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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