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Page No 145:

Question 1:

Answer:

(i) 3x + 5= 7.5
This can be expressed in the form ax + by + c = 0 as 3x+5y+-7.5=0.
(ii) 2x-y5+6=0
This can be expressed in the form ax + by + c = 0 as 2x+-15y+6=0
(iii) 3y – 2x = 6
This can be expressed in the form ax + by + c = 0 as 2x+-3y+6=0.
(iv) 4x = 5y
This can be expressed in the form ax + by + c = 0 as 4x-5y+0=0
(v) x5-y6=1
This can be expressed in the form ax + by + c = 0 as 6x-5y=30
(vi) 2x+3y=5
This can be expressed in the form ax + by + c = 0 as 2x+3y+-5=0



Page No 146:

Question 2:

(i) 3x + 5= 7.5
This can be expressed in the form ax + by + c = 0 as 3x+5y+-7.5=0.
(ii) 2x-y5+6=0
This can be expressed in the form ax + by + c = 0 as 2x+-15y+6=0
(iii) 3y – 2x = 6
This can be expressed in the form ax + by + c = 0 as 2x+-3y+6=0.
(iv) 4x = 5y
This can be expressed in the form ax + by + c = 0 as 4x-5y+0=0
(v) x5-y6=1
This can be expressed in the form ax + by + c = 0 as 6x-5y=30
(vi) 2x+3y=5
This can be expressed in the form ax + by + c = 0 as 2x+3y+-5=0

Answer:

(i) x = 6
In the form of ax + by + c = 0 we have x+0y+(-6)=0 where a = 1, b = 0 and c=-6.
(ii) 3x – y = x – 1
In the form of ax + by + c = 0 we have 2x+-1y+1=0 where a = 2, b = -1 and c = 1
(iii) 2x + 9 = 0
In the form of ax + by + c = 0 we have 2x+-1y+1=0 where a = 2, b = -1 and c = 1
(iv) 4y = 7
In the form of ax + by + c = 0 we have 0x+4y+-7=0 where a = 0, b = 4 and c = -7
(v) x + y = 4
In the form of ax + by + c = 0 we have 1x+1y+-4=0 where a=1,b=1,c=-4
(vi) x2-y3=16+y
In the form of ax + by + c = 0 we have 3x-2y=1+6y3x-8y+-1=0 where a=3,b=-8,c=-1.

Page No 146:

Question 3:

(i) x = 6
In the form of ax + by + c = 0 we have x+0y+(-6)=0 where a = 1, b = 0 and c=-6.
(ii) 3x – y = x – 1
In the form of ax + by + c = 0 we have 2x+-1y+1=0 where a = 2, b = -1 and c = 1
(iii) 2x + 9 = 0
In the form of ax + by + c = 0 we have 2x+-1y+1=0 where a = 2, b = -1 and c = 1
(iv) 4y = 7
In the form of ax + by + c = 0 we have 0x+4y+-7=0 where a = 0, b = 4 and c = -7
(v) x + y = 4
In the form of ax + by + c = 0 we have 1x+1y+-4=0 where a=1,b=1,c=-4
(vi) x2-y3=16+y
In the form of ax + by + c = 0 we have 3x-2y=1+6y3x-8y+-1=0 where a=3,b=-8,c=-1.

Answer:

The equation given is 5x – 4y = 20.
(i) (4, 0) 
Putting the value in the given equation we have 
LHS:54-40=20RHS:20LHS=RHS
Thus, (4, 0) is a solution of the given equation.
(ii) (0, 5)
Putting the value in the given equation we have 
LHS:50-45=0-20=-20RHS:20LHSRHS
Thus, (0, 5) is not a solution of the given equation.
(iii) -2, 52
Putting the value in the given equation we have 
LHS:5-2-452=-10-10=-20RHS:20LHSRHS
Thus, -2, 52 is not a solution of the given equation.
(iv) (0, –5)
Putting the value in the given equation we have 
LHS:50-4-5=0+20=20RHS:20LHS=RHS
Thus, (0, –5) is a solution of the given equation.
(v) 2, -52
Putting the value in the given equation we have 
LHS:52-4-52=10+10=20RHS:20LHS=RHS
Thus, 2, -52 is a solution of the given equation.








 

Page No 146:

Question 4:

The equation given is 5x – 4y = 20.
(i) (4, 0) 
Putting the value in the given equation we have 
LHS:54-40=20RHS:20LHS=RHS
Thus, (4, 0) is a solution of the given equation.
(ii) (0, 5)
Putting the value in the given equation we have 
LHS:50-45=0-20=-20RHS:20LHSRHS
Thus, (0, 5) is not a solution of the given equation.
(iii) -2, 52
Putting the value in the given equation we have 
LHS:5-2-452=-10-10=-20RHS:20LHSRHS
Thus, -2, 52 is not a solution of the given equation.
(iv) (0, –5)
Putting the value in the given equation we have 
LHS:50-4-5=0+20=20RHS:20LHS=RHS
Thus, (0, –5) is a solution of the given equation.
(v) 2, -52
Putting the value in the given equation we have 
LHS:52-4-52=10+10=20RHS:20LHS=RHS
Thus, 2, -52 is a solution of the given equation.








