Question

Which of the following equations have a unique solution or infinitely many solutions.

(i) $x-3y-3=0;3x-9y-2=0$

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

• A. option (ii) is correct.
• B. option (i) is correct.

Answer 1

The correct option is A

option (ii) is correct.

Explanation for the correct option:

Option (ii): Determine whether the equations have a unique solution or infinitely many solutions.

Consider $2x+y=5;3x+2y=8$.

The conditions for the pair of equations $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ are as follows:

1. Infinite solutions: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
2. No solutions: $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
3. Unique solutions: $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Compare the given equations with pair of general equations $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$.

$a_1=2,b_1=1,c_1=-5,a_2=3,b_2=2,c_2=-8$

$\begin{array}{rcl}\frac{2}{3}& =& \frac{1}{2}=\frac{-5}{-8}\\ & \Rightarrow & \frac{2}{3}\ne \frac{1}{2}\ne \frac{5}{8}\end{array}$

Thus, $2x+y=5;3x+2y=8$ has uniques solutions.

Hence, the given equations have a unique solution.

Explanation for the incorrect options:

Option (i): Consider $x-3y-3=0;3x-9y-2=0$.

Compare the given equations with pair of general equations $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$.

$a_1=1,b_1=-3,c_1=-3,a_2=3,b_2=-9,c_2=-2$

$\begin{array}{rcl}\frac{1}{3}& =& \frac{-3}{-9}=\frac{-3}{-2}\\ & \Rightarrow & \frac{1}{3}=\frac{1}{3}\ne \frac{3}{2}\end{array}$

Thus, $x-3y-3=0;3x-9y-2=0$ has no solutions.

Therefore, option (i) is incorrect.

Hence, option (ii) is correct.