**1. ** Let the cost of a notebook be x.

Let the cost of a pen be Rs. y .

We need to write a linear equation in two variables to represent the statement, “Cost of a notebook is twice the cost of a pen”.

Therefore, we can conclude that the required statement will be x = 2y.

We need to express the linear equation 2x + 3y = in the form ax + by + c = 0 and indicate the values of a, b and c.

2x + 3y = can also be written as 2x + 3y – = 0.

We need to compare the equation 2x + 3y – = 0 with the general equation ax + by + c = 0, to get the values of a, b and c.

Therefore, we can conclude that a = 2, b= 3 and c = –

We need to express the linear equation in the form ax + by + c = 0 and indicate the values of a, b and c.

We need to compare the equation 1 · with the general equation ax + by + c = 0, to get the values of a, b and c.

Therefore, we can conclude that a = 1, and c = -10.

(iii) -2x + 3y = 6

We need to express the linear equation -2x + 3y = 6 in the form ax + by + c = 0 and indicate the values of a, b and c.

-2x + 3y = 6 can also be written as – 2x +3y – 6 = 0.

We need to compare the equation -2x + 3y – 6 = 0 with the general equation ax + by + c = 0, to get the values of a, b and c.

Therefore, we can conclude that a = -2, b = 3 and c = -6.

(iv) x = 3y

We need to express the linear equation x = 3y in the form ax + by + c = 0 and

indicate the values of a, b and c.

x = 3y can also be written as x – 3y + 0 = 0.

We need to compare the equation x – 3y + 0 = 0 with the general equation ax + by + c = 0, to get the values of a, b and c.

Therefore, we can conclude that a = 1, b = -3 and c = 0.

(v) 2x = -5y

We need to express the linear equation 2x = -5y in the form ax + by + c = 0 and

indicate the values of a, b and c.

2x = -5y can also be written as 2x + 5y + 0 = 0.

We need to compare the equation 2x + 5y + 0 = 0 with the general equation ax + by + c = 0, to get the values of a, b and c.

Therefore, we can conclude that a = 2, b = 5 and c = 0 .

(vi) 3x + 2 = 0

We need to express the linear equation 3x + 2 = 0 in the form ax + by + c = 0 and indicate the values of a, b and c.

3x + 2 = 0 can also be written as 3x + 0 · y + 2 = 0.

We need to compare the equation 3x + 0 · y + 2 = 0 with the general equation ax + by + c = 0, to get the values of a, b and c.

Therefore, we can conclude that a = 3, b = 0 and c = 2.

(vii) y – 2 = 0

We need to express the linear equation y – 2 = 0 in the form ax + by + c = 0 and

indicate the values of a, b and c.

y – 2 = 0 can also be written as 0 · x + 1 · y – 2 = 0.

We need to compare the equation 0 · x + 1 · y – 2 = 0 with the general equation ax + by + c = 0, to get the values of a, b and c.

Therefore, we can conclude that a = 0, b = 1 and c = -2 .

(viii) 5 = 2x

We need to express the linear equation 5 = 2x in the form ax + by + c = 0 and indicate the values of a, b and c.

5= 2x can also be written as – 2x + 0 y + 5 = 0.

We need to compare the equation -2x + 0 · y + 5 = 0 with the general equation ax + by + c = 0, to get the values of a, b and c.

Therefore, we can conclude that a = -2, b = 0 and c = 5.

**3. ** We need to the number of solutions of the linear equation y = 3x + 5.

We know that any linear equation has infinitely many solutions. Justification:

If x = 0 then y = 3 × 0 + 5 = 5

If x = 1 then y = 3 × 1 + 5 = 8

If x – 2 then y = 3 × (-2) + 5 = -1

Similarly we can find infinite many solutions by putting the values of x.

**4. ** (i) 2x + y = 7

We know that any linear equation has infinitely many solutions. Let us put x = 0 in the linear equation 2x + y = 7 , to get

2 (0) + y = 7 ⇒ y = 7.

Thus, we get first pair of solution as (0,7).

Let us put x = 2 in the linear equation 2x + y = 7 , to get

2(2) + y = 7 ⇒ y + 4 = 7 ⇒ y = 3.

Thus, we get second pair of solution as (2,3).

Let us put x = 4 in the linear equation 2x + y = 7 , to get

2(4) + y = 7 ⇒ y + 8 = 7 ⇒ y = -1.

Thus, we get third pair of solution as (4, -1).

