Question

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions.
(i) $x-3y-3=0;3x-9y-2=0$

(ii) $2x+y=5;3x+2y=8$

(iii) $3x-5y=20;6x-10y=40$

(iv) $x-3y-7=0;3x-3y-15=0$

Answer 1

(i) $x–3y–3=0$

$3x–9y–2=0$

Comparing these with pair of equations of the form $a_{1}x+b_{1}y+c_1=0$

and $a_{2}x+b_{2}y+c_2=0,$

$\frac{a_1}{a_2}=\frac{1}{3},\frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3},$

$\frac{c_1}{c_2}=\frac{3}{2}$

$\because \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$, the lines have no solution.

(ii) $2x+y=5$

$3x+2y=8$

$\frac{a_1}{a_2}=\frac{2}{3},\frac{b_1}{b_2}=\frac{1}{2},\frac{c_1}{c_2}=\frac{5}{8}$

$\because \frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, the lines have unique solution.

(iii) $3x-5y=20$

$6x-10y=40$

$\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2},\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2},\frac{c_1}{c_2}=\frac{20}{40}=\frac{1}{2}$

$\because \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, the lines have infinitely many solution.

(iv) $x-3y-7=0;3x-3y-15=0$

$\frac{a_1}{a_2}=\frac{1}{3},\frac{b_1}{b_2}=\frac{-3}{-3}=1,\frac{c_1}{c_2}=\frac{7}{15}$

$\because \frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, the lines have unique solution.