Class 8 Maths Solution | Lesson 12 Equation and Graph | Curriculum Development Centre (CDC)

Equation and Graph

Exercise 12.1

1. Solve the following pairs of equations using the graphical method and check the answer.

(a) x + y = 5, x - y = 3

Solution:

Here,

x + y = 5 ........(i)

and x - y = 3 ......(ii)

From equation (i),

x + y = 5

or, y = 5 - x

Substituting x = 3, 4, 5 in y = 5 - x,

| x | 3 | 4 | 5 |

| y | 2 | 1 | 0 |

Thus, points of the line are (3, 2), (4, 1), (5, 0).

Similarly, from equation (ii),

x - y = 3

or, y = x - 3

Substituting x = 3, 4, 5 in y = x - 3,

| x | 3 | 4 | 5 |

| y | 0 | 1 | 2 |

Thus, points of the line are (3, 0), (4, 1), (5, 2).

Hence, intersection point is (4, 1).

(b) 3x + y = 7, x = 2y

Solution:

Here,

3x + y = 7 ........(i)

and x = 2y ......(ii)

From equation (i),

3x + y = 7

or, y = 7 - 3x

Substituting x = 1, 2, 3 in y = 7 - 3x,

| x | 1 | 2 | 3 |

| y | 4 | 1 | -2 |

Thus, points of the line are (1, 4), (2, 1), (3, -2).

From equation (ii),

x = 2y

or, y = x / 2

Substituting x = 2, 4, 6 in y = x / 2,

| x | 2 | 4 | 6 |

| y | 1 | 2 | 3 |

Thus, points of the line are (2, 1), (4, 2), (6, 3).

Hence, intersection point is (2, 1).

(c) x + y = 13, 2x - y = 8

From equation x + y = 13:

Substituting x = 5, 6, 7 in y = 13 - x:

| x | 5 | 6 | 7 |

| y | 8 | 7 | 6 |

Points of the line: (5, 8), (6, 7), (7, 6).

From equation 2x - y = 8:

Substituting y = 8, 7, 6 in x = (y + 8) / 2:

| x | 8 | 7 | 7 |

| y | 8 | 7 | 6 |

Points of the line: (8, 8), (7, 7), (7, 6).

Intersection point: x = 6, y = 7.

(d) x + y = 6, x - y = 2

Solution:

Here,

x + y = 6 ........(i)

and x - y = 2 ......(ii)

From equation (i),

x + y = 6

or, y = 6 - x

Substituting x = 2, 3, 4 in y = 6 - x,

| x | 2 | 3 | 4 |

| y | 4 | 3 | 2 |

Thus, points of the line are (2, 4), (3, 3), (4, 2).

From equation (ii),

x - y = 2

or, y = x - 2

Substituting x = 3, 4, 5 in y = x - 2,

| x | 3 | 4 | 5 |

| y | 1 | 2 | 3 |

Thus, points of the line are (3, 1), (4, 2), (5, 3).

Hence, intersection point is (4, 2).

(e) x + y = 8, x – y = 4

Solution:

Here,

x + y = 8 ........(i)

and x - y = 4 ......(ii)

From equation (i),

x + y = 8

or, y = 8 - x

Substituting x = 4, 5, 6 in y = 8 - x,

| x | 4 | 5 | 6 |

| y | 4 | 3 | 2 |

Thus, points of the line are (4, 4), (5, 3), (6, 2).

From equation (ii),

x - y = 4

or, y = x - 4

Substituting x = 5, 6, 7 in y = x - 4,

| x | 5 | 6 | 7 |

| y | 1 | 2 | 3 |

Thus, points of the line are (5, 1), (6, 2), (7, 3).

Hence, intersection point is (6, 2).

(f) 4x + y = 2, 3x – 2y = 7

Solution:

Here,

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

and 3x – 2y = 7 ......(ii)

From equation (i),

4x + y = 2

or, y = 2 - 4x

Substituting x = 0, 1, -1 in y = 2 - 4x,

| x | 0 | 1 | -1 |

| y | 2 | -2 | 6 |

Thus, points of the line are (0, 2), (1, -2), (-1, 6).

From equation (ii),

3x – 2y = 7

or, y = (3x - 7) / 2

Substituting x = 1, 3, 5 in y = (3x - 7) / 2,

| x | 1 | 3 | 5 |

| y | -2 | 1 | 4 |

Thus, points of the line are (1, -2), (3, 1), (5, 4).

Hence, intersection point is (1, -2).

(g) x + 2y = 6, 2y – x = 2

Solution:

Here,

x + 2y = 6 ........(i)

and 2y – x = 2 ......(ii)

From equation (i),

x + 2y = 6

or, y = (6 - x) / 2

Substituting x = 0, 2, 4 in y = (6 - x) / 2,

| x | 0 | 2 | 4 |

| y | 3 | 2 | 1 |

Thus, points of the line are (0, 3), (2, 2), (4, 1).

