Question

Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions?
\[\left( {\text{i}} \right)\]\[x - 3y - 3 = 0;{\text{ }}3x - 9y - 2 = 0\]
\[\left( {{\text{ii}}} \right)\] \[2x + y = 5;{\text{ }}3x + 2y = 8\]
\[\left( {{\text{iii}}} \right)\]\[3x - 5y = 20;{\text{ }}6x - 10y = 40\]
\[\left( {{\text{iv}}} \right)\]\[x - 3y - 7 = 0;{\text{ }}3x - 3y - 15 = 0\]

Answer 1

Hint: The 3 conditions for a pair of linear equations are as follows:
1. A system of linear equations \[ax + by + c = 0\] and \[dx + ey + f = 0\] will have a unique solution if the 2 lines represented by these equations intersect at a point. In simple words their slopes shouldn't be equal. In mathematical form \[\dfrac{a}{d} \ne \dfrac{b}{e}\]
2. A system of linear equations \[ax + by + c = 0\] and \[dx + ey + f = 0\] will have no solution if the 2 lines represented by these equations are parallel i.e., they never intersect. Their slopes will be equal i.e., \[\] \[\dfrac{a}{d} = \dfrac{b}{e} \ne \dfrac{c}{f}\]
3. A system of linear equations \[ax + by + c = 0\] and \[dx + ey + f = 0\] will have infinitely many solutions if the 2 lines represented by these equations are coincident. Here \[\dfrac{a}{d} = \dfrac{b}{e} = \dfrac{c}{f}\]
We compare each part of the question with the standard equation \[ax + by + c = 0\] and \[dx + ey + f = 0\], then we will mathematically find what solution it gives.

Complete step-by-step answer:
\[\left( {\text{i}} \right)\] It is given that the linear equations are, \[x - 3y - 3 = 0;{\text{ }}3x - 9y - 2 = 0\]
Here, \[a = 1,{\text{ }}b = - 3,{\text{ }}c = - 3;{\text{ }}d = 3,{\text{ }}e = - 9,{\text{ }}f = - 2\]
We have to find out some condition mentioned in the hint part so we get,
 \[\dfrac{a}{d} = \dfrac{1}{3}\]
\[\dfrac{b}{e} = \dfrac{{ - 3}}{{ - 9}} = \dfrac{1}{3}\]
\[\dfrac{c}{f} = \dfrac{{ - 3}}{{ - 2}} = \dfrac{3}{2}\]
Therefore, \[\dfrac{a}{d} = \dfrac{d}{e} \ne \dfrac{c}{f}\]
Hence the given pair of linear equations does not have a unique solution.
\[\left( {{\text{ii}}} \right)\]It is given that the linear equations are \[2x + y = 5;{\text{ }}3x + 2y = 8\]
Here we have to convert this equation to the standard form.
\[2x + y - 5 = 0\;;{\text{ }}3x + 2y - 8 = 0\]
Here \[a = 2,{\text{ }}b = 1,{\text{ }}c = - 5{\text{ }};{\text{ }}d = 3,{\text{ }}e = 2,{\text{ }}c = - 8\]
Now we have to find the condition mentioned in the hint part, we get
\[\dfrac{a}{d} = \dfrac{2}{3}\]
\[\dfrac{b}{e} = \dfrac{1}{2}\]
Therefore, \[\dfrac{a}{d} \ne \dfrac{b}{e}\]
Hence the given pair of linear equations has a unique solution.
\[\left( {{\text{iii}}} \right)\]It is given that the linear equations are \[3x - 5y = 20;{\text{ }}6x - 10y = 40\]
We have to convert this equation to the standard form.
\[3x - 5y - 20 = 0\;;{\text{ }}6x - 10y - 40 = 0\]
Here \[a = 3,{\text{ }}b = - 5,{\text{ }}c = - 20{\text{ }};{\text{ }}d = 6,{\text{ }}e = - 10,{\text{ }}c = - 40\]
Now we have to find the condition mentioned in the hint part, we get
\[\dfrac{a}{d} = \dfrac{3}{6} = \dfrac{1}{2}\]
\[\dfrac{b}{e} = \dfrac{{ - 5}}{{ - 10}} = \dfrac{1}{2}\]
\[\dfrac{c}{f} = \dfrac{{ - 20}}{{ - 40}} = \dfrac{1}{2}\]
Therefore, \[\dfrac{a}{d} = \dfrac{b}{e} = \dfrac{c}{f}\]
Hence the given pair of linear equations has infinitely many solutions.
\[\left( {{\text{iv}}} \right)\]It is given that the linear equations are \[x - 3y - 7 = 0;{\text{ }}3x - 3y - 15 = 0\]
Here \[a = 1,{\text{ }}b = - 3,{\text{ }}c = - 7{\text{ }};{\text{ }}d = 3,{\text{ }}e = - 3,{\text{ }}f = - 15\]
Now we have to find the condition mentioned in the hint part, we get
\[\dfrac{a}{d} = \dfrac{1}{3}\]
\[\dfrac{b}{e} = \dfrac{{ - 3}}{{ - 3}} = \dfrac{1}{1}\]
Therefore, \[\dfrac{a}{d} \ne \dfrac{b}{e}\]
Hence the given pair of linear equations has a unique solution.

Note: In these types of problems the system of linear equations must be converted to the standard form.
Also with the help of imagining lines represented from these equations we know what type of solution it may give.