Let us put x = 6 in the linear equation 2x + y = 7 , to get

2(6) + y = 7 ⇒ y + 12 = 7 ⇒ y = -1.

Thus, we get fourth pair of solution as (6,-5).

Therefore, we can conclude that four solutions for the linear equation 2x + y = 7 are (0, 7), (2, 3), (4, -1) and (6, -5) .

(ii) πx + y = 9

We know that any linear equation has infinitely many solutions. Let us put x = 0 in the linear equation πx + y = 9, to get

π(0) + y = 9 ⇒ y = 9

Thus, we get first pair of solution as (0, 9).

Let us put y = 0 in the linear equation πx + y = 9, to get

Thus, we get second pair of solution as

Let us put x = 1 in the linear equation πx + y = 9 , to get

Thus, we get third pair of solution as

Let us put y = 2 in the linear equation zx + y = 9, to get

Thus, we get fourth pair of solution as

Therefore, we can conclude that four solutions for the linear equation πx + y = 9 are

(iii) x = 4y

We know that any linear equation has infinitely many solutions.

Let us put y = 0 in the linear equation x = 4y , to get

x = 4 (0) ⇒ x = 0

Thus, we get first pair of solution as (0, 0).

Let us put y = 2 in the linear equation x = 4y, to get

x = 4 (2) ⇒ x = 8

Thus, we get second pair of solution as (8, 2).

Let us put y = 4 in the linear equation x = 4y, to get

x = 4 (4) ⇒ x =16

Thus, we get third pair of solution as (16, 4).

Let us put y = 6 in the linear equation x = 4y, to get

x = 4 (6) ⇒ x = 24

Thus, we get fourth pair of solution as (24, 6).

Therefore, we can conclude that four solutions for the linear equation x = 4y are (0, 0), (8, 2), (16, 4) and (24, 6).

**5. **(i) (0,2)

We need to put x = 0 and y = 2 in the L.H.S. of linear equation x – 2y = 4, to get

(0) – 2(2) = -4

∴ L.H.S. ≠ R.H.S.

Therefore, we can conclude that (0, 2) is not a solution of the linear equation

x – 2y = 4.

(ii) (2, 0)

We need to put x = 2 and y = 0 in the L.H.S. of linear equation x – 2y = 4 , to get

(2) – 2(0) = 2

∴ L.H.S. ≠ R.H.S.

Therefore, we can conclude that (2,0) is not a solution of the linear equation

x – 2y = 4 .

(iii) (4, 0)

We need to put x = 4 and y = 0 in the linear equation x – 2y = 4 , to get

(4) – 2(0) = 4

∴ L.H.S. = R.H.S.

Therefore, we can conclude that (4, 0) is a solution of the linear equation

x – 2y = 4.

(iv) (√2, 4√2)

We need to put x = 5 and y = 4-5 in the linear equation x -2y = 4, to get

(√2) – 2(4√2) = -7√2

∴ L.H.S. ≠ R.H.S.

Therefore, we can conclude that (√2, 4√2) is not a solution of the linear equation

x – 2y = 4.

(v) (1, 1)

We need to put x = 1 and y = 1 in the linear equation x – 2y = 4 , to get

(1) – 2(1) = -1

∴ L.H.S. ≠ R.H.S.

Therefore, we can conclude that (1, 1) is not a solution of the linear equation

x – 2y = 4.

**6. ** We know that, if x = 2 and y = 1 is a solution of the linear equation 2x + 3y = k , then on substituting the respective values of x and y in the linear equation 2x + 3y = k, the LHS and RHS of the given linear equation will not be effected.

∴ 2 (2) + 3 (1) = k ⇒ k = 4 + 3 ⇒ k = 7

Therefore, we can conclude that the value of k, for which the linear equation 2x + 3y = k has x = 2 and y = 1 as one of its solutions is 7.

**7. **(i) x + y = 4

We can conclude that x = 0, y = 4; x = 1, y = 3 and x = 2, y = 2 are the solutions of the linear equation x + y = 4.

We can optionally consider the given below table for plotting the linear equation

x + y = 4 on the graph.

x 0 1 2

y 4 3 2

(ii) x – y = 2

We can conclude that x = 0, y = -2; x = 1, y = -1 and x = 2, y = 0 are the solutions of the linear equation x – y = 2.