 

Answer:

(i) 2x – 3y = 6
 

x 0 3 -3 92 2
y -2 0 -4 1 -23

(ii) 2x5+3y10=3
x 0 152 5 10 3
y 10 0 103 -103 6


(iii) 3y = 4x
 
x 3 -3 -6 6 0
y 4 -4 -8 8 0

Page No 146:

Question 5:

(i) 2x – 3y = 6
 

x 0 3 -3 92 2
y -2 0 -4 1 -23

(ii) 2x5+3y10=3
x 0 152 5 10 3
y 10 0 103 -103 6


(iii) 3y = 4x
 
x 3 -3 -6 6 0
y 4 -4 -8 8 0

Answer:

Given: 5x – 3y = k
Since x = 3 and = 4 is a solution of the given equation so, it should satisfy the equation.
53-34=k15-12=k3=k 
 

Page No 146:

Question 6:

Given: 5x – 3y = k
Since x = 3 and = 4 is a solution of the given equation so, it should satisfy the equation.
53-34=k15-12=k3=k 
 

Answer:

Given: 4– 3+ 1 = 0                                 .....(1)
x = 3k + 2 and y = 2k – 1
Putting these values in the equation (1) we get
43k+2-32k-1+1=012k+8-6k+3+1=06k+12=0k+2=0k=-2

Page No 146:

Question 7:

Given: 4– 3+ 1 = 0                                 .....(1)
x = 3k + 2 and y = 2k – 1
Putting these values in the equation (1) we get
43k+2-32k-1+1=012k+8-6k+3+1=06k+12=0k+2=0k=-2

Answer:

Let cost of a pencil to be ₹ x and that of a ballpoint to be ₹ y.
Cost of 5 pencils = 5x
Cost of 2 ballpoints = 2y
Cost of 5 pencils = cost of 2 ballpoints
5x=2y5x-2y=0



Page No 161:

Question 1:

Let cost of a pencil to be ₹ x and that of a ballpoint to be ₹ y.
Cost of 5 pencils = 5x
Cost of 2 ballpoints = 2y
Cost of 5 pencils = cost of 2 ballpoints
5x=2y5x-2y=0

Answer:


(i) The equation of given line is x = 4. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (4, 0).

 

(ii) The equation of given line is x + 4 = 0 or x = –4. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (–4, 0).



(iii) The equation of given line is y = 3. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, 3).



(iv) The equation of given line is y = –3. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, –3).



(v) The equation of given line is x = –2. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (–2, 0).



(vi) The equation of given line is x = 5. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (5, 0).



(vii) The equation of given line is y + 5 = 0 or y = –5. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, –5).



(viii) The equation of given line is y = 4. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, 4).

Page No 161:

Question 2:


(i) The equation of given line is x = 4. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (4, 0).

 

(ii) The equation of given line is x + 4 = 0 or x = –4. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (–4, 0).



(iii) The equation of given line is y = 3. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, 3).



(iv) The equation of given line is y = –3. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, –3).



(v) The equation of given line is x = –2. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (–2, 0).



(vi) The equation of given line is x = 5. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (5, 0).



(vii) The equation of given line is y + 5 = 0 or y = –5. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, –5).



(viii) The equation of given line is y = 4. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, 4).

Answer:

Given equation:  y = 3x
   
When x = -2, y = -6.
When x = -1, y = -3.
Thus, we have the following table:

      x     -2    -1
      y     -6     -3

Now plot the points (-2,-6), (-1, -3) on a graph paper.
Join the points and extend the line in both the directions.
The line segment is the required  graph of the equation.
From the graph we can see that when x = 2, y = 6



Also, when x = -2, y = -6.

Page No 161:

Question 3:

Given equation:  y = 3x
   
When x = -2, y = -6.
When x = -1, y = -3.
Thus, we have the following table:

      x     -2    -1
      y     -6     -3

Now plot the points (-2,-6), (-1, -3) on a graph paper.
Join the points and extend the line in both the directions.
The line segment is the required  graph of the equation.
From the graph we can see that when x = 2, y = 6



Also, when x = -2, y = -6.

Answer:

Given equation: x + 2y - 3 = 0
        Or,  x + 2y = 3
When y = 0, x + 0 = 3 ⇒ x = 3
 When y = 1, x + 2 = 3 ⇒ x = 3-2 = 1
 When y = 2, x + 4 = 3 ⇒ x = 3 - 4 = -1
Thus, we have the following table:

      x      3      1     -1
      y      0      1      2

Now plot the points (3,0) ,(1,1) and (-1,2) on the graph paper.
Join the points and extend the line in both the directions.
The line segment is the required graph of the equation.



When x = 5,

y = 3 -x2y = 3 -52y = -1

Similarly, from the graph we can see that when x = −5, y = 4.