From equation (ii),

2y – x = 2

or, y = (x + 2) / 2

Substituting x = 0, 2, 4 in y = (x + 2) / 2,

| x | 0 | 2 | 4 |

| y | 1 | 2 | 3 |

Thus, points of the line are (0, 1), (2, 2), (4, 3).

Hence, intersection point is (2, 2).

(h) 3x + 2y = 4, x – 3y = 5

Solution:

Here,

3x + 2y = 4 ........(i)

and x – 3y = 5 ......(ii)

From equation (i),

3x + 2y = 4

or, y = (4 - 3x) / 2

Substituting x = 0, 2, 4 in y = (4 - 3x) / 2,

x | 0 | 2 | 4 |

---|---|---|---|

y | 2 |-1 |-4 |

Thus, points of the line are (0, 2), (2, -1), (4, -4).

From equation (ii),

x – 3y = 5

or, 3y = x - 5

or, y = (x - 5)/3

Substituting x = 2, 5, 8 in y = (x - 5)/3,

x | 2 | 5 | 8 |

---|---|---|---|

y |-1 | 0 | 1 |

Thus, points of the line are (2, -1), (5, 0), (8, 1).

Hence, intersection point is (2, -1).

(i) 2x = 5 + 3y, 5y = 2x – 3

Solution:

Here,

2x = 5 + 3y ........(i)

and 5y = 2x – 3 ......(ii)

From equation (i),

2x = 5 + 3y

or, x = (5 + 3y) / 2

Substituting y = 1, 3, 5 in x = (5 + 3y) / 2,

x | 4 | 7 | 10 |

---|---|---|----|

y | 1 | 3 | 5 |

Thus, points of the line are (4, 1), (7, 3), (10, 5).

From equation (ii),

5y = 2x – 3

or, y = (2x – 3) / 5

Substituting x = 4, 7, 10 in y = (2x – 3) / 5,

x | 4 | 7 | 10 |

---|----|----|----|

y | 1 | 3 | 5 |

Thus, points of the line are (4, 1), (7, 3), (10, 5).

Hence, intersection point is (4, 1).

(j) 2x – 1 = y, 3x – 2y = 0

Solution:

Here,

2x – 1 = y ........(i)

and 3x – 2y = 0 ......(ii)

From equation (i),

2x – 1 = y

or, y = 2x – 1

Substituting x = 1, 2, 3 in y = 2x – 1,

x | 1 | 2 | 3 |

---|---|---|---|

y | 1 | 3 | 5 |

Thus, points of the line are (1, 1), (2, 3), (3, 5).

From equation (ii),

3x – 2y = 0

or, 2y = 3x

or, y = (3x) / 2

Substituting x = 2, 4, 6 in y = (3x) / 2,

x | 2 | 4 | 6 |

---|---|---|---|

y | 3 | 6 | 9 |

Thus, points of the line are (2, 3), (4, 6), (6, 9).

Hence, intersection point is (2, 3).

(k) x + 3 = 2y, 2x + y = 14

Solution:

Here,

x + 3 = 2y ........(i)

and 2x + y = 14 ......(ii)

From equation (i),

x + 3 = 2y

or, x = 2y - 3

Substituting y = 1, 2, 3, 4 in x = 2y - 3,

x | -1 | 1 | 3 | 5 |

---|----|---|---|---|

y | 1 | 2 | 3 | 4 |

Thus, points of the line are (-1, 1), (1, 2), (3, 3), (5, 4).

From equation (ii),

2x + y = 14

or, y = 14 - 2x

Substituting x = 4, 5, 6 in y = 14 - 2x,

x | 4 | 5 | 6 |

---|---|---|---|

y | 6 | 4 | 2 |

Thus, points of the line are (4, 6), (5, 4), (6, 2).

Hence, intersection point is (5, 4).

(l) x – 2y = 5, 2x + 3y = 10

Solution:

Here,

x – 2y = 5 ........(i)

and 2x + 3y = 10 ......(ii)

From equation (i),

x – 2y = 5

or, x = 2y + 5

Substituting y = 0, 1, 2 in x = 2y + 5,

| x | 5 | 7 | 9 |

|---|---|---|---|

| y | 0 | 1 | 2 |

Thus, points of the line are (5, 0), (7, 1), (9, 2).

From equation (ii),

2x + 3y = 10

or, y = (10 - 2x) / 3

Substituting x = 1, 2, 4 in y = (10 - 2x) / 3,

| x | 1 | 2 | 4 |

|---|---|---|---|

| y | 2 | 2 | 0 |

Thus, points of the line are (1, 2), (2, 2), (4, 0).

