Question

Question 1 (iv)
Which of the following pairs of linear equations has a unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.
x - 3y - 7 = 0 ; 3x - 3y - 15= 0

Answer 1

x - 3y - 7 = 0
3x - 3y - 15 = 0

Compare the given equations with

$a_{1}x+b_{1}y+c_1=0and{a}_{2}x+b_{2}y+c_2=0$.

$\frac{a_1}{a_2}=\frac{1}{3}$
$\frac{b_1}{b_2}=\frac{-3}{-3}=1$
$\frac{c_1}{c_2}=\frac{-7}{-15}=\frac{7}{15}$
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
So, they will intersect each other at a unique point.
Therefore, there will be a unique solution for these equations.
By cross-multiplication,
$\frac{x}{45-\left(21\right)}=\frac{y}{-21-(-15)}=\frac{1}{-3-(-9)}$
$\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$
$\frac{x}{24}=\frac{1}{6}and\frac{y}{-6}=\frac{1}{6}$
x = 4 and y = -1
$\therefore$ x = 4, y = -1