## AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables InText Questions

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

### 10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables InText Questions and Answers

Question 1.

Solve the following systems of equations: i) x – 2y = 0; 3x + 4y = 20 (Page No. 79)

Answer:

i) x – 2y = 0; 3x + 4y = 20

-2y = -x 4y = 20 – 3x

y = \(\frac{x}{2}\) y = \(\frac{20-3x}{4}\)

The two lines meet at (4, 2).

The solution set is {(4, 2)}

ii) x + y = 2

2x + 2y = 4

Answer:

x + y = 2

2x + 2y = 4

These two are coincident lines.

∴ There are infinitely many solutions.

iii) 2x – y = 4

4x – 2y = 6

Answer:

2x – y = 4 4x – 2y = 6

⇒ y = 2x – 4 ⇒ 2y = 4x – 6 ⇒ y = 2x – 3

These two are parallel lines.

∴ The pair of linear equations has no solution.

Question 2.

Two rails of a railway track are represented by the equations.

x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation graphically. (Page No. 79)

Answer:

x + 2y – 4 = 0 2x + 4y – 12 = 0

2y = 4 – x 4y = 12 – 2x (or) 4y = 2 (6 – x)

y = \(\frac{4-x}{2}\) y = \(\frac{6-x}{2}\)

x + 2y – 4 = 0 2x + 4y – 12 = 0

These lines are parallel and hence no solution.

Question 3.

Check each of the given systems of equations to see if it has a unique solution, infinitely many solutions or no solution. Solve them graphically. (Page No. 83)

i) 2x + 3y = 1

3x – y = 7

Answer:

Let a_{1}x + b_{1}y – c_{1} = 0 ≃ 2x + 3y – 1 = 0

a_{2}x + b_{2}y + c_{2} = 0 ≃ 3x – y – 7 = 0

Now comparing their coefficients i.e., \(\frac{a_{1}}{a_{2}}\) and \(\frac{b_{1}}{b_{2}}\)

⇒ \(\frac{2}{3}\) ≠ \(\frac{3}{-1}\)

The given lines are intersecting lines.

2x + 3y = 1 3x – y = 7

3y = 1 – 2x y – 3x = 7

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

The system of equations has a unique solution (2, – 1).

ii) x + 2y = 6

2x + 4y = 12

Answer:

From the given pair of equations,

\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);

\(\frac{c_{1}}{c_{2}}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\)

∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)

∴ The lines are dependent and have infinitely many solutions.

x + 2y = 6 2x + 4y = 12

2y = 6 – x 4y = 12 – 2x (or) 4y = 2(6 – x)

y = \(\frac{6-x}{2}\) y = \(\frac{12-2x}{4}\) or y = \(\frac{6-x}{2}\)

iii) 3x + 2y = 6

6x + 4y = 18

Answer:

From the given pair of equations,

\(\frac{a_{1}}{a_{2}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);

\(\frac{c_{1}}{c_{2}}\) = \(\frac{6}{18}\) = \(\frac{1}{3}\)

∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\)

∴ The lines are parallel and hence no solution.

3x + 2y = 6 6x + 4y = 18

2y = 6 – 3x 4y = 18 – 6x

y = \(\frac{6-3x}{2}\) y = \(\frac{18-6x}{4}\)

Try these

(Page No. 75, 76)

Question 1.

Mark the correct option in the following questions:

Which of the following equations is not a linear equation?

a) 5 + 4x = y + 3

b) x + 2y = y – x

c) 3 – x = y^{2} + 4

d) x + y = 0

Answer:

[ c ]

Question 2.

Which of the following is a linear equation in one variable?

a) 2x + 1 = y – 3

b) 2t – 1= 2t + 5

c) 2x – 1 = x^{2}

d) x^{2} – x + 1 =0

Answer:

[ b ]

Question 3.

Which of the following numbers is a solution for the equation 2(x + 3) = 18?

a) 5

b) 6

c) 13

d) 21

Answer:

[b]

Question 4.

The value of x which satisfies the equation 2x – (4 – x) = 5 – x is

a) 4.5

b) 3

c) 2.25

d) 0.5

Answer:

[ c ]

Question 5.

The equation x – 4y = 5 has

a) no solution

b) unique solution

c) two solutions

d) infinitely many solutions

Answer:

[ d ]

Question 6.

In the example given above, can you find the cost of each bat and ball? (Page No. 79)

Answer:

We can’t find the exact values for the costs of bat and ball as there are infinitely many possibilities.

Question 7.

