Question

For what value of $k$, the equations $5x-2y+8=0$ & $15x+ky=-24$ have an unique solution ?

Answer 1

Step 1: Estimate the constants of the equations

The given equations are,

$5x-2y+8=0$

$15x+ky=-24$

The equations are of the form $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$

Here, $a_1=5,b_1=-2,c_1=8$

$a_2=15,b_2=k,c_2=24$

Step 2: Calculate the value of $k$

Here, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ i.e., the given system of equations has a unique solution.

$\therefore \frac{5}{15}\ne \frac{-2}{k}$

$\Rightarrow$ $5k\ne -30$

$\Rightarrow$ $k\ne -\frac{30}{5}$

$\Rightarrow$ $k\ne -6$

Thus, $k$ can have any real value except $-6$.