Question

Given that $kx-3y=4$and $4x-5y=7$
In the system of equations above, $k$ is a constant and $x$ and $y$ are variables. For what value of $k$ will the system of equations have no solution?

• A. $\frac{12}{5}$
• B. $\frac{16}{7}$
• C. $-\frac{16}{7}$
• D. $-\frac{12}{5}$

Answer 1

The correct option is A $\frac{12}{5}$

Let $kx-3y=4$ ..(1)

and $4x-5y=7$ ..(2)

Multiplying equation (1) by $5$ and equation (2) by $3$, we have

$5kx-15y=20$ ..(3)

$12x-15y=21$ ..(4)

Since the coefficient of $y$ is same in equations (3) and (4), and the terms on right hand side of equality are different, the coefficients of $x$ cannot be the same.

$\therefore 5k\ne 12$

$\therefore k\ne \frac{12}{5}$