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;