For the following system of equation determine the value of k for which the given system of equation has a unique solution.

$$2x-3y=1$$

$$kx+5y=7$$

#### Solution

The given system of equations is

$$2x-3y-1=0$$

$$kx+5y-7=0$$

It is of the form $$a_1x+b_1y+c_1=0$$

$$a_2x+b_2y+c_1=0$$

where, $$a_1=2, b_1=-3, c_1=-1$$ and $$a_2=k, b_2=5, c_2=-7$$

For a unique solution, we must have

$$\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}$$ i.e., $$\dfrac{2}{k}\neq \dfrac{-3}{5}\Rightarrow k\neq \dfrac{-10}{3}$$

So, the given system of equations is consistent with a unique solution for all values of k other than $$-10/3$$.

$$2x-3y=1$$

$$kx+5y=7$$

$$2x-3y-1=0$$

$$kx+5y-7=0$$

It is of the form $$a_1x+b_1y+c_1=0$$

$$a_2x+b_2y+c_1=0$$

where, $$a_1=2, b_1=-3, c_1=-1$$ and $$a_2=k, b_2=5, c_2=-7$$

For a unique solution, we must have

$$\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}$$ i.e., $$\dfrac{2}{k}\neq \dfrac{-3}{5}\Rightarrow k\neq \dfrac{-10}{3}$$

So, the given system of equations is consistent with a unique solution for all values of k other than $$-10/3$$.