Students can practice 10th Class Maths Solutions Telangana Chapter 4 Pair of Linear Equations in Two Variables InText Questions to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables InText Questions

Think – Discuss

Question 1.

Two situations are given below. (Page No. 73)

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

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

Identify the unknowns in each situation. We observe that there are two unknowns in each case.

Solution:

i) Let the cost of 1 kg potatoes be ₹ x.

The cost of 1 kg tomatoes be ₹ y.

Given the cost of 1 kg potatoes and 2 kg tomatoes = ₹ 30

∴ x + 2y = 30

Given the cost of 2 kg potatoes and 4 kg tomatoes = ₹ 66

∴ 2x + 4y = 66

The pair of linear equations in two variables are x + 2y = 30 and 2x + 4y = 66.

We observe that there are two unknowns in this case.

ii) Let the cost of one bat be ₹ x.

The cost of one ball be ₹ y.

Given the cost of 3 bats and 6 balls = ₹ 3,900

3x + 6y = 3,900

Given 4 bats and 2 balls cost = ₹ 1,300

∴ 4x + 2y = 1,300

The pair of linear equations in two variables are 3x + 6y = 3,900 and 4x + 2y = 1300

We observe that there are two unknowns in this case.

Question 2.

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

Answer:

If the lines intersect at a point gives the unique solution of the equations.

If the lines coincide then there are infinitely many solutions each point on the line being a solution. No, a dependent pair of linear equations are always consistent.

Try This

Mark the correct option in the following questions :

Question 1.

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 (Page No. 75)

Solution:

a) 5 + 4x = y + 3

4x + 5 = y + 3

4x + 5 – y – 3 = 0

4x – y + 2 = 0

This equation is in the form of ax + by + c = 0 where a, b, c are real numbers.

So, the given equation is a linear equation.

b) x + 2y = y – x

x + 2y – y + x = 0

2x + y = 0

This equation is in the form of ax + by + c = 0 where a, b, c are real numbers.

So, the given equation is a linear equation.

c) 3 – x = y^{2} + 4

3 – x – y^{2} – 4 = 0

– x – y^{2} – 1 = 0

x + y^{2} + 1 = 0

This equation is not in the form of ax + by + c = 0 where a, b, c are real numbers.

∴ The given equation is not a linear equation.

d) x + y = 0

This equation is in the form of ax + by + c = 0 where a, b, c are real numbers.

So, the given equation is a linear equation.

Question 2.

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

a) 2x + 1 = y – 3

b) 2t – 1 = 2t + 5

c) 2x – 1 = x^{2}

d) x^{2} – x + 1 = 0 (Page No. 76)

Solution:

a) 2x + 1 = y – 3

⇒ 2x + 1 – y + 3 = 0

⇒ 2x – y + 4 = 0

Is a linear equation in two variables.

They are x and y.

b) 2t – 1 = 2t + 5

⇒ 2t – 1 – 2t – 5 = 0

⇒ -6 = 0 (False)

Is not a linear equation in one variable.

c) 2x – 1 = x^{2} ⇒ x^{2} – 2x + 1 = 0

Is a linear equation in one variable i.e., ‘x’.

d) x^{2} – x + 1 = 0

Is a linear equation in one variable.

i.e., ‘x’.

Question 3.

Which of the following numbers is a solution for the equation

2(x + 3) = 18 ? ( b )

a) 5

b) 6

c) 13

d) 21 (Page No. 76)

Solution:

Given equation 2(x + 3) = 18

a) At x = 5; 2 (5 + 3) = 18

2 × 8 = 18

16 = 18 (False) Not a solution.

b) At x = 6; 2(6 + 3) = 18

2 × 9 = 18

18 = 18 (True)

x = 6 is a solution for the given equation.

c) At x = 13; 2(13 + 3) = 18

2 × 16 = 18

32 = 18 (False) Not a solution.

d) At x = 21; 2(21 + 3) = 18

2 × 24 = 18

48 = 18 (False) Not a solution.

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 ( c ) (Page No. 76)

Solution:

Given equation 2x – (4 – x) = 5 – x

a) At x = 4.5; 2 (4.5) – (4 – 4.5) = 5 – 4.5

9 – (-0.5) = (0.5)

9 + 0.5 = 0.5

9.5 = 0.5 (False)

∴ Value of x does not satisfies the equation.

b) At x = 3 ; 2(3) – (4 – 3) = 5 – 3

6 – 1 = 2

5 = 2 (False)

∴ Value of x does not satisfies the equation.

c) At x = 2.25 ; 2(2.25) – (4 – 2.25)

= 5 – 2.25

4.50-1.75 = 2.75

2.75 = 2.75 (True)

∴ Value of x satisfies the equation.

d) At x = 0.5 ; 2(0.5) – (4 – 0.5) = 5 – 0.5

1 – 3.5 = 4.5

– 2.5 = 4.5 (False)

∴ Value of x does not satisfies the equation.

