# Solvability of linear equations

In this chapter we will learn about the solvability of linear equations with solved examples.

Here for a given system of equation, we will check if the equations provides unique solution, multiple solution or no solution.

## Solvability of simultaneous equations

Let two equations are given to us;

\mathtt{a_{1} x\ +b_{1} y\ +\ c_{1} =0}\\\ \\ \mathtt{a_{2} x\ +b_{2} y\ +\ c_{2} =0}

Here, three conditions are possible;

(a) \mathtt{\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}}

The above condition tells that the linear equation intersect at one point. Hence, on solving we get only one solution.

(b) \mathtt{\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}}}

The above condition tells that both lines are coincident. Hence, there are infinite many solution.

(c) \mathtt{\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}}

The above condition tells that both the given lines are parallel to each other which do not intersect at any point. Hence, the equations have no solution.

I hope you understood the above three conditions. Let us solve some problems for further clarity.

## Linear equation solvability conditions – solved problems

Example 01
Find the solvability of given simultaneous equations.

2x + 3y – 7 = 0
3x + 2y – 8 = 0

Solution
Comparing \mathtt{\frac{a_{1}}{a_{2}} ,\ \frac{b_{1}}{b_{2}} \ and\ \frac{c_{1}}{c_{2}}} \\\ \\

\mathtt{\frac{2}{3} \neq \frac{7}{2} \neq \frac{7}{8}}

Since \mathtt{\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}} , the system of equation intersect at one point and offers unique solution.

The graphical representation of above equation is shown below.

Here both the lines intersect at point A (2, 1).

Hence, (2, 1) is the unique solution for given set of equation.

Example 02
Check the solvability of given equations.

x + 5y – 7 = 0
x + 5y -11 = 0

Solution
Comparing value of \mathtt{\frac{a_{1}}{a_{2}} ,\ \frac{b_{1}}{b_{2}} \ and\ \frac{c_{1}}{c_{2}}} \\\ \\

\mathtt{\frac{a_{1}}{a_{2}} =1\ }\\\ \\ \mathtt{\frac{b_{1}}{b_{2}} =\frac{5}{5} =1\ }\\\ \\ \mathtt{\ \frac{c_{1}}{c_{2}} =\frac{-7}{-11} =\frac{7}{11}}

Here, \mathtt{\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}}

Hence, the above system of equation is a parallel line which offers no solution.

Given below is the graph of above equations.

Example 03
Check the solvability of given equations.

x – 2y + 5 = 0
3x + 4y – 20 = 0

Solution
Finding value of \mathtt{\frac{a_{1}}{a_{2}} ,\ \frac{b_{1}}{b_{2}} \ and\ \frac{c_{1}}{c_{2}}}\\\ \\

\mathtt{\frac{a_{1}}{a_{2}} =\frac{1}{3}}\\\ \\ \mathtt{\frac{b_{1}}{b_{2}} =\frac{2}{4} =\frac{1}{2} \ }

Since \mathtt{\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}} , both the lines intersect at one point offering unique solution.

Below is the graphical representation of above equations.

Here the red point is the solution of given equation.

Example 04
Find solvability of below equations.

5x + 3y – 10 = 0
10x + 6y – 20 = 0

Solution

Finding values of \mathtt{\frac{a_{1}}{a_{2}} ,\ \frac{b_{1}}{b_{2}} \ and\ \frac{c_{1}}{c_{2}}}

\mathtt{\frac{a_{1}}{a_{2}} =\frac{5}{10} =\frac{1}{2}}\\\ \\ \mathtt{\frac{b_{1}}{b_{2}} =\frac{3}{6} =\frac{1}{2} \ }\\\ \\ \mathtt{\ \frac{c_{1}}{c_{2}} =\frac{-10}{-20} =\frac{1}{2}}

Here, \mathtt{\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}}}

So both the linear equations are coincident lines. Hence, the system offers infinite solutions.

Given below is the graphical representation.

Here both the equations are represented by same line.

You cannot copy content of this page