Hence, intersection point is (5, 0).

2. Solve the following pair of equations using the graphical method and check the answer.

a. If the sum of two numbers is 15 and the difference between them is 5, find the numbers.

Solution:

Let the two numbers be x and y.

According to the question:

x + y = 15 .................. (i)

x - y = 5 ................... (ii)

From equation (i):

x + y = 15

or, x = 15 - y

Substituting y = 0, 5, and 10 into x = 15 - y:

| x | 15 | 10 | 5 |

|---|----|----|---|

| y | 0 | 5 | 10 |

Thus, the points of the line for equation (i) are: (15, 0), (10, 5), (5, 10).

From equation (ii):

x - y = 5

or, x = y + 5

Substituting y = 0, 5, and 10 into x = y + 5:

| x | 5 | 10 | 15 |

|---|----|----|----|

| y | 0 | 5 | 10 |

Thus, the points of the line for equation (ii) are: (5, 0), (10, 5), (15, 10).

By solving the equations, we find the intersection point:

The first number (x) = 10

The second number (y) = 5.

b. The sum of two numbers is 12. If the greater is 3 times more than the smaller, find the numbers.

Solution:

Let the two numbers be x (greater) and y (smaller).

According to the question:

x + y = 12 .................. (i)

x = 3y ...................... (ii)

From equation (i):

x + y = 12

or, x = 12 - y

Substituting y = 0, 3, and 6 into x = 12 - y:

| x | 12 | 9 | 6 |

|---|----|----|----|

| y | 0 | 3 | 6 |

Thus, the points of the line for equation (i) are: (12, 0), (9, 3), and (6, 6).

From equation (ii):

x = 3y

Substituting y = 0, 3, and 6 into x = 3y:

| x | 0 | 9 | 18 |

|---|----|----|----|

| y | 0 | 3 | 6 |

Thus, the points of the line for equation (ii) are: (0, 0), (9, 3), and (18, 6).

By solving the equations, we find the intersection point:

The greater number (x) = 9

The smaller number (y) = 3.

c. The difference between the two numbers is 5. If five times the smaller is equal to 4 times the greater, then find the numbers.

Solution:

Let the two numbers be x (greater) and y (smaller).

According to the question:

x - y = 5 ................... (i)

5y = 4x ..................... (ii)

From equation (i):

x - y = 5

or, x = y + 5

Substituting y = 10, 15, and 20 into x = y + 5:

| x | 15 | 20 | 25 |

|---|----|----|----|

| y | 10 | 15 | 20 |

Thus, the points of the line for equation (i) are: (15, 10), (20, 15), and (25, 20).

From equation (ii):

5y = 4x

or, y = (4/5)x

Substituting x = 5, 10, and 25 into y = (4/5)x:

| x | 5 | 10 | 25 |

|---|----|----|----|

| y | 4 | 8 | 20 |

Thus, the points of the line for equation (ii) are: (5, 4), (10, 8), and (25, 20).

By solving the equations, we find the intersection point:

The greater number (x) = 25

The smaller number (y) = 20.

d. The cost of buying 3 notebooks and 4 pens is Rs. 200. The cost of buying 5 notebooks and 2 pens is Rs. 240. Find the cost of a notebook and a pen.

Solution:

Let the cost of a notebook be x and the cost of a pen be y.

According to the question:

3x + 4y = 200 ............... (i)

5x + 2y = 240 ............... (ii)

From equation (i):

3x + 4y = 200

or, x = (200 - 4y) / 3

Substituting y = 20, 40, and 60 into x = (200 - 4y) / 3:

| x | 40 | 32 | 24 |

|---|----|----|----|

| y | 20 | 40 | 60 |

Thus, the points of the line for equation (i) are: (40, 20), (32, 40), and (24, 60).

From equation (ii):

5x + 2y = 240

or, x = (240 - 2y) / 5

Substituting y = 20, 40, and 60 into x = (240 - 2y) / 5:

| x | 40 | 32 | 24 |

|---|----|----|----|

| y | 20 | 40 | 60 |

Thus, the points of the line for equation (ii) are: (40, 20), (32, 40), and (24, 60).

By solving the equations, we find the intersection point:

The cost of a notebook (x) = Rs. 40

The cost of a pen (y) = Rs. 20.

e. Father’s present age is 3 less than three times the present age of the daughter. If the difference between the age of the father and daughter is 37 years, find their present age.

Solution:

Let the father's age be x and the daughter's age be y.

According to the question:

x = 3y - 3 ................... (i)

x - y = 37 .................. (ii)

From equation (i):
Substituting y = 10, 20, and 30 into x = 3y - 3:
If y = 10, x = 3(10) - 3 = 27
If y = 20, x = 3(20) - 3 = 57
If y = 30, x = 3(30) - 3 = 87

Thus, the points of the line for equation (i) are: (27, 10), (57, 20), and (87, 30).