Page No 161:

Question 4:

Given equation: x + 2y - 3 = 0
        Or,  x + 2y = 3
When y = 0, x + 0 = 3 ⇒ x = 3
 When y = 1, x + 2 = 3 ⇒ x = 3-2 = 1
 When y = 2, x + 4 = 3 ⇒ x = 3 - 4 = -1
Thus, we have the following table:

      x      3      1     -1
      y      0      1      2

Now plot the points (3,0) ,(1,1) and (-1,2) on the graph paper.
Join the points and extend the line in both the directions.
The line segment is the required graph of the equation.



When x = 5,

y = 3 -x2y = 3 -52y = -1

Similarly, from the graph we can see that when x = −5, y = 4.

Answer:

Given equation :

 2x - 3y = 5 2x = 3y + 5 x = 3y + 52
When, y = -1,  x = -3+52 = 22 = 1

When, y = -3,  x = -9+52 = -42 =-2
Thus, we have the following table:

   x     1    -2  
   y    -1    -3
Plot the points (-2, -3) , (1,-1) on the graph paper and extend the line in both directions.

(i) When x = 4:
          4 = 3y+52  8 = 3y+5                                       3y = 8 - 5 = 3                                     3y = 3                                     y = 1
(ii) When y = 3:
      x = 3y+52 = 142 = 7



Page No 162:

Question 5:

Given equation :

 2x - 3y = 5 2x = 3y + 5 x = 3y + 52
When, y = -1,  x = -3+52 = 22 = 1

When, y = -3,  x = -9+52 = -42 =-2
Thus, we have the following table:

   x     1    -2  
   y    -1    -3
Plot the points (-2, -3) , (1,-1) on the graph paper and extend the line in both directions.

(i) When x = 4:
          4 = 3y+52  8 = 3y+5                                       3y = 8 - 5 = 3                                     3y = 3                                     y = 1
(ii) When y = 3:
      x = 3y+52 = 142 = 7

Answer:


Given equation:
                   2x +y = 6 y = 6 - 2x
When, x = 0y = 6 - 0 = 6.
When, x =1y = 6 - 2 = 4.
When, x = 2y = 6 - 4 = 2.
Thus, we have the following table:

    x      0      1     2
    y      6      4     2

Plot the points (0,6), (1,4) and (2,2) on the graph paper. Join these points and extend the line.



Clearly, the graph cuts the x-axis at P(3,0).

Page No 162:

Question 6:


Given equation:
                   2x +y = 6 y = 6 - 2x
When, x = 0y = 6 - 0 = 6.
When, x =1y = 6 - 2 = 4.
When, x = 2y = 6 - 4 = 2.
Thus, we have the following table:

    x      0      1     2
    y      6      4     2

Plot the points (0,6), (1,4) and (2,2) on the graph paper. Join these points and extend the line.



Clearly, the graph cuts the x-axis at P(3,0).

Answer:

Given equation: 3x + 2y = 6. Then,

   2y = 6 - 3x  y = 6 - 3x2

When x = 2, y = 6 - 62 = 0

When x = 4, y = 6 - 122 = -3
Thus, we get the following table:

      x        2       4
      y       0       -3

Plot the points (2,0) ,(4,-3) on the graph paper. Join the points and extend the graph in both the directions. 





Clearly, the graph cuts the y-axis at P(0,3).

Page No 162:

Question 7:

Given equation: 3x + 2y = 6. Then,

   2y = 6 - 3x  y = 6 - 3x2

When x = 2, y = 6 - 62 = 0

When x = 4, y = 6 - 122 = -3
Thus, we get the following table:

      x        2       4
      y       0       -3

Plot the points (2,0) ,(4,-3) on the graph paper. Join the points and extend the graph in both the directions. 





Clearly, the graph cuts the y-axis at P(0,3).

Answer:


3x-2y=42y=3x-4y=3x-42
When x = 0, y=3×0-42=0-42=-42=-2
When x = 2, y=3×2-42=6-42=22=1
When x = –2, y=3×-2-42=-6-42=-102=-5
Thus, the points on the line 3x – 2y = 4 are as given in the following table:

x 0 2 –2
y –2 1 –5

Plotting the points (0, –2), (2, 1) and (–2–5) and drawing a line passing through these points, we obtain the graph of of the line 3x – 2y = 4.

x+y-3=0y=-x+3
When x = 0, y=-0+3=3
When x = 1, y=-1+3=2
When x = –1, y=--1+3=1+3=4
Thus, the points on the line x + y – 3 = 0 are as given in the following table:
x 0 1 –1
y 3 2 4

Plotting the points (0, 3), (1, 2) and (–14) and drawing a line passing through these points, we obtain the graph of of the line x + y – 3 = 0.



It can be seen that the lines 3x – 2y = 4 and x + y – 3 = 0 intersect at the point (2, 1).

Page No 162:

Question 8:


3x-2y=42y=3x-4y=3x-42
When x = 0, y=3×0-42=0-42=-42=-2
When x = 2, y=3×2-42=6-42=22=1
When x = –2, y=3×-2-42=-6-42=-102=-5
Thus, the points on the line 3x – 2y = 4 are as given in the following table:

x 0 2 –2
y –2 1 –5

Plotting the points (0, –2), (2, 1) and (–2–5) and drawing a line passing through these points, we obtain the graph of of the line 3x – 2y = 4.

x+y-3=0y=-x+3
When x = 0, y=-0+3=3
When x = 1, y=-1+3=2
When x = –1, y=--1+3=1+3=4
Thus, the points on the line x + y – 3 = 0 are as given in the following table:
x 0 1 –1
y 3 2 4

Plotting the points (0, 3), (1, 2) and (–14) and drawing a line passing through these points, we obtain the graph of of the line x + y – 3 = 0.



