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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

Solution:

2x – y = 3 → (1)

3x + y = 7 → (2)

By adding (1) and (2)

5x + 0 = 10

x = \(\frac{10}{5}\)

x = 2

Substitute the value of x = 2 in (1)

2(2) – y = 3

4 – y = 3

-y = 3 – 4

-y = -1

y = 1

The value of x = 2 and y = 1

(ii) x – y = 5; 3x + 2y = 25

Solution:

x – y = 5 → (1)

3x + 2y = 25 → (2)

(1) × 2 ⇒ 2x – 2y = 10 → (3)

(2) × 1 ⇒ 3x + 2y = 25 → (2)

(3) + (2) ⇒ 5x + 0 = 35

x = \(\frac{35}{5}\)

= 7

Substitute the value of x = 7 in (1)

x – y = 5

7 – y = 5

-y = 5 – 7

-y = -2

y = 2

∴ The value of x = 7 and y = 2

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

Solution:

\(\frac{x}{10}\) + \(\frac{y}{5}\) = 14

LCM of 10 and 5 is 10

Multiply by 10

x + 2y = 140 → (1)

\(\frac{x}{8}\) + \(\frac{y}{6}\) = 15

LCM of 8 and 6 is 24

3x + 4y = 360 → (2)

(1) × 2 ⇒ 2x + 4y = 280 → (3)

(2) × 1 ⇒ 3x + 4y = 360 → (2)

(3) – (2) ⇒ -x + 0 = -80

∴ x = 80

Substitute the value of x = 80 in (1)

x + 2y = 140

80 + 2y = 140

2y = 140 – 80

2y = 60

y = \(\frac{60}{2}\)

y = 30

∴ The value of x = 80 and y = 30

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

Solution:

3(2x + y) = 7xy

6x + 3y = 7xy

Divided by xy

3a + 6b = 7 → (1)

3(x + 3y) = 11xy

3x + 9y = 11xy

Divided by xy

9a + 3b = 11 → (2)

(1) × 3 ⇒ 9a + 18b = 21 → (3)

(2) × 1 ⇒ 9a + 3b = -11 → (2)

(3) – (2) ⇒ 15b = 10

b = \(\frac{10}{15}\) = \(\frac{2}{3}\)

Substitute the value of b = \(\frac{2}{3}\) in (1)

3a + 6 × \(\frac{2}{3}\) = 7

3a + 4 = 7

3a = 7 – 4

3a = 3

a = \(\frac{3}{3}\)

= 1

But \(\frac{1}{x}\) = a

\(\frac{1}{x}\) = 1

x = 1

But \(\frac{1}{y}\) = b

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

2y = 3

y = \(\frac{3}{2}\)

∴ The value of x = 1 and y = \(\frac{3}{2}\)

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

Solution:

Let \(\frac{1}{x}\) = a

4a + 5y = 7 → (1)

3a + 4y = 5 → (2)

(1) × 4 ⇒ 16a + 20y = 28 →(3)

(2) × 5 ⇒ 15a + 20y = 25 → (4)

(3) – (4) ⇒ a + 0 = 3

a = 3

Substitute the value of a = 3 in (1)

4(3) + 5y = 7

5y = 7 – 12

5y = -5

5y = \(\frac{-5}{5}\) = -1

But \(\frac{1}{x}\) = a

\(\frac{1}{x}\) = 3

3x = 1 ⇒ x = \(\frac{1}{3}\)

The value of x = \(\frac{1}{3}\) and y = -1

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

Solution:

13x + 11y = 70 → (1)

11x + 13y = 74 → (2)

(1) + (2) ⇒ 24x + 24y = 144

x + y = 6 → (3) (Divided by 24)

(1) – (2) ⇒ 2x – 2y = -4

x – y = -2 → (4) (Divided by 2)

(4) + (3) ⇒ 2x = 4

x = \(\frac{4}{2}\)

= 2

Substitute the value x = 2 in (3)

2 + y = 6

y = 6 – 2

= 4

∴ The value of x = 2 and 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 Rs 5,000 per month, find the monthly income of each.

Solution:

Let the income of “A” be “x” and the income of “B” be “y”.

By the given first condition

x : y = 3 : 4

4x = 3y (Product of the extreme is equal to the product of the means)

4x – 3y = 0 → (1)

Expenditure of A = x – 5000

Expenditure of B = y – 5000

By the given second condition

x – 5000 : y – 5000 = 5 : 7

7(x – 5000) = 5(y – 5000)

7x – 35000 = 5y – 25000

7x – 5y = -25000 + 35000

7x – 5y = 10000 → (2)

(1) × 5 ⇒ 20x – 15y = 0 → (3)

(2) × 3 ⇒ 21x – 15y = 30000 → (4)

(3) – (4) ⇒ x + 0 = 30000

x = 30000

Substitute the value of x in (1)

4 (30000) – 3y = 0

120000 = 3y

y = \(\frac{120000}{3}\) = 40000

∴ Monthly income of A is Rs 30,000

Monthly income of B is Rs 40,000

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 age of a man be “x” and the age of a son be “y”

5 years ago

Age of a man = x – 5 years

Age of his son = y – 5 years

By the given first condition

x – 5 = 7(y – 5)

x – 5 = 7y – 35

x – 7y = -35 + 5

x – 7y = -30 → (1)

Five years hence

Age of a man = x + 5 years

Age of his son = y + 5 years

By the given second condition

x + 5 = 4 (y + 5)

x + 5 = 4y + 20

x – 4y = 20 – 5

x – 4y = 15 → (2)

(1) – (2) ⇒ -3y = -45

3y = 45

y = \(\frac{45}{3}\)

= 15

Substitute the value of y = 15 in (1)

x – 7(15) = -30

x – 105 = -30

x = -30 + 105

= 75

Age of the man is 75 years

Age of his son is 15 years