Question

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method. $x-3y-3=0$ and $3x-9y-2=0$.

Answer 1

Solution.

We have equation: $x-3y-3=0$ and $3x-9y-2=0$.

Here,

$$\begin{gathered} a_1=1,b_1=-3,c_1=-3 \\ {a}_2=3,b_2=-9,c_2=-2 \end{gathered}$$

Now,

$$\begin{gathered} \frac{a_1}{a_2}=\frac{1}{3}, \\ \frac{b_1}{b_2}=\frac{-3}{-9} \\ =\frac{1}{3} \\ \frac{c_1}{c_2}=\frac{-3}{-2} \\ =\frac{3}{2} \\ \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2} \end{gathered}$$

Since, the given set of lines are parallel to each other they will not intersect each other.

Hence, there is no solution for the given equations.