AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Optional Exercise

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Optional Exercise Textbook Questions and Answers

Question 1.
i) \(\frac{2x}{a}\) + \(\frac{y}{b}\) = 2
\(\frac{x}{a}\) – \(\frac{y}{b}\) = 4
Answer:
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 1
Substituting x = 2a in the equation (1) we get
\(\frac{2}{a}\)(2a) + \(\frac{y}{b}\) = 2
⇒ 4 + \(\frac{y}{b}\) = 2
⇒ \(\frac{y}{b}\) = -2
⇒ y = -2b
∴ The solution (x, y) = (2a, -2b)

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

ii) \(\frac{x+1}{2}\) + \(\frac{y-1}{3}\) = 8
\(\frac{x-1}{3}\) + \(\frac{y+1}{2}\) = 9
Answer:
Given: \(\frac{x+1}{2}\) + \(\frac{y-1}{3}\) = 8
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 2
⇒ 3x + 2y + 1 = 48
⇒ 3x + 2y = 47 …… (1)
and \(\frac{x-1}{3}\) + \(\frac{y+1}{2}\) = 9
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 3
⇒ 2x + 3y + 1 = 54
⇒ 2x + 3y = 53 …… (2)
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 4
⇒ y = \(\frac{-65}{-5}\) = 13
Substituting y = 13 in equation (1) we get
3x + 2(13) = 47
⇒ 3x = 47 – 26
⇒ 3x = 21
⇒ x = \(\frac{21}{3}\) = 7
∴ The solution (x, y) = (7, 13)

iii) \(\frac{x}{7}\) + \(\frac{y}{3}\) = 5
\(\frac{x}{2}\) – \(\frac{y}{9}\) = 6
Answer:
Given: \(\frac{x}{7}\) + \(\frac{y}{3}\) = 5 and \(\frac{x}{2}\) – \(\frac{y}{9}\) = 6
⇒ \(\frac{3x+7y}{21}\) = 5 and \(\frac{9x-2y}{18}\) = 6
⇒ 3x + 7y = 105 …….. (1) and
9x – 2y = 108 …….. (2)
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 5
⇒ y = \(\frac{207}{23}\) = 9
Substituting y = 9 in equation (1) we get
3x + 7(9) = 105
⇒ 3x = 105 – 63
⇒ 3x = 42
⇒ x = \(\frac{42}{3}\) = 14
∴ The solution (x, y) = (14, 9)

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

iv) √3x + √2y = √3
√5x + √3y= √3
Answer:
Given that √3x + √2y = √3 …… (1)
√5x + √3y = √3 …… (2)
By following elimination method
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 6
Now again following elimination method
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 7
∴ The solution x = \(\frac{3-\sqrt{6}}{3-\sqrt{10}}\) and y = \(\frac{3-\sqrt{15}}{3-\sqrt{10}}\)

v) \(\frac{ax}{b}\) + \(\frac{by}{a}\) = a + b
ax – by = 2ab
Answer:
Given: \(\frac{ax}{b}\) + \(\frac{by}{a}\) = a + b ……. (1)
ax – by = 2ab …….. (2)
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 8
Substituting y = – a in equation (2)
we get ax – b(-a) = 2ab
ax + ab = 2ab
ax = 2ab – ab = ab
⇒ x = \(\frac{ab}{a}\) = b
∴ (x, y) = (b, -a)

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

vi) 2x + 3y = 17
2x+2 – 3y+1 = 5
Answer:
Given: 2x + 3y = 17 and
2x+2 – 3y+1 = 5
Take 2x = a and 3y = b then the given equations reduce to
2x + 3y = 17 ⇒ a + b = 17 …… (1)
2x . 22 – 3y . 3 = 5 ⇒ 4a – 3b = 5 …… (2)
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 9
Substituting b = 9 in equation (1) we get
a + 9 = 17 ⇒ a = 17 – 9 = 8
But a = 2x – 8 and b = 3y = 9
⇒ 2x = 23 and 3y = 32
⇒ x = 3 and y = 2
∴ The solution (x, y) is (3, 2)

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

Question 2.
Animals in an experiment are to be kept on a strict diet. Each animal is to receive among other things 20g of protein and 6g of fat. The laboratory technicians purchased two food mixes, A and B. Mix A has 10% protein and 6% fat. Mix B has 20% protein and 2% fat. How many grams of each mix should be used?
Answer:
Let x gms of mix A and y gms of mix B are to be mixed, then
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 10
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise 11
Substituting y = 60 in equation (1)
we get x + 2 × 60 = 200
⇒ x + 120 = 200
⇒ x = 200 – 120 = 80 gm
∴ Quantity of mix. A = 80 gms.
Quantity of mix. B = 60 gms.