Question 5.

The equation x – 4y = 5 has

a) no solution

b) unique solution

c) two solutions

d) infinitely many solutions (d) (Page No. 76)

Solution:

Only one equation with two unknowns (variables) we can find many solutions.

Question 6.

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

Solution:

No, we cannot find the cost of each bat and ball because the equations are geometrically shown by a pair of coincident lines. Every point on the line is a common solution to both the equations.

Question 7.

For what value of ‘p’ for the following pair of equations has a unique solution.

2x + py = -5 and 3x + 3y = -6 (Page No. 83)

Solution:

Given equations are 2x + py = -5 and 3x + 3y = -6

a_{1} = 2 ; b_{1} = p; c_{1} = -5

a_{2} = 3; b_{2} = 3 ; c_{2} = -6

\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{2}{3}\) ; \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{p}{3}\) ; \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{-5}{-6}\)

Given the pair of equations has a unique solution.

∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) ≠ \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) (∵ p ≠ 2)

If p = except 2 then the given pair of equations has a unique solution.

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 pair of equations

2x – ky + 3 = 0 and 4x + 6y – 5 = 0

a_{1} = 2; b_{1} = -k; c_{1} = 3

a_{2} = 4; b_{2} = 6; c_{2} = -5

Given the pair of lines are parallel.

∴ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)

\(\frac{a_1}{a_2}\) = \(\frac{\mathrm{b}_1}{\mathrm{~b}_2}\) ⇒ \(\frac{2}{4}\) = \(\frac{-\mathrm{k}}{6}\)

-4k = 2 × 6

-4k = 12

∴ k = \(\frac{12}{-4}\) = -3

Question 9.

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

Solution:

Given pair of equations 3x + 4y + 2 = 0 and 9x + 12y + k = 0 (A.P.Mar.’16) (A.P. Jun.’15)

Given the pair of lines are coincident.

∴ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)

\(\frac{a_1}{a_2}\) = \(\frac{c_1}{c_2}\) ⇒ \(\frac{3}{9}\) = \(\frac{2}{\mathrm{k}}\) ⇒ \frac{1}{3}\(\) = \(\frac{2}{k}\)

∴ k = 3 × 2 = 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

Solution:

Given pair of equations are px + 3y – (p – 3) = 0 and 12x + py – p = 0

a_{1} = p; b_{1} = 3; c_{1} = -(p – 3);

a_{2} = 12; b_{2} = p; c_{2} = -p

Given equations has infinitely many solutions.

∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\)

⇒ \(\frac{\mathrm{p}}{12}\) = \(\frac{3}{p}\) = \(\frac{-(p-3)}{-p}\)

⇒ \(\frac{\mathrm{p}}{12}\) = \(\frac{3}{p}\) = \(\frac{-\mathrm{p}+3}{-\mathrm{p}}\) ⇒ \(\frac{\mathrm{p}}{12}\) = \(\frac{3}{p}\) = \(\frac{\mathrm{p}-3}{\mathrm{p}}\)

⇒ p × p = 3 × 12

⇒ p^{2} = 36

⇒ p = \(\sqrt{36}\) = 6

Do This

Question 1.

Solve the following systems of equations : (Page No. 79)

i) x – 2y = 0

3x + 4y = 20

Solution:

i) x – 2y = 0

-2y = -x

y = \(\frac{x}{2}\)

3x+ 4y = 20

4y = 20 – 3x

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

Scale :X-axis – 1 unit = 1 cm

Y – axis – 1 unit = 1 cm

The two lines are intersecting lines, meet at (4, 2).

The solution set is {(4, 2)}.

ii)

x + y = 2

2x + 2y = 4

Solution:

x + y = 2

y = 2 – x

2x + 2y = 4

2y = 4 – 2x

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

Scale : X – axis – 1 unit = 1 cm

Y – axis – 1 unit = 1 cm

Scale : X – axis – 1 unit = 1 cm

Y – axis – 1 unit = 1 cm

These two lines are coincident lines.

∴ There are infinitely many solutions.

iii) 2x – y = 4; 4x – 2y = 6

Solution:

2x – y = 4 ⇒ y = 2x – 4

4x – 2y = 6 ⇒ 2y = 4x – 6

⇒ y = 2x – 3

These two are parallel lines.

∴ The pair of linear equations has no solution

Scale: X – axis – 1 unit = 1 cm

Y – axis – 1 unit = 1 cm

Question 2.

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

x + 2y – 4 = 0 and 2x + 4y – 12 = 0.

Represent this situation graphically. (Page No. 79)

Solution:

x + 2y – 4 = 0; 2y = 4 – x

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

x + 2y – 4 = 0

2x + 4y – 12 = 0

⇒ 4y = 12 – 2x ⇒ 4y = 2(6 – x)

⇒ y = \(\frac{6-x}{2}\)

∴ 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. (A.P. Mar.16’) (Page No. 83)

i) 2x + 3y = 1

3x – y = 7

Solution:

Let a_{1}x + b_{1}y + c_{1} = 0 \(\simeq\) 2x + 3y – 1 = 0

a_{2}x + b_{2}y + c_{2} = 0 \(\simeq\) 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.

