1. An equation of the form , where and are real numbers, such that and are not both zero, is called a linear equation in two variables.
2. A linear equation in two variables has infinitely many solutions.
3. The graph of every linear equation in two variables is a straight line.
i.e.,
4. is the equation of the -axis and is the equation of the -axis.
5. The graph of is a straight line parallel to the -axis.
i.e.,
6. The graph of is a straight line parallel to the -axis.
i.e.,
7. An equation of the type represents a line passing through the origin.
i.e.,
8. Every solution of the linear equation is a point on the graph of the linear equation.
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Solution: Let the cost of a notebook to be Rs. and a pen to be Rs. respectively .
A/Q ,
(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
Solution: (i)
Here, and
(ii)
Here, and
(iii)
Here, and
(iv)
Here, and
(v)
Here, and
(vi)
Here, and
(vii)
Here, and
(viii)
Here, and
Solution: (iii) has infinitely many solutions . Because , a linear equation in two variables has infinitely many solutions .
2. Write four solutions for each of the following equations :
(i) (ii) (iii)
Solution : (i)
Taking x , then we get .
Taking x , then we get .
Taking x , then we get .
Taking x , then we get .
Therefore, the four solutions are and .
(ii)
Taking , then we get y .
Taking , then we get y .
Taking , then we get y .
Taking , then we get y .
Therefore, the four solutions are and .
(iii)
Taking x , then we get .
Taking x , then we get .
Taking x , then we get .
Taking x , then we get .
Therefore , the four solutions are and .
Solution: (i)
Here ,
L.H.S :
So, is not a solution of .
(ii)
Here ,
L.H.S :
So, is not a solution of .
(iii)
Here ,
L.H.S :
So, is a solution of .
(iv)
Here ,
L.H.S:
So, is a solution of .
(v)
Here ,
L.H.S:
So, is a solution of .
Solution: Here,
We have ,
Solution: We have,
If ,then
If , then
If , then
|
0 |
3 |
2 |
|
4 |
1 |
2 |
Graph :
(ii) Solution: We have,
If ,then
If , then
If , then
|
2 |
0 |
4 |
|
0 |
– 2 |
2 |
Graph :
(iii) Solution: We have,
If ,then
If , then
If , then
|
1 |
– 1 |
2 |
|
3 |
– 3 |
6 |
Graph:
(iv) Solution: We have,
If ,then
If , then
If , then
|
0 |
1 |
2 |
|
3 |
1 |
– 1 |
Graph:
Solution: Since, (2 , 14) is a solution of a linear equation .
So, the equation of the two lines passing through are and .
There are infinitely many lines are satisfied by the coordinates of the point (2,14) .
Solution : Here , ,
We have,
Therefore, the value of is .
Solution: let the distance covered as km and total fare as Rs .
A/Q ,
For graph : We have ,
If , then
If , then
If , then
|
0 |
– 1 |
– 2 |
|
3 |
– 2 |
– 7 |
Graph:
Solution: In figure 4.6 ,
Given the points of the line are :
|
0 |
1 |
– 1 |
|
0 |
– 1 |
1 |
(i) We have,
If ,then
Therefore , is not the point of the line .
(ii) We have,
If , then
If , then
Therefore, and are the points of the line .
So, the linear equation of the given figure is .
(iii)We have,
If , then
Therefore, is not the point of the line .
(iv) We have,
If , then
Therefore, is not the point of the line .
For Fig. 4.7 :
In figure 4.7 ,
Given the points of the line are :
|
2 |
0 |
– 1 |
|
0 |
2 |
3 |
(i) We have,
If , then
Therefore, is not the point of the line .
(ii) We have,
If , then
If , then
Therefore , is not the point of the line .
(iii) We have,
If , then
If , then
If , then
Therefore , and is the point of the line .
So, the linear equation of the given figure is .
(iv) We have,
If , then
Therefore , is not the point of the line .
Solution: let be the work done and be the distance travelled by the body .
We know that, the work done the constant forcethe distance
A/Q,
For graph : We have,
If , then
If , then
If , then
|
5 |
– 5 |
10 |
|
1 |
– 1 |
2 |
Graph:
(i) Given, units
We have,
units
(ii) Given, units
We have,
units
Solution: let and (in Rs)be contributions of Yamini and Fatima respectively .
A/Q,
For graph : We have,
If ,then
If then
If then
|
0 |
60 |
40 |
|
100 |
40 |
60 |
Solution: (i) We have ,
If , then
If , then
If , then
For graph :
(Celsious) |
0 |
5 |
10 |
(Farenheit) |
32 |
41 |
50 |
(ii) We have ,
Given ,
So,
(iii) We have ,
Given ,
So,
(iv) We have,
Given,
Again ,
(v) We have ,
Given,
So,
Therefore , and .
Solution: (i) We have ,
The representation of the solution on the number line is given below :
Where is an equation in one variable .
(ii) We have,
So, is a linear equation in the variables and . This is represented by a line . Hence , three solutions of the given equation are :
|
1 |
2 |
3 |
|
3 |
3 |
3 |
Solution: (i) We have,
The representation of the solution on the number line is given below :
Where is an equation in one variable .
(ii) We have ,
So, is a linear equation in the variables and . This is represented by a line . Hence , three solutions of the given equation are :
|
– 4.5 |
– 4.5 |
– 4.5 |
|
1 |
2 |
3 |