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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.12

Question 1.

Solve by the method of elimination

(i) 2x – y = 3; 3x + y = 7

(ii) x – y = 5; 3x + 2y = 25

(iii) \(\frac{x}{10}+\frac{y}{5}\) = 14; \(\frac{x}{8}+\frac{y}{6}\) = 15

(iv) 3(2x + y) = 7xy; 3(x + 3y) = 11xy

(v) \(\frac{4}{x}\) + 5y = 7; \(\frac{3}{x}\) + 4y = 5

(vi) 13x + 11y = 70; 11x + 13y = 74

Solution:

(i) 2x – y = 3 ………….. (1)

3x + y = 7 ………… (2)

Substitute x = 2 in (1)

2(2) – y = 3

4 – y = 3

-y = 3 – 4

-y = -1

∴ Solution: x = 2; y = 1

Verification:

Substitute x = 2, y = 1 in (2)

3(2) + 1 = 7 = RHS

∴ Verified.

Substitute y = 2 in (1)

x – 2 = 5

x = 5 + 2

x = 7

∴ Solution: x = 7, y = 2

Verification:

Substitute x = 7, y = 2 in (2)

3(7) + 2(2) = 21 + 4 = 25 = RHS

∴ Verified.

Substitute y = 30 in (1)

x + 2 (30) = 140

x + 60 = 140

x = 140 – 60

x = 80

∴ Solution: x = 80; y = 30

Verification:

Substitute x = 80, y = 30 in (2)

3(80) + 4(30) = 240 + 120 = 360 = RHS

∴ Verified.

(iv) 3(2x +y) = 7xy ⇒ 6x + 3y = 7xy ………. (1)

3(x + 3y) = 11xy ⇒ 3x + 9y = 11xy ………….. (2)

Substitute a = 1 in (5)

6b + 3(1) = 7

6b + 3 = 7

6b = 7 – 3

b = \(\frac{4}{6}=\frac{2}{3}\)

∴a = \(\frac{1}{x}\) = 1 ⇒ x = 1

b = \(\frac{1}{y}=\frac{2}{3}\) ⇒ y = \(\frac{3}{2}\)

∴ Solution: x = 1; y = \(\frac{3}{2}\)

Substitute y = 4 in (1)

13x + 11 (4) = 70

13x + 44 = 70

13x = 70 – 44 = 26

x = \(\frac{26}{13}\) = 2

∴ Solution: x = 2; y = 4

Question 2.

The monthly income of A and B are in the ratio 3:4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 5,000 per month, find the monthly income of each.

Solution:

Let the monthly income of A and B be 3x and 4x respectively.

Let the monthly expenditure of A and B be 5y and 7y respectively.

∴ 3x – 5y = 5000 ……… (1)

4x – 7y = 5000 ……….. (2)

Substitute y = 5000 in (1)

3x – 5 (5000) = 5000

3x – 25000 = 5000

3x = 5000 + 25000

3x = 30000

x = 10000

∴ Monthly income of A is 3x = 3 × 10000 = ₹ 30000

Monthly income of B is 4x = 4 × 10000 = ₹ 40000

Question 3.

Five years ago, a man was seven times as old as his son, while five year hence, the man will be four times as old as his son. Find their present age.

Solution:

Let the man’s present age = x

Five years ago his age is = x – 5

Let his son’s age be = y

5 years ago his son’s age = y – 5

∴ x – 5 = 7(y – 5)

x – 5 = 7y – 35

x – 7y = -35 + 5

x – 7y = – 30 ……….. (1)

After 5 years, man’s age will be = x + 5

His son’s age will be = y + 5

∴ x + 5 = 4(y + 5)

x + 5 = 4y + 20

x – 4y = 20 – 5

⇒ x – 4y = 15 ………….. (2)

Substitute y = 15 in (1)

x – 7 (15) = -30

x – 105 – 30

x = – 30 + 105

x = 75

∴ Man’s Age = 75, His son’s Age =15