From equation (ii):
Substituting y = 10, 20, and 30 into x = y + 37:
If y = 10, x = 10 + 37 = 47
If y = 20, x = 20 + 37 = 57
If y = 30, x = 30 + 37 = 67

Thus, the points of the line for equation (ii) are: (47, 10), (57, 20), and (67, 30).

By solving the equations, we find the intersection point:
Father's age (x) = 57
Daughter's age (y) = 20.

f. The present age of Kamala is 5 years more than that of Bimala. The age of Kamala after five years will be two times the present age of Bimala. Find their present age.

Solution:

Let Kamala's present age be x and Bimala's present age be y.

According to the question:

x = y + 5 .................... (i)

x + 5 = 2y ............. (ii)

From equation (i):

x = y + 5

Substituting y = 0, 10, and 20 into x = y + 5:

| x | 5 | 15 | 25 |

|---|----|----|----|

| y | 0 | 10 | 20 |

Thus, the points of the line for equation (i) are: (5, 0), (15, 10), and (25, 20).

From equation (ii):

x + 5 = 2y

or, x = 2y - 5

Substituting y = 0, 10, and 20 into x = 2y - 5:

| x | -5 | 15 | 35 |

|---|----|----|----|

| y | 0 | 10 | 20 |

Thus, the points of the line for equation (ii) are: (-5, 0), (15, 10), and (35, 20).

By solving the equations, we find the intersection point:

Kamala's age (x) = 15

Bimala's age (y) = 10.

g.

Solution:

Let Bipin's present age be x and Bipana's present age be y.

According to the question:

x = y + 6 .................... (i)

x - 2 = 2(y - 2) ............. (ii)

From equation (i):

x = y + 6

Substituting x = 6, 16, 26, and 14 into y = x - 6:

| x | 6 | 16 | 26 | 14 |

|----|----|----|----|----|

| y | 0 | 10 | 20 | 8 |

Thus, the points of the line for equation (i) are: (6, 0), (16, 10), (26, 20), and (14, 8).

From equation (ii):

x - 2 = 2(y - 2)

or, x = 2y - 2

Substituting x = 6, 16, 26, and 14 into y = (x + 2) / 2:

| x | 6 | 16 | 26 | 14 |

|----|----|----|----|----|

| y | 4 | 9 | 14 | 8 |

Thus, the points of the line for equation (ii) are: (6, 4), (16, 9), (26, 14), and (14, 8).

By solving the equations, we find the intersection point:

Bipin's age (x) = 14

Bipana's age (y) = 8.

h. The age difference between Kusum and her father is 20 years. If the father’s age is more than two times the age of Kusum by 4 years, then find their age.

Solution:

Let the father's age be x and Kusum's age be y.

According to the question:

x - y = 20 ................... (i)

x = 2y + 4 ................... (ii)

From equation (i):

x = y + 20

Substituting y = 0, 10, 20, and 16 into x = y + 20:

| x | 20 | 30 | 40 | 36 |

|----|----|----|----|----|

| y | 0 | 10 | 20 | 16 |

Thus, the points of the line for equation (i) are: (20, 0), (30, 10), (40, 20), and (36, 16).

From equation (ii):

x = 2y + 4

Substituting y = 0, 10, 20, and 16 into x = 2y + 4:

| x | 4 | 24 | 44 | 36 |

|----|----|----|----|----|

| y | 0 | 10 | 20 | 16 |

Thus, the points of the line for equation (ii) are: (4, 0), (24, 10), (44, 20), and (36, 16).

By solving the equations, we find the intersection point:

Father's age (x) = 36

Kusum's age (y) = 16.

i. In an exam, Ram scored 20 marks more than Shyam. If Ram's score is two times Shyam's score, then find their scores.

Solution:

Let Ram's score be x and Shyam's score be y.

According to the question:

x = y + 20 ................... (i)

x = 2y ....................... (ii)

From equation (i):

x = y + 20

Substituting y = 0, 10, 20, and 20 into x = y + 20:

| x | 20 | 30 | 40 | |----|----|----|----| | y | 0 | 10 | 20 |

Thus, the points of the line for equation (i) are: (20, 0), (30, 10), and (40, 20).

From equation (ii):

x = 2y

Substituting y = 0, 10, 20, and 20 into x = 2y:

| x | 0 | 20 | 40 | |----|----|----|----| | y | 0 | 10 | 20 |

Thus, the points of the line for equation (ii) are: (0, 0), (20, 10), and (40, 20).

By solving the equations, we find the intersection point:

Ram's score (x) = 40

Shyam's score (y) = 20.