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. (iv) $x–3y–7=0$ and $3x–3y15=0$.

Answer 1

Solution.

Given $x–3y–7=0$ and $3x–3y15=0$.

$$\begin{gathered} \frac{a_1}{a_2}=\frac{1}{3} \\ \frac{b_1}{b_2}=\frac{-3}{-3} \\ =1 \\ \frac{c_1}{c_2}=\frac{-7}{-15} \\ \therefore \frac{a_1}{a_2}\ne \frac{b_1}{b_2} \end{gathered}$$

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

By cross multiplication,

$$\begin{gathered} \frac{x}{(45-21)}=\frac{y}{(-21+15)}=\frac{1}{(-3+9)} \\ \Rightarrow \frac{x}{24}=\frac{y}{-6}=\frac{1}{6} \\ \Rightarrow \frac{x}{24}=\frac{1}{-6}\mathrm{and}\frac{y}{-6}=\frac{1}{6} \\ \therefore x=4\mathrm{and}y=1 \end{gathered}$$

Hence, given equations have a unique solution and unique solution is $x=4,y=1$.