For what value of ‘p’ the following pair of equations has a unique solution. (Page No. 83)

2x + py = – 5 and 3x + 3y = – 6

Answer:

Given: 2x + py = – 5

3x 4- 3y = – 6

To have an unique solution we should have

∴ The pair has unique solution when p ≠ 2.

Question 8.

Find the value of ‘k’ for which the pair of equations 2x – ky + 3 = 0, 4x + 6y – 5 = 0 represent parallel lines. (Page No. 83)

Answer:

Given: 2x – ky + 3 = 0

4x + 6y – 5 = 0

If the above lines are to be parallel, then

∴ k = – 3 is the required value for which lines are parallel.

Question 9.

For what value of ‘k’, the pair of equation 3x + 4y + 2 = 0 and 9x + 12y + k = 0 represent coincident lines. (Page No. 83)

Answer:

Given: 3x + 4y + 2 = 0

9x + 12y + k = 0

If the lines are to be coincident with each other, then

∴ k = 2 × 3 = 6

Question 10.

For what positive values of ‘p’ the following pair of linear equations have infinitely many solutions? (Page No. 83)

px + 3y – (p – 3) = 0

12x + py – p = 0

Answer:

Given: px + 3y – (p – 3) = 0

12x + py – p = 0

The above equations to have infinitely many solutions

p.p = 12 × 3

⇒ p^{2} = 36

⇒ p = ±6

Think & Discuss

Question 1.

Two situations are given below:

i) The cost of 1 kg potatoes and 2 kg tomatoes was Rs. 30 on a certain day. After two days, the cost of 2 kg potatoes and 4 kg tomatoes was found to be Rs. 66.

ii) The coach of a cricket team of M.K. Nagar High School buys 3 bats and 6 balls for Rs. 3900. Later he buys one more bat and 2 balls for Rs. 1300.

Identify the unknowns in each situation. We observe that there are two unknowns in each case. (Page No. 73)

Answer:

i) The unknowns in the first problem are

a) cost of 1 kg tomatoes

b) cost of 1 kg potatoes

ii) In the second problem, the unknowns are

a) cost of each bat

b) cost of each ball

Question 2.

Is a dependent pair of linear equations always consistent? Why or why not? (Page No. 79)

Answer:

Reason: \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\) always holds. In other words, they have infinitely many solutions.

Do these

Solve each pair of equations by using the substitution method. (Page No. 88)

Question 1.

3x – 5y = -1

x – y = -1

Answer:

Given: 3x – 5y = -1 ……. (1)

x – y = -1 …….. (2)

From equation (2), x – y = – 1

x = y – 1

Substituting x = y – 1 in equation (1)

we get

3 (y – 1) – 5y = – 1

⇒ 3y – 3 – 5y = r 1

⇒ – 2y = – 1 + 3

⇒ 2y = – 2

⇒ y = -1

Substituting y = – 1 in equation (1) we get

3x – 5 (- 1) = -1

3x + 5 = – 1

3x = – 1 – 5

x = \(\frac{-6}{3}\) = -2

∴ The solution is (-2, -1)

Question 2.

x + 2y = – 1

2x – 3y = 12

Answer:

Given: x + 2y = -1 ……. (1)

2x – 3y = 12 …….. (2)

From equation (1)x + 2y = -l

⇒ x = – 1 – 2y

Substituting x = – 1 – 2y in equation (2), we get

2 (- 1 – 2y) – 3y = 12

– 2 – 4y – 3y = 12

– 2 – 7y = 12

7y = – 2 – 12

∴ y = \(\frac{-14}{7}\) = -2

Substituting y = – 2 in equation (1), we get

x + 2 (- 2) = – 1

x = – 1 + 4

x = 3

∴ The solution is (3, – 2)

Question 3.

2x + 3y = 9

3x + 4y = 5

Answer:

Given: 2x + 3y – 9 …….. (1)

3x + 4y = 5 ……. (2)

From equation (1);

2x = 9 – 3y

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

Substituting x = \(\frac{9-3y}{2}\) in equation (2)

we get

Substituting y = + 17 in equation (1) we get

2x + 3 (+ 17) = 9

⇒ 2x = 9 – 51

⇒ 2x = -42

⇒ x = -21

∴ The solution is (-21, 17)

Question 4.

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

3x – \(\frac{8}{y}\) = 5

Answer:

Given:

x + \(\frac{6}{y}\) = 6 …….. (1)

3x – \(\frac{8}{y}\) = 5 …….. (2)

From equation (1) x = 6 – \(\frac{6}{y}\)

Substituting x = 6 – \(\frac{6}{y}\) in equation (2)

we get

Substituting y = 2 in equation (1) we get

x + \(\frac{6}{2}\) = 6 ⇒ x + 3 = 6

∴ x = 3

∴ The solution is (3, 2)

Question 5.

