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:

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)

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

⇒ 3x + 2y + 1 = 48

⇒ 3x + 2y = 47 …… (1)

and \(\frac{x-1}{3}\) + \(\frac{y+1}{2}\) = 9

⇒ 2x + 3y + 1 = 54

⇒ 2x + 3y = 53 …… (2)

⇒ 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)

⇒ 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)

iv) √3x + √2y = √3

√5x + √3y= √3

Answer:

Given that √3x + √2y = √3 …… (1)

√5x + √3y = √3 …… (2)

By following elimination method

Now again following elimination method

∴ 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)

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)

vi) 2^{x} + 3^{y} = 17

2^{x+2} – 3^{y+1} = 5

Answer:

Given: 2^{x} + 3^{y} = 17 and

2^{x+2} – 3^{y+1} = 5

Take 2^{x} = a and 3^{y} = b then the given equations reduce to

2^{x} + 3^{y} = 17 ⇒ a + b = 17 …… (1)

2^{x} . 2^{2} – 3^{y} . 3 = 5 ⇒ 4a – 3b = 5 …… (2)

Substituting b = 9 in equation (1) we get

a + 9 = 17 ⇒ a = 17 – 9 = 8

But a = 2^{x} – 8 and b = 3^{y} = 9

⇒ 2^{x} = 2^{3} and 3^{y} = 3^{2}

⇒ x = 3 and y = 2

∴ The solution (x, y) is (3, 2)

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

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.