We can optionally consider the given below table for plotting the linear equation

x – y = 2 on the graph.

x 0 1 2

y -2 -1 0

(iii) y = 3x

We can conclude that x = 0, y = 0; x = 1, y = 3 and x = 2, y = 6 are the solutions of the linear equation y = 3x .

We can optionally consider the given below table for plotting the linear equation y = 3x on the graph.

x 0 1 2

y 0 3 6

(iv) 3 = 2x + y

We can conclude that x = 0, y = 3; x = 1, y = 1 and x = 2, y = -1 are the solutions of the linear equation 3 = 2x + y.

We can optionally consider the given below table for plotting the linear equation 3 = 2x + y on the graph.

x 0 1 2

y 3 1 -1

**8. **We need to give the two equations of the line that passes through the point (2, 14).

We know that infinite number of lines can pass through any given point.

We can consider the linear equations 7x – y = 0 and 2x + y =18 .

We can conclude that on putting the values x = 2 and y = 14 in the above mentioned linear equations, we get LHS = RHS.

Therefore, we can conclude that the line of the linear equations 7x – y = 0 and 28x – 4y = 0 will pass through the point (2, 14).

**9. ** We know that if any point lie on the graph of any linear equation, then that point is the solution of that linear equation.

We can conclude that (3, 4) is a solution of the linear equation 3y = ax + 7.

We need to substitute x = 3 and y = 4 in the linear equation 3y = ax + 7, to get

Therefore, we can conclude that the value of a will be

**10. **From the given situation, we can conclude that the distance covered at the rate Rs 5 per km will be (x – 1), as first kilometer is charged at Rs 8 per km.

We can conclude that the linear equation for the given situation will be:

8 + 5 (x – 1) = y = 8 + 5x – 5 = y = 3 + 5x = y.

We need to draw the graph of the linear equation 3 + 5x = y.

We can conclude that x = 0, y = 3; x = 1, y = 1 and x = 2, y = -1 are the solutions of the linear equation 3 + 5x = y.

We can optionally consider the given below table for plotting the linear equation 3 + 5x = y on the graph.

x 0 -1 -2

y 3 -1 -7

**11.** **For First figure**

(i) y = x

We know that if any point lie on the graph of any linear equation, then that point is the solution of that linear equation.

Let us check whether x = -1, y = 1; x = 0, y = 0 and x = 1, y = -1 are the solutions of

the linear equation y = x.

For x = -1, y = 1, we get

y = x ⇒ -1 ≠ 1

Therefore, the given graph does not belong to the linear equation y = x .

(ii) x + y = 0

We know that if any point lie on the graph of any linear equation, then that point is the solution of that linear equation.

For x = -1, y = 1 , we get

-1 + 1 = 0 ⇒ 0 = 0.

For x = 0, y = 0, we get

0+ 0 = 0 ⇒ 0 = 0.

For x = 1, y = -1, we get

1 + (-1) = 0 ⇒ 1 – 1 = 0 ⇒ 0 = 0.

Therefore, the given graph belongs to the linear equation x + y = 0.

(iii) y = 2x

We know that if any point lie on the graph of any linear equation, then that point is the solution of that linear equation.

For x = -1, y = 1 , we get

y = 2x ⇒ -1 = 2(1) -1≠ 2.

Therefore, the given graph does not belong to the linear equation y = 2x.

(iv) 2 + 3y = 7x

For x = -1, y = 1 , we get

2 + 3(1) = 7(-1) ⇒ 2 + 3 = -7 ⇒ 5 ≠ – 7.

Therefore, the given graph does not belong to the linear equation 2 + 3y = 7x.

**For Second figure **

(i) y = x + 2

We know that if any point lie on the is the solution of that linear equation.

For x = -1, y = 3 , we get

3 = -1 + 2 ⇒ 3 ≠ 1.

Therefore, the given graph does not

(ii) y = x – 2

We know that if any point lie on the is the solution of that linear equation.

For x – 1, y = 3, we get

3 = -1 -2 ⇒ 3 ≠ -3.

Therefore, the given graph does not

(iii) y = -x + 2

We know that if any point lie on the is the solution of that linear equation.

For x = -1**, **y = 3, we get

3 = (-1) + 2 ⇒ 3 = 1 + 2

For x = 0, y = 2, we get

2 -(0) + 2 ⇒ 2 = 2.

For x = 2, y = 0, we get

0 = -(2) **+ **2 ⇒ 0 = 0.

Therefore, hat the given graph belongs to the linear equation y = -x + 2.

