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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.
(i) $x – 3y – 3 = 0$
$3x – 9y – 2 = 0$
(ii) $2x + y = 5$
$3x + 2y = 8$
(iii) $3x – 5y = 20$
$6x – 10y = 40$
(iv) $x – 3y – 7 = 0$
$3x – 3y – 15 = 0$.
To do:
We have to solve the system of equations and determine if the system has a unique, infinite or no solution.
Solution:
(i) The given system of equations can be written as:
$x – 3y – 3 = 0$
$x-3y=3$.....(i)
$3x – 9y – 2 = 0$
$3x-9y=2$
The given system of equations is of the form:
$a_{1} x+b_{1} y=c_{1}\ and\ a_{2} x+b_{2} y=c_{2}$
Here,
$a_1 = 1, b_1=-3, c_1=3 \ and \ a_2=3, b_2=-9, c_2=2 $
We have,
$\frac{a_1}{a_2}=\frac{1}{3}$
$\frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3}$
$\frac{c_1}{c_2}=\frac{3}{2} $
Clearly,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}$
This implies the given system of equations has no solution.
(ii) The given system of equations can be written as:
$2x\ +\ y\ =\ 5$
$3x\ +\ 2y\ =\ 8$
The given system of equations is of the form:
$a_{1} x+b_{1} y=c_{1}\ and\ a_{2} x+b_{2} y=c_{2}$
Where, $a_1 = 2, b_1=1, c_1=5 \ and \ a_2=3, b_2=2, c_2=8 $
We have,
$\frac{a_1}{a_2}=\frac{2}{3}$
$\frac{b_1}{b_2}=\frac{1}{2}$
$\frac{c_1}{c_2}=\frac{5}{8}$
Clearly, $\frac{a_1}{a_2}≠\frac{b_1}{b_2}$
This implies,
The given system of equations has one solution.
 By cross multiplication method, we get,
$\frac{x}{1(8)-2(5)}=\frac{y}{5(3)-2(8)}=\frac{-1}{2(2)-3(1)}$
$\frac{x}{8-10}=\frac{y}{15-16}=\frac{-1}{4-3}$
$\frac{x}{-2}=\frac{-1}{1}$ and $\frac{y}{-1}=\frac{-1}{1}$
$x=-1(-2)$ and $y=-1(-1)$
$x=2$ and $y=1$
(iii) The given system of equations can be written as:
$3x\ –\ 5y\ =\ 20$
$6x\ –\ 10y\ =\ 40$
The given system of equations is of the form:
$a_{1} x+b_{1} y=c_{1}\ and\ a_{2} x+b_{2} y=c_{2}$
Where, $a_1 = 3, b_1=-5, c_1=20 \ and \ a_2=6, b_2=-10, c_2=40 $
We have,
$\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}$
$\frac{c_1}{c_2}=\frac{20}{40}=\frac{1}{2}$
Clearly, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
This implies the given system of equations has infinite solutions. 
(iv) The given system of equations can be written as:
$x\ -\ 3y\ =\ 7$
$3x\ -\ 3y\ =\ 15$
The given system of equations is of the form:
$a_{1} x+b_{1} y=c_{1}\ and\ a_{2} x+b_{2} y=c_{2}$
Where, $a_1 = 1, b_1=-3, c_1=7 \ and \ a_2=3, b_2=-3, c_2=15$
We have,
$\frac{a_1}{a_2}=\frac{1}{3}$
$\frac{b_1}{b_2}=\frac{-3}{-3}$
$\frac{c_1}{c_2}=\frac{7}{15}$
Clearly,
$\frac{a_1}{a_2}≠\frac{b_1}{b_2}$
This implies,
The given system of equations has one solution.
 By cross multiplication method, we get,
$\frac{x}{-3(-15)-(-3)(-7)}=\frac{y}{-7(3)-(-15)(1)}=\frac{1}{1(-3)-3(-3)}$
$\frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}$
$\frac{x}{24}=\frac{-1}{6}$ and $\frac{y}{-6}=\frac{1}{6}$
$x=\frac{24}{6}$ and $y=\frac{-6}{6}$
$x=4$ and $y=-1$
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