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.