(iv) x + 2y = 6

For x = -1, y = 3, we get

(-1) + 2(3) = 6 ⇒ -1 + 6 = 6 ⇒ 5 ≠ 6.

Therefore, the given graph does not belong to the linear equation x + 2y = 6.

**12.** We are given that the work done by a body on application of a constant force is directly proportional to the distance travelled by the body.

Let the work done be W and let constant force be F.

Let distance travelled by the body be D.

According to the question,

W ∝ D ⇒ W = F · D.

We need to draw the graph of the linear equation W = F · D, when the force is constant as 5 units, i.e., W = 5D.

We can conclude that x = 0, y = 0; x = 5, y = 1 and x = 10, y = 2 are the solutions of the linear equation W = 5D.

x 0 5 10

y 0 1 2

Therefore, we can conclude from the above mentioned graph, the work done by the body, when the distance is 2 units will be 10 units and when the distance is 0 units, the work done will be 0 unit.

**13. ** The contribution made by Yamini is Rs x and the contribution made by Fatime is Rs y.

We are given that together they both contributed Rs 100.

We get the given below linear equation from the given situation.

x + y = 100.

We need to consider any 3 solutions of the linear equation x + y = 100, to plot the graph of the linear equation x + y = 100.

We can conclude that x = 0, y = 3; x = 1, y = 1 and x = 2, y = -1 are the solutions of the linear equation x + y = 100 .

We can optionally consider the given below table for plotting the linear equation x + y = 100 on the graph.

x 0 50 100

y 100 50 0

**14. ** We are given a linear equation that converts the temperature in Fahrenheit into degree Celsius.

(i) We need to consider any 3 solutions of the linear equation , to plot the graph of the linear equation

We can conclude that x = 0, y = 3; x = 1, y = 1 and x = 2, y = -1 are the solutions of

C -40 0 40

F -40 32 104

(ii) We need to find the temperature in Fahrenheit, when the temperature in degree Celsius is 30^{0}.

Therefore, we can conclude that the temperature in Fahrenheit will be 86^{0}F.

(iii) We need to find the temperature in degree Celsius, when the temperature in Fahrenheit is 95^{0}.

Therefore, we can conclude that the temperature in degree Celsius will be 35^{0}.

(iv) We need to find the temperature in Fahrenheit, when the temperature in degree Celsius is 0^{0}.

Therefore, we can conclude that the temperature in Fahrenheit will be 32^{0}.

We need to find the temperature in degree Celsius, when the temperature in Fahrenheit is 0^{0}.

Therefore, we can conclude that the temperature in degree Celsius will be -17.77^{0}

(v) We need to find a temperature that is numerically same in both Fahrenheit and degree Celsius.

Therefore, we can conclude that the temperature that is numerically same in Fahrenheit and degree Celsius will be – 40^{0}.

**15. ** (i) We need to represent the linear equation y = 3 geometrically in one variable.

We can conclude that in one variable, the geometric representation of the linear equation y = 3 will be same as representing the number 3 on a number line.

Given below is the representation of number 3 on the number line.

(ii) We need to represent the linear equation y = 3 geometrically in two variables. We know that the linear equation y = 3 can also be written as 0 · x + y =

We can conclude that in two variables, the geometric representation of the linear equation y = 3 will be same as representing the graph of linear equation

0 · x + y = 3.

Given below is the representation of the linear equation 0 · x + y = 3 on a graph.

We can optionally consider the given below table for plotting the linear equation

0 · x + y = 3 on the graph.

x 1 0

y 3 3

**16.** (i) We need to represent the linear equation 2x + 9 = 0 geometrically in one variable.

We know that the linear equation 2x + 9 = 0 can also be written as

We can conclude that in one variable, the geometric representation of the linear equation 2x + 9 = 0 will be same as representing the number -4.5 on a number line. Given below is the representation of number 3 on the number line.

(ii) We need to represent the linear equation 2x + 9 = 0 geometrically in two variables.

We know that the linear equation 2x + 9 = 0 can also be written as 2x + 0 · y = 9.

We can conclude that in two variables, the geometric representation of the linear equation 2x + 9 = 0 will be same as representing the graph of linear equation 2x + 0 · y = 9.

Given below is the representation of the linear equation 2x + 0 · y = 9 on a graph.

We can optionally consider the given below table for plotting the linear equation 2x + 0 · y = 9 on the graph.

x 1 0

y 4.5 4.5