It can be seen that the lines 3x – 2y = 4 and x + y – 3 = 0 intersect at the point (2, 1).

Answer:


4x+3y=243y=-4x+24y=-4x+243
When x = 0, y=-4×0+243=0+243=243=8
When x = 3, y=-4×3+243=-12+243=123=4
When x = 6, y=-4×6+243=-24+243=03=0
Thus, the points on the line 4x + 3y = 24 are as given in the following table:

x 0 3 6
y 8 4 0

Plotting the points (0, 8), (3, 4) and (60) and drawing a line passing through these points, we obtain the graph of of the line 4x + 3y = 24.



(i) It can be seen that the line 4x + 3y = 24 intersects the x-axis at (6, 0) and y-axis at (0, 8).

(ii) The triangle formed by the line and the coordinate axes is a right triangle right angled at the origin.

∴ Area of the triangle = 12×6×8 = 24 square units

Page No 162:

Question 9:


4x+3y=243y=-4x+24y=-4x+243
When x = 0, y=-4×0+243=0+243=243=8
When x = 3, y=-4×3+243=-12+243=123=4
When x = 6, y=-4×6+243=-24+243=03=0
Thus, the points on the line 4x + 3y = 24 are as given in the following table:

x 0 3 6
y 8 4 0

Plotting the points (0, 8), (3, 4) and (60) and drawing a line passing through these points, we obtain the graph of of the line 4x + 3y = 24.



(i) It can be seen that the line 4x + 3y = 24 intersects the x-axis at (6, 0) and y-axis at (0, 8).

(ii) The triangle formed by the line and the coordinate axes is a right triangle right angled at the origin.

∴ Area of the triangle = 12×6×8 = 24 square units

Answer:


2x+y=6y=-2x+6
When x = 0, y=-2×0+6=0+6=6
When x = 1, y=-2×1+6=-2+6=4
When x = 2, y=-2×2+6=-4+6=2

Thus, the points on the line 2x + y = 6 are as given in the following table:

x 0 1 2
y 6 4 2

Plotting the points (0, 6), (1, 4) and (22) and drawing a line passing through these points, we obtain the graph of of the line 2x + y = 6.

2x-y+2=0y=2x+2
When x = 0, y=2×0+2=0+2=2
When x = 1, y=2×1+2=2+2=4
When x = –1, y=2×-1+2=-2+2=0

Thus, the points on the line 2x – y + 2 = 0 are as given in the following table:
x 0 1 –1
y 2 4 0

Plotting the points (0, 2), (1, 4) and (–10) and drawing a line passing through these points, we obtain the graph of of the line 2x – y + 2 = 0.



The shaded region represents the area bounded by the lines 2x + y = 6, 2x – y + 2 = 0 and the x-axis. This represents a triangle.

It can be seen that the lines intersect at the point C(1, 4). Draw CD perpendicular from C on the x-axis.

Height = CD = 4 units

Base = AB = 4 units

∴ Area of the shaded region = Area of âˆ†ABC = 12×AB×CD=12×4×4 = 8 square units

Page No 162:

Question 10:


2x+y=6y=-2x+6
When x = 0, y=-2×0+6=0+6=6
When x = 1, y=-2×1+6=-2+6=4
When x = 2, y=-2×2+6=-4+6=2

Thus, the points on the line 2x + y = 6 are as given in the following table:

x 0 1 2
y 6 4 2

Plotting the points (0, 6), (1, 4) and (22) and drawing a line passing through these points, we obtain the graph of of the line 2x + y = 6.

2x-y+2=0y=2x+2
When x = 0, y=2×0+2=0+2=2
When x = 1, y=2×1+2=2+2=4
When x = –1, y=2×-1+2=-2+2=0

Thus, the points on the line 2x – y + 2 = 0 are as given in the following table:
x 0 1 –1
y 2 4 0

Plotting the points (0, 2), (1, 4) and (–10) and drawing a line passing through these points, we obtain the graph of of the line 2x – y + 2 = 0.



The shaded region represents the area bounded by the lines 2x + y = 6, 2x – y + 2 = 0 and the x-axis. This represents a triangle.

It can be seen that the lines intersect at the point C(1, 4). Draw CD perpendicular from C on the x-axis.

Height = CD = 4 units

Base = AB = 4 units

∴ Area of the shaded region = Area of âˆ†ABC = 12×AB×CD=12×4×4 = 8 square units

Answer:


x-y=1y=x-1
When x = 0, y=0-1=-1
When x = 1, y=1-1=0
When x = 2, y=2-1=1

Thus, the points on the line x – y = 1 are as given in the following table:

x 0 1 2
y –1 0 1

Plotting the points (0, –1), (1, 0) and (21) and drawing a line passing through these points, we obtain the graph of of the line x – y = 1.

