You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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)
Samacheer Kalvi 9th Maths Chapter 3 Algebra Ex 3.12 1
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.

Samacheer Kalvi 9th Maths Chapter 3 Algebra Ex 3.12 2
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.

Samacheer Kalvi 9th Maths Chapter 3 Algebra Ex 3.12 3
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)
Samacheer Kalvi 9th Maths Chapter 3 Algebra Ex 3.12 4
Samacheer Kalvi 9th Maths Chapter 3 Algebra Ex 3.12 5
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 = \(\frac{3}{2}\)

Samacheer Kalvi 9th Maths Chapter 3 Algebra Ex 3.12 6
Samacheer Kalvi 9th Maths Chapter 3 Algebra Ex 3.12 7
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

Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.12

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)
Samacheer Kalvi 9th Maths Chapter 3 Algebra Ex 3.12 8
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)
Samacheer Kalvi 9th Maths Chapter 3 Algebra Ex 3.12 9
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

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