- What is Linear equations
- Linear equations Solutions
- Graphical Representation of Linear equation in one and two variable
- Steps to Draw the Given line on Cartesian plane

$ax+b=0$

a and b are constant

$ax+by+c=0$

a,b,c are constants

So, $ax^2 +b=0$ is not a linear equation

Similary $ \frac {a}{x}+b=0$ is not a linear equation

Linear equations are straight lines when plotted on Cartesian plane

S.no |
Type of equation |
Mathematical representation |
Solutions |

1 |
Linear equation in one Variable |
$ax+b=0 \; , \; a \neq 0 $ a and b are real number |
One solution |

2 |
Linear equation in two Variables |
$ax+by+c=0 \; , \; a \neq 0 \; and \; b \neq 0$ i,e a and b both should not be zero. This condition is often represented as $a^2 + b^2 \ne 0$ a, b and c are real number |
Infinite solution possible |

3 |
Linear equation in three Variables |
$ax+by+cz+d=0 \; , \; a \neq 0 \;, \; b \neq 0 \; and \; c \neq 0$ i,e a , b and c all should not be zero. This condition is often represented as $a^2 + b^2 +c^2 \ne 0$ a, b, c, d are real number |
Infinite solution possible |

1. Find four different solutions of the equation $x + 2y = 6$.

**Solution**

We can get many solutions of the linear equation by just picking a value of your choice for x

Here We can find the solutions by putting x=0,1,2,3

For x= 0

$0 +2y=6$ or $2y=6$ or y=3

So (0,3) is a solution

For x= 1

$1 +2y=6$ or $2y=5$ or y=2.5

So (1,2.5) is a solution

For x= 2

$2 +2y=6$ or $2y=4$ or y=2

So (2,2) is a solution

For x= 3

$3 +2y=6$ or $2y=3$ or y=1.5

So (3,1.5) is a solution

Hence the four solutions are

(0,3) , (1,2.5) ,(2,2) and (3,1.5)

We can get many solutions of the linear equation by just picking a value of your choice for x

Here We can find the solutions by putting x=0,1,2,3

For x= 0

$0 +2y=6$ or $2y=6$ or y=3

So (0,3) is a solution

For x= 1

$1 +2y=6$ or $2y=5$ or y=2.5

So (1,2.5) is a solution

For x= 2

$2 +2y=6$ or $2y=4$ or y=2

So (2,2) is a solution

For x= 3

$3 +2y=6$ or $2y=3$ or y=1.5

So (3,1.5) is a solution

Hence the four solutions are

(0,3) , (1,2.5) ,(2,2) and (3,1.5)

- Linear equation in two variables is represented by straight line the Cartesian plane.
- Every point on the line is the solution of the equation.
- Infact Linear equation in one variable can also be represented on Cartesian plane, it will be a straight line either parallel to x –axis or y –axis

$x-2=0$ , (straight line parallel to y axis). It means ( 2,<any value on y axis ) will satisfy this line

$y-2=0$, ( straight line parallel to x axis ). It means ( <any value on x-axis ),2 ) will satisfy this line

- Suppose the equation given is

$ax+by+c=0 \; , \; a \neq 0 \; and \; b \neq 0$ - Find the value of y for x=0

$y=\frac {-c}{b}$

This point will lie on Y –axis. And the coordinates will be $(0,\frac{-c}{b})$ - Find the value of x for y=0

$x=\frac {-c}{a}$

This point will lie on X –axis. And the coordinates will be $(\frac {-c}{a}, 0)$ - Now we can draw the line joining these two points
- It is easy to draw the points if the values are integers,So if you don't get integer in previous step you may choose point where the values are integers

Draw the graph of each of the equations given below. Also, find the co-ordinates of the point where the graph cuts the co-ordinates axis:

(a) 2x + y = 6

(b) 3x + 2y + 6 = 0

To draw the graph, we need at least two solutions of the equation We can get two points by putting x=0 and y=0. And then draw the graph. Corrdinates will be given by (x,0) and (0,y)

a.

Coordinates are (3,0) and (0,6)

b.

Coordinates are (-2,0) and (0,-3)

Any point on the y-axis is of the form

(a) (x, 0)

(b) (x, y)

(c) (0, y)

(d) ( y, y)

Answer is C

At what point does the graph of the linear equation x + y = 5 meet a line which is parallel to the y-axis, at a distance 2 units from the origin and in the positive direction of x-axis.

The coordinates of the points lying on the line parallel to the y-axis, at a distance 2 units from the origin and in the positive direction of the x-axis are of the form (2, a).

Putting x = 2, y = a in the equation x + y = 5, we get a = 3. Thus, the required point is (2, 3).

Class 10 Maths Class 10 Science