# Graphing simultaneous equations

In this chapter we will learn to graph simultaneous linear equations and try to find the common solutions using graph.

Before plotting the equations, let us first revise the basics.

## What are simultaneous linear equations ?

Two or more linear equations with common variable are called simultaneous equations.

Here we will discuss simultaneous linear equations with two variables.

Two given simultaneous equations can have following relation;

(a) Unique solution
(b) Infinite solution
(c) No solution

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}

Consider the below three conditions;

(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.

## Plotting simultaneous equations

To plot the simultaneous equation, follow the below steps;

(a) Express both the equation in form of y = mx + c

(b) For both the equation, find two points using hit and trial and join them using straight line.

(c) Now observe both the lines.

If the lines intersect each other, the simultaneous equation has unique solution.

If the line superimpose each other then we have infinite solution.

If the lines are parallel then we have no solution.

Using the above three points we can solve any given simultaneous equations.

## Graphing simultaneous equations – Solved examples

Example 01
Graph the following simultaneous equation.

x – 3y – 7 = 0
3x – 3y – 15 = 0

Solution
Comparing \mathtt{\frac{a_{1}}{a_{2}} \ \&\ \frac{b_{1}}{b_{2}}}

Since \mathtt{\frac{1}{3} \ \neq \frac{-3}{-3}} , the equations offers unique solution.

Now plotting both the equations on graph.

(a) x – 3y – 7 = 0

Representing equation in form of y = mx + c;

\mathtt{x-3y-7=0}\\\ \\ \mathtt{3y\ =\ x-7}\\\ \\ \mathtt{y=\frac{x-7}{3}}

Finding first point.
Put x = 0

\mathtt{y=\frac{0-7}{3}}\\\ \\ \mathtt{y\ =\ \frac{-7}{3} =-2.33}

Here (0, -2.33) is the first point.

Finding second point
Put x = 1

\mathtt{y=\frac{1-7}{3}}\\\ \\ \mathtt{y\ =\ \frac{-6}{3} =-2}

We got (1, -2) as second point.

Plot the points (0, -2.33) and (1, -2) on graph and then join with the straight line.

Given below is the graph of equation x – 3y – 7 = 0

(b) Now let’s plot the second equation, 3x – 3y – 15 = 0 .

Express in format y = mx + c

\mathtt{3x-3y-15=0}\\\ \\ \mathtt{3y=\ 3x-15}\\\ \\ \mathtt{y\ =\ x-5}

Now find the two points using hit and trial method.

First Point
Put x = 0

\mathtt{y\ =\ 0-5}\\\ \\ \mathtt{y\ =\ -5}

We get the first point (0, -5)

Finding second point
Put x = 1

\mathtt{y\ =\ 1-5}\\\ \\ \mathtt{y\ =\ -4}

We got (1, -4) as second point.

Now plot (0,-5) and (1, -4) on graph and then join them using straight line.

Now plotting both the equations together, we get the following graph.

The point of intersection of both the lines is the solution of both the equations.

Example 02
Plot the below simultaneous equation in graph

x + 2y – 4 = 0
2x + 4y – 12 = 0

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

Comparing the above equation, we get;

\mathtt{\frac{1}{2} =\ \frac{2}{4} \ \neq \ \frac{-4}{12}}

Hence, the given equations are parallel to each other.

(a) Graphing the equation x + 2y – 4 = 0

Represent equation in form of y = mx + c

\mathtt{x+\ 2y\ -\ 4\ =\ 0}\\\ \\ \mathtt{2y=-x+4}\\\ \\ \mathtt{y\ =\frac{-x+4}{2} \ }

Finding two points using hit and trial.

First Point
Put x = 0

\mathtt{y=\frac{0+4}{2}}\\\ \\ \mathtt{y\ =\ 2}

First point is (0, 2)

Second Point
Put x = 1 in equation.

\mathtt{y=\frac{-1+4}{2}}\\\ \\ \mathtt{y\ =\ 1.5}

Second point is (1, 1.5).

Plot the two point in the graph and join them with straight line.

(b) Now graph the equation 2x + 4y – 12 = 0

Express the equation form of y = mx + c

\mathtt{4y=\ 12-2x}\\\ \\ \mathtt{y=\frac{12-2x}{4}}

Finding two points using hit and trial.

First Point
Put x = 0

\mathtt{y\ =\frac{12-0}{4}}\\\ \\ \mathtt{y\ =\ 3}

The first point is (0, 3)

Second Point
Put x= 2

\mathtt{y\ =\frac{12-4}{4}}\\\ \\ \mathtt{y\ =\ 2}

The second point is (2, 2)

Plotting the two points and joining them with straight line.

Now plotting the two graph together, we get;