Scale: X – axis – 1 unit = 1 cm

Y – axis – 1 unit = 1 cm

2x + 3y = 1

⇒ 3y = 1 – 2x

⇒ y = \(\frac{1-2 x}{3}\)

∴ The system of equation has a unique solution (2, -1).

ii) x + 2y = 6

2x + 4y = 12

Solution:

From the given pair of equations,

\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{1}{2}\) ; \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{2}{4}\) ; \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\)

∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\)

∴ The lines are dependent and have infinitely many solutions.

x + 2y = 6 ⇒ 2y = 6 – x

⇒ y = \(\frac{6-x}{2}\)

2x + 4y = 12

⇒ 4y = 12 – 2x ⇒ y = \(\frac{12-2 \mathrm{x}}{4}\)

Scale: X – axis – 1 unit = 1 cm

Y – axis – 1 unit = 1 cm

iii) 3x + 2y = 6

6x + 4y = 18

Solution:

From the given equations

\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\) ; \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\); \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{6}{18}\) = \(\frac{1}{3}\)

∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\)

⇒ The lines are parallel and hence no solution.

3x + 2y = 6 ⇒ 2y = 6 – 3x

⇒ y = \(\frac{6-3 x}{2}\)

6x + 4y = 18 ⇒ 4y = 18 – 6x

⇒ y = \(\frac{18-6 \mathrm{x}}{4}\)

Scale : X – axis – 1 unit = 1 cm

Y – axis – 1 unit = 1 cm

Do Thiš

Question 1.

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

i) 3x – 5y = -1

x – y = -1

Solution:

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 = -1

⇒ -2y = -1 + 3

⇒ 2y = -2

⇒ y = -1

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

3x – 5(-1) = -1

3x + 5 = -1

3x = -6

x = -2

∴ The solution is (-2, -1)

ii) x + 2y = -1

2x – 3y = 12

Solution:

Given x + 2y = -1 —- (1)

2x – 3y = 12 —– (2)

From Equation (1), x + 2y = -1

⇒ 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).

iii) 2x + 3y = 9

3x + 4y = 5

Solution:

Given : 2x + 3y = 9 —– (1)

3x + 4y = 5 —– (2)

From equation (1),

2x = 9 – 3y

⇒ x = \(\frac{9-3 y}{2}\)

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

3(\(\frac{9-3 y}{2}\)) + 4y = 5

⇒ \(\frac{27-9 y+2 \times 4 y}{2}\) = 5

⇒ 27 – 9y + 8y = 5 × 2

⇒ -y = 10 – 27

∴ y = 17

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

2x + 3 (+ 17) = 9

⇒ 2x = 9 – 51 ⇒ 2x = -42

⇒ x = -21

∴ The solution is (-21, 17).

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

3x – \(\frac{8}{y}\) = 5

Solution:

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).

v) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

Solution:

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

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

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

⇒ 26 – 6y + 5y = 23

⇒ -y + 26 = 23

⇒ y = 26 – 23 = 3

Substituting y = 3 in equation (1). we get

⇒ 2x + 3(3) = 13

⇒ 2x + 9 = 13

⇒ 2x = 13 – 9

⇒ 2x = 4

⇒ x = 2

∴ The solution is (2, 3).

vi) \(\sqrt{2}\)x + \(\sqrt{3}\)y = 0

\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0

Solution:

Given:

\(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 —–(1)

\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0 —— (2)

∴ The solution is x = 0, y = 0.

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

a_{2}x + b_{2}y + c_{2} = 0

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

Question 2.

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

i) 8x + 5y = 9

3x + 2y = 4

Solution:

Given : 8x + 5y = 9 —- (1)

3x + 2y = 4 —- (2)

Substituting y = 5 in equation (1), we get

8x + 5 × 5 = 9

⇒ 8x = 9 – 25

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

∴ The solution is (-2, 5).

ii) 2x + 3y = 8

4x + 6y = 7

Solution:

Given: 2x + 3y = 8 —- (1)

4x + 6y = 7 —- (2)

The lines are parallel.

∴ The pair of lines has no solution.

iii) 3x + 4y = 25

5x – 6y = -9

Solution:

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 3.

In a competitive exam, 3 marks are to be 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). Use the elimination method. (Page No. 91)

Solution:

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 4.

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 by the substitution method.

(Page No. 92)

Solution:

The Equation formed are

x – 7y + 42 = 0 —- (1)

x – 3y – 6 = 0 — (2)

From (1), x = -42 + 7y

Substituting in equation (2), we get

x = -42 + 7y

⇒ -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