2x+y=8y=-2x+8
When x = 1, y=-2×1+8=-2+8=6
When x = 2, y=-2×2+8=-4+8=4
When x = 3, y=-2×3+8=-6+8=2

Thus, the points on the line  2x + y = 8 are as given in the following table:
x 1 2 3
y 6 4 2

Plotting the points (1, 6), (2, 4) and (32) and drawing a line passing through these points, we obtain the graph of of the line  2x + y = 8.



The shaded region represents the area bounded by the lines x – y = 1, 2x + y = 8 and the y-axis. This represents a triangle.

It can be seen that the lines intersect at the point C(3, 2). Draw CD perpendicular from C on the y-axis.

Height = CD = 3 units

Base = AB = 9 units

∴ Area of the shaded region = Area of âˆ†ABC = 12×AB×CD=12×9×3 = 272 square units

Page No 162:

Question 11:


x-y=1y=x-1
When x = 0, y=0-1=-1
When x = 1, y=1-1=0
When x = 2, y=2-1=1

Thus, the points on the line x – y = 1 are as given in the following table:

x 0 1 2
y –1 0 1

Plotting the points (0, –1), (1, 0) and (21) and drawing a line passing through these points, we obtain the graph of of the line x – y = 1.

2x+y=8y=-2x+8
When x = 1, y=-2×1+8=-2+8=6
When x = 2, y=-2×2+8=-4+8=4
When x = 3, y=-2×3+8=-6+8=2

Thus, the points on the line  2x + y = 8 are as given in the following table:
x 1 2 3
y 6 4 2

Plotting the points (1, 6), (2, 4) and (32) and drawing a line passing through these points, we obtain the graph of of the line  2x + y = 8.



The shaded region represents the area bounded by the lines x – y = 1, 2x + y = 8 and the y-axis. This represents a triangle.

It can be seen that the lines intersect at the point C(3, 2). Draw CD perpendicular from C on the y-axis.

Height = CD = 3 units

Base = AB = 9 units

∴ Area of the shaded region = Area of âˆ†ABC = 12×AB×CD=12×9×3 = 272 square units

Answer:


x+y=6y=-x+6
When x = 0, y=-0+6=6
When x = 1, y=-1+6=5
When x = 3, y=-3+6=3

Thus, the points on the line x + y = 6 are as given in the following table:

x 0 1 3
y 6 5 3

Plotting the points (0, 6), (1, 5) and (33) and drawing a line passing through these points, we obtain the graph of of the line x + y = 6.

x-y=2y=x-2
When x = 0, y=0-2=-2
When x = 2, y=2-2=0
When x = –1, y=-1-2=-3

Thus, the points on the line x – y = 2 are as given in the following table:
x 0 2 –1
y –2 0 –3

Plotting the points (0, –2), (2, 0) and (–1–3) and drawing a line passing through these points, we obtain the graph of of the line x – y = 2.



It can be seen that the lines x + y = 6 and x – y = 2 intersect at the point (4, 2).

Page No 162:

Question 12:


x+y=6y=-x+6
When x = 0, y=-0+6=6
When x = 1, y=-1+6=5
When x = 3, y=-3+6=3

Thus, the points on the line x + y = 6 are as given in the following table:

x 0 1 3
y 6 5 3

Plotting the points (0, 6), (1, 5) and (33) and drawing a line passing through these points, we obtain the graph of of the line x + y = 6.

x-y=2y=x-2
When x = 0, y=0-2=-2
When x = 2, y=2-2=0
When x = –1, y=-1-2=-3

Thus, the points on the line x – y = 2 are as given in the following table:
x 0 2 –1
y –2 0 –3

Plotting the points (0, –2), (2, 0) and (–1–3) and drawing a line passing through these points, we obtain the graph of of the line x – y = 2.



It can be seen that the lines x + y = 6 and x – y = 2 intersect at the point (4, 2).

Answer:


Let the contribution of A and B be â‚¹ and â‚¹ y, respectively.

Total contribution of A and B = â‚¹ + â‚¹ = â‚¹ (y)

It is given that the total contribution of A and B is ₹ 100.

∴ y = 100

This is the linear equation satisfying the the given data.

= 100 

⇒ y = 100 – x

When x = 10, y = 100 – 10 = 90

When x = 40, y = 100 – 40 = 60

When x = 60, y = 100 – 60 = 40

Thus, the points on the line = 100  are as given in the following table:

x 10 40 60
y 90 60 40

Plotting the points (10, 90), (40, 60) and (6040) and drawing a line passing through these points, we obtain the graph of of the line = 100.

Page No 162:

Question 1:


Let the contribution of A and B be â‚¹ and â‚¹ y, respectively.

Total contribution of A and B = â‚¹ + â‚¹ = â‚¹ (y)

It is given that the total contribution of A and B is ₹ 100.

∴ y = 100

This is the linear equation satisfying the the given data.

= 100 

⇒ y = 100 – x

When x = 10, y = 100 – 10 = 90

When x = 40, y = 100 – 40 = 60

When x = 60, y = 100 – 60 = 40

Thus, the points on the line = 100  are as given in the following table:

x 10 40 60
y 90 60 40

Plotting the points (10, 90), (40, 60) and (6040) and drawing a line passing through these points, we obtain the graph of of the line = 100.