0.2x + 0.3y =1.3

0.4x + 0.5y = 2.3

Answer:

Given:

0.2x + 0.3y = 1.3

⇒ 2x + 3y = 13 …… (1)

0.4x + 0.5y = 2.3

⇒ 4x + 5y = 23 …… (2)

From equation (1)

2x = 13 – 3y

⇒ x = \(\frac{13-3y}{2}\)

Substituting x = \(\frac{13-3y}{2}\) equation (2) we get

\(\frac{13-3y}{2}\) + 5y = 23

⇒ 26 – 6y + 5y = 23

⇒ -y + 26 = 23

⇒ y = 26 — 23 = 3

Substituting y = 3 in equaion (1) we get

2x + 3(3) = 13

⇒ 2x + 9 = 13

⇒ 2x = 13 – 9

⇒ 2x = 4

⇒ x = \(\frac{4}{2}\) = 2

∴ The solution is (2, 3)

Question 6.

√2x + √3y = 0

√3x – √8y = 0

Answer:

Given:

√2x + √3y = 0 ……. (1)

√3x – √8y = 0 ……. (2)

Substitute x = 0 in (1),

√2(0) + √3y = 0

√3y = 0

∴ y = 0

∴ The solution is x = 0, y = 0

Note: a_{1}x + b_{1}y + c_{1} = 0

a_{2}x + b_{2}y + c_{2} = 0, if c_{1} = c_{2} = 0

then, x = 0, y = 0 is a solution.

Solve each of the following pairs of equations by the elimination method. (Page No. 89)

Question 7.

8x + 5y = 9

3x + 2y = 4

Answer:

Given: 8x + 5y = 9 ……. (1)

3x + 2y = 4 …….. (2)

∴ y = 5

Substituting y = 5 in equation (1) we get

8x + 5 × 5 = 9

⇒ 8x = 9 – 25

x = \(\frac{-16}{8}\) = -2

∴ The solution is (- 2, 5)

Question 8.

2x + 3y = 8

4x + 6y = 7

Answer:

Given: 2x + 3y = 8 ……. (1)

4x + 6y = 7 …….. (2)

The lines are parallel.

∴ The pair of lines has no solution.

Question 9.

3x + 4y = 25

5x – 6y = -9

Answer:

Given: 3x + 4y = 25 ……. (1)

5x – 6y = -9 …….. (2)

Substituting y = 4 in equation (1) we get

3x + 4 × 4 = 25

3x = 25 – 16

⇒ x = \(\frac{9}{3}\) = 3

∴ (3,4) is the solution for given pair of lines.

Question 10.

In a competitive exam, 3 marks are awarded for every correct answer and for every wrong answer, 1 mark will be deducted. Madhu scored 40 marks in this exam. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer, Madhu would have scored 50 marks. How many questions were there in the test? (Madhu attempted all the questions) .

Now use the elimination method to solve the above example – 9.

Answer:

The equations formed are

3x – y = 40 ……. (1)

4x – 2y = 50 …….. (2)

Substituting y = 5 in equation (1) we get

3x – 5 = 40

⇒ 3x = 40 + 5

⇒ x = \(\frac{45}{3}\) = 15

Total number of questions = Number of correct questions + Number of wrong answers

= x + y

= 15 + 5 = 20

Question 11.

Mary told her daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Find the present age of Mary and her daughter. Solve example – 10 by the substitution method.

Answer:

The equations formed are

The equations formed are

x – 7y + 42 = 0 ……. (1)

x – 3y – 6 = 0 …….. (2)

From (1), x = – 42 + 7y

Substituting x = – 42 + 7y in equation (2) we get

-42 + 7y – 3y – 6 = 0

⇒ 4y – 48 = 0

⇒ y = \(\frac{48}{4}\) = 12

Substituting y = 12 in equation (2) we get

x – 3 × 12 – 6 = 0

x – 36 – 6 = 0

x = 42

Try this

Question 1.

Solve the given pair of linear equations, (a – b)x + (a + b)y = a^{2} – 2ab – b^{2}

(a + b) (x + y) = a^{2} + b^{2} (Page No. 89)

Answer:

x(-2b) = – 2b(a + b)

⇒ x = (a + b)

Put this value of ‘x’ in eq (1) we get

(a – b) (a + b) + (a + b)y = a^{2} – 2ab – b^{2}

a^{2} – b^{2} + (a + b)y = a^{2} – 2ab – b^{2}

⇒ y = \(\frac{-2ab}{a+b}\)

∴ Solution to given pair of linear equations x = a + b, y = \(\frac{-2ab}{a+b}\)