

We know that, for two linear equations

$a_{1}x+b_1=c_1$ and $a_{2}x+b_{2}y=c_2$:

(a) If $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, the system is consistent and has infinitely many solutions.

(b) If $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$, the system has no solution and is inconsistent.

In the given equations, $3x-4y=5$ and $12x-16y=20$, we have:

$\frac{3}{12}=\frac{-4}{-16}=\frac{5}{20}=\frac{1}{4}$

Hence, the system is consistent and has infinitely many solutions.