Answer:

Since, the y-coordinate of any point on x-axis is always 0.

So, the equation of the x-axis is y = 0.

Hence, the correct option is (b).

Page No 162:

Question 2:

Since, the y-coordinate of any point on x-axis is always 0.

So, the equation of the x-axis is y = 0.

Hence, the correct option is (b).

Answer:

Since, the x-coordinate of any point on y-axis is always 0.

So, the equation of the y-axis is x = 0.

Hence, the correct option is (a).



Page No 163:

Question 3:

Since, the x-coordinate of any point on y-axis is always 0.

So, the equation of the y-axis is x = 0.

Hence, the correct option is (a).

Answer:

(c) the line y = x
Given, a point of the form (a, a), where a≠ 0 .
When a = 1, the point is (1,1)
When a = 2, the point is (2,2).....and so on.
Plot the points (1,1) and (2,2).......and so on. Join the points and extend them in both the direction. You will get the equation of the line y = x.

Page No 163:

Question 4:

(c) the line y = x
Given, a point of the form (a, a), where a≠ 0 .
When a = 1, the point is (1,1)
When a = 2, the point is (2,2).....and so on.
Plot the points (1,1) and (2,2).......and so on. Join the points and extend them in both the direction. You will get the equation of the line y = x.

Answer:

(d) the line x + y = 0
Given, a point of the form (a, -a), where a ≠ 0.
When a = 1, the point is (1,-1).
When a = 2, the point is (2,-2).
When a = 3, the point is (3,-3).......and so on.
Plot these points on a graph paper. Join these points and extend them in both the directions. 
You will get the equation of the line x + y = 0.

Page No 163:

Question 5:

(d) the line x + y = 0
Given, a point of the form (a, -a), where a ≠ 0.
When a = 1, the point is (1,-1).
When a = 2, the point is (2,-2).
When a = 3, the point is (3,-3).......and so on.
Plot these points on a graph paper. Join these points and extend them in both the directions. 
You will get the equation of the line x + y = 0.

Answer:

(c) infinitely many solutions
Given linear equation: 3x - 5y = 15
Or, x = 5y + 153

When y = 0, x = 153 = 5.

When  y = 3, x = 303 = 10.

When y = -3, x = 03 = 0.

Thus, we have the following table:

     x       5       10     0
     y       0       3     -3

Plot the points A(5,0) , B(10,3) and C(0,-3). Join the points and extend them in both the directions.
We get infinite points that satisfy the given equation. 
Hence, the linear equation has infinitely many solutions.

Page No 163:

Question 6:

(c) infinitely many solutions
Given linear equation: 3x - 5y = 15
Or, x = 5y + 153

When y = 0, x = 153 = 5.

When  y = 3, x = 303 = 10.

When y = -3, x = 03 = 0.

Thus, we have the following table:

     x       5       10     0
     y       0       3     -3

Plot the points A(5,0) , B(10,3) and C(0,-3). Join the points and extend them in both the directions.
We get infinite points that satisfy the given equation. 
Hence, the linear equation has infinitely many solutions.

Answer:

Since, every point on the line represented by the equation 2x + 5= 7 is its solution.

Therefore, there are infinite solutions of the equation the equation 2x + 5= 7 in which the values of x and y are rational numbers, positive real numbers or real numbers.

But, as 2 + 5 = 7, i.e. x = 1 and y = 1 are the only pair of natural numbers that are the solution of the equation the equation 2x + 5= 7.

So, the equation 2x + 5= 7 has a unique solution, if x and are both are natural numbers.

Hence, the correct option is (a).

Page No 163:

Question 7:

Since, every point on the line represented by the equation 2x + 5= 7 is its solution.

Therefore, there are infinite solutions of the equation the equation 2x + 5= 7 in which the values of x and y are rational numbers, positive real numbers or real numbers.

But, as 2 + 5 = 7, i.e. x = 1 and y = 1 are the only pair of natural numbers that are the solution of the equation the equation 2x + 5= 7.

So, the equation 2x + 5= 7 has a unique solution, if x and are both are natural numbers.

Hence, the correct option is (a).

Answer:

As, the graph of y = 5 is a line parallel to x-axis i.e. y = 0.

The line represented by the equation y = 5 is parallel to x-axis and intersects y-axis at y = 5.

So, the graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin making an intercept 5 on the y-axis.

Hence, the correct answers are options (b) and (c).
Disclaimer: In this question, there are two correct answers.

Page No 163:

Question 8:

As, the graph of y = 5 is a line parallel to x-axis i.e. y = 0.

The line represented by the equation y = 5 is parallel to x-axis and intersects y-axis at y = 5.

So, the graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin making an intercept 5 on the y-axis.

Hence, the correct answers are options (b) and (c).
Disclaimer: In this question, there are two correct answers.

Answer:

As, the graph of x = 4 is a line parallel to y-axis i.e. x = 0.

 The line represented by the equation x = 4 is parallel to y-axis and intersects x-axis at x = 4.

So, the graph of x = 4 is parallel to y-axis at a distance of 4 units from the origin making an intercept 4 on the x-axis.

Hence, the correct options are (a) and (d).

Disclaimer: In this question, there are two correct answers.

Page No 163:

Question 9:

As, the graph of x = 4 is a line parallel to y-axis i.e. x = 0.

 The line represented by the equation x = 4 is parallel to y-axis and intersects x-axis at x = 4.

So, the graph of x = 4 is parallel to y-axis at a distance of 4 units from the origin making an intercept 4 on the x-axis.

Hence, the correct options are (a) and (d).

Disclaimer: In this question, there are two correct answers.

Answer:

As, the graph of x + 3 = 0 or x = -3 is a line parallel to y-axis i.e. x = 0.

 The line represented by the equation x = -3 is parallel to y-axis and intersects x-axis at x = -3.

So, the graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis making an intercept -3 on the x-axis.

Hence, the correct options are (a) and (c).

Disclaimer: In this question, there are two correct answers.

Page No 163:

Question 10:

As, the graph of x + 3 = 0 or x = -3 is a line parallel to y-axis i.e. x = 0.

 The line represented by the equation x = -3 is parallel to y-axis and intersects x-axis at x = -3.

So, the graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis making an intercept -3 on the x-axis.

Hence, the correct options are (a) and (c).

Disclaimer: In this question, there are two correct answers.

Answer:

As, the graph of y + 2 = 0 or y = −2 is a line parallel to x-axis i.e. y = 0.

⇒ The line represented by the equation y = −2 is parallel to x-axis and intersects y-axis at y = −2.

So, the graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis making an intercept −2 on the y-axis.

Hence, the correct options are (b) and (c).

Disclaimer: In this question, there are two correct answers.

Page No 163:

Question 11:

As, the graph of y + 2 = 0 or y = −2 is a line parallel to x-axis i.e. y = 0.

⇒ The line represented by the equation y = −2 is parallel to x-axis and intersects y-axis at y = −2.

So, the graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis making an intercept −2 on the y-axis.

Hence, the correct options are (b) and (c).

Disclaimer: In this question, there are two correct answers.

Answer:

If he graph of the linear equation 2x + 3= 6 meets the y-axis, then x = 0.

Substituting the value of x = 0 in equation 2x + 3= 6, we get

20+3y=63y=6y=63y=2

So, the point of meeting is (0, 2).

Hence, the correct answer is option (c).



Page No 164:

Question 12:

If he graph of the linear equation 2x + 3= 6 meets the y-axis, then x = 0.

Substituting the value of x = 0 in equation 2x + 3= 6, we get

20+3y=63y=6y=63y=2

So, the point of meeting is (0, 2).

Hence, the correct answer is option (c).

Answer:

If he graph of the linear equation 2x + 5= 10 meets the x-axis, then y = 0.

Substituting the value of y = 0 in equation 2x + 5= 10, we get

2x+50=102x=10x=102x=5
​
So, the point of meeting is (5, 0).

Hence, the correct option is (c).

Page No 164:

Question 13:

If he graph of the linear equation 2x + 5= 10 meets the x-axis, then y = 0.

Substituting the value of y = 0 in equation 2x + 5= 10, we get

2x+50=102x=10x=102x=5
​
So, the point of meeting is (5, 0).

Hence, the correct option is (c).

Answer:

(c) (3, 2)
The graph of line x = 3 is a line parallel to the y-axis.
Hence, its passes through (3,2), satisfying x =3.

Page No 164:

Question 14:

(c) (3, 2)
The graph of line x = 3 is a line parallel to the y-axis.
Hence, its passes through (3,2), satisfying x =3.

Answer:

Since, the graph of the line y = 3 is parallel to x-axis at a distance of 3 units from the x-axis.

Or, the y-coordinate of every point on the line is always equal to 3.

So, the graph of the line y = 3 passes through the point (2, 3).

Hence, the correct option is (c).

Page No 164:

Question 15:

Since, the graph of the line y = 3 is parallel to x-axis at a distance of 3 units from the x-axis.

Or, the y-coordinate of every point on the line is always equal to 3.

So, the graph of the line y = 3 passes through the point (2, 3).

Hence, the correct option is (c).

Answer:

(d) (−3, 2)

The graph of the line y = -3 does not pass through (-3,2) since (-3,2) does not satisfy y = -3.

Page No 164:

Question 16:

(d) (−3, 2)

The graph of the line y = -3 does not pass through (-3,2) since (-3,2) does not satisfy y = -3.

Answer:

(d) (1, 1)
Given equation: x - y = 0 or, x = y
When x = 1, y = 1
When x = 2, y =2... and so on
Thus, we get the following table:

      x       1       2
      y       1       2

​Plot the points on the graph paper. Join the points and extend them in both the directions.
We can see the linear equation x - y = 0 passes through the point (1,1).

Page No 164:

Question 17:

(d) (1, 1)
Given equation: x - y = 0 or, x = y
When x = 1, y = 1
When x = 2, y =2... and so on
Thus, we get the following table:

      x       1       2
      y       1       2

​Plot the points on the graph paper. Join the points and extend them in both the directions.
We can see the linear equation x - y = 0 passes through the point (1,1).

Answer:

(b) x + y = 0
Given points: (-2, 2) , (0, 0) and (2, -2)
We have to check which equation satisfies the given points.
Let us check for (a) x - y = 0
Substituting (-2, 2) in the equation, we get: x - y = -2 -2 = -4
Substituting (0, 0) in the equation, we get: x - y = 0 - 0 = 0
Substituting (2, - 2) in the equation, we get: x - y = 2 + 2 = 4
So, the given points do not satisfy the equation.
Now, let us check (b) x + y = 0
Substituting (-2, 2) in the equation, we get: x + y = -2 + 2 = 0
Substituting (0, 0) in the equation, we get: x + y = 0 + 0 = 0
Substituting (2,-2) in the equation, we get:  x + y = 2 - 2= 0
So, the given points satisfy the equation.
Similarly, we can check other equations and find that the given points will not satisfy the equations.
Hence, each of (-2, 2), (0, 0) and (2, -2) is a solution of the linear equation x + y = 0.

Page No 164:

Question 18:

(b) x + y = 0
Given points: (-2, 2) , (0, 0) and (2, -2)
We have to check which equation satisfies the given points.
Let us check for (a) x - y = 0
Substituting (-2, 2) in the equation, we get: x - y = -2 -2 = -4
Substituting (0, 0) in the equation, we get: x - y = 0 - 0 = 0
Substituting (2, - 2) in the equation, we get: x - y = 2 + 2 = 4
So, the given points do not satisfy the equation.
Now, let us check (b) x + y = 0
Substituting (-2, 2) in the equation, we get: x + y = -2 + 2 = 0
Substituting (0, 0) in the equation, we get: x + y = 0 + 0 = 0
Substituting (2,-2) in the equation, we get:  x + y = 2 - 2= 0
So, the given points satisfy the equation.
Similarly, we can check other equations and find that the given points will not satisfy the equations.
Hence, each of (-2, 2), (0, 0) and (2, -2) is a solution of the linear equation x + y = 0.

Answer:

(c) Infinitely many
Infinite linear equations are satisfied by x = 2, y = 3.

Page No 164:

Question 19:

(c) Infinitely many
Infinite linear equations are satisfied by x = 2, y = 3.

Answer:

(a) a ≠ 0, b ≠ 0
A linear equation in two variables x and y is of the form ax + by + c= 0, where a ≠ 0 and b ≠ 0.

Page No 164:

Question 20:

(a) a ≠ 0, b ≠ 0
A linear equation in two variables x and y is of the form ax + by + c= 0, where a ≠ 0 and b ≠ 0.

Answer:

Since, (2, 0) is a solution of the linear equation 2x + 3y = k.

Substituting the values x = 2 and y = 0 in the equation 2x + 3y = k, we get

22+30=k4=kor, k=4

Hence, the correct option is (d).

Page No 164:

Question 21:

Since, (2, 0) is a solution of the linear equation 2x + 3y = k.

Substituting the values x = 2 and y = 0 in the equation 2x + 3y = k, we get

22+30=k4=kor, k=4

Hence, the correct option is (d).

Answer:

(c) (x, 0),
Any point on the x-axis is of the form (x,0), where x ≠ 0.

Page No 164:

Question 22:

(c) (x, 0),
Any point on the x-axis is of the form (x,0), where x ≠ 0.

Answer:

(b) (0, y),
Any point on the y-axis is of the form (0,y), where y ≠ 0.

Page No 164:

Question 23:

(b) (0, y),
Any point on the y-axis is of the form (0,y), where y ≠ 0.

Answer:

Substituting the values x = 5, y = 2 in

(a) x + 2y = 7, we get

LHS=5+22=5+4=97=RHSi.e. LHS  RHS

(b) 5x + 2y = 7, we get

LHS=55+22=25+4=297=RHSi.e. LHS  RHS

(c) x + y = 7, we get

LHS=5+2=7=RHSi.e. LHS = RHS

(d) 5x + y = 7, we get

LHS=55+2=25+2=277=RHSi.e. LHS  RHS

Hence, the correct option is (c).

Page No 164:

Question 24:

Substituting the values x = 5, y = 2 in

(a) x + 2y = 7, we get

LHS=5+22=5+4=97=RHSi.e. LHS  RHS

(b) 5x + 2y = 7, we get

LHS=55+22=25+4=297=RHSi.e. LHS  RHS

(c) x + y = 7, we get

LHS=5+2=7=RHSi.e. LHS = RHS

(d) 5x + y = 7, we get

LHS=55+2=25+2=277=RHSi.e. LHS  RHS

Hence, the correct option is (c).

Answer:

Given equation: 3y = ax + 7.
Also, (3, 4) lies on the graph of the equation.
Putting x = 3, y = 4 in the equation, we get:
  3×4 = 3a + 7
⇒  12 = 3a + 7
⇒   3a = 12 - 7 = 5
⇒    a = 53
Hence, the correct answer is option (b).



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