Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 5 Linear Equations in Two Variables.

## Practice Set 5.1 Algebra 9th Std Maths Part 1 Answers Chapter 5 Linear Equations in Two Variables

Question 1.

By using variables x and y form any five linear equations in two variables.

Answer:

The general form of a linear equation in two variables x and y is ax + by + c = 0,

where a, b, c are real numbers and a ≠ 0, b ≠ 0.

Five linear equations in two variables are as follows:

i. 3x + 4y – 12 = 0

ii. 3x – 4y + 12 = 0

iii. 5x + 5y – 6 = 0

iv. 7x + 12y – 11 = 0

v. x – y + 5 = 0

Question 2.

Write five solutions of the equation x + y = 1.

Answer:

i. x = 1, y = 6

ii. x = -1, y = 8

iii. x = 5, y = 2

iv. x = 0, y = 7

v. x = 10, y = -3

Question 3.

Solve the following sets of simultaneous equations.

i. x + y = 4 ; 2x – 5y = 1

ii. 2x + y = 5 ; 3x – y = 5

iii. 3x – 5y = 16; x – 3y= 8

iv. 2y – x = 0; 10x + 15y = 105

v. 2x + 3y + 4 = 0; x – 5y = 11

vi. 2x – 7y = 7; 3x + y = 22

Solution:

i. Substitution Method:

x + y = 4

∴ x = 4 – y …(i)

2x – 5y = 1 ……(ii)

Substituting x = 4 – y in equation (ii),

2(4 – y) – 5y = 1

∴ 8 – 2y – 5y = 1

∴ 8 – 7y = 1

∴ 8 – 1 = 7y

∴ 7 = 7y

∴ y = \(\frac { 7 }{ 7 }\)

∴ y = 1

Substituting y = 1 in equation (i),

x = 4 – 1 = 3

∴ (3,1) is the solution of the given equations.

Alternate method:

Elimination Method:

x + y = 4 …(i)

2x – 5y = 1 ……(ii)

Multiplying equation (i) by 5,

5x + 5y = 20 … (iii)

Adding equations (ii) and (iii),

2x – 5y = 1

+ 5x + 5y = 20

7 = 21

∴ x = \(\frac { 21 }{ 7 }\)

∴ x = 3

Substituting x = 3 in equation (i),

3 + y = 4

∴ y = 4 – 3 = 1

(3,1) is the solution of the given equations.

ii. 2x + y = 5 …(i)

3x – y = 5 …(ii)

Adding equations (i) and (ii),

2x + y = 5

+ 3x – y = 5

5x = 10

∴ x = \(\frac { 10 }{ 5 }\)

∴ x = 2

Substituting x = 2 in equation (i),

2(2) + y = 5

4 + y = 5

∴ y = 5 – 4 = 1

∴ (2, 1) is the solution of the given equations.

iii. 3x – 5y = 16 …(i)

x – 3y = 8

∴x = 8 + 3y …..(ii)

Substituting x = 8 + 3y in equation (i),

3(8 + 3y) – 5y = 16

24 + 9y- 5y = 16

∴4y= 16 – 24

∴ 4y = -8

∴ y = \(\frac { -8 }{ 4 }\)

y = -2

Substituting y = -2 in equation (ii),

x = 8 + 3 (-2)

∴ x = 8 – 6 = 2

∴ (2, -2) is the solution of the given equations.

iv. 2y – x = 0

∴ x = 2y …(i)

10x + 15y = 105 …(ii)

Substituting x = 2y in equation (ii),

10(2y) + 15y = 105

∴ 20y + 15y = 105

∴ 35y = 105

∴ y = \(\frac { 105 }{ 35 }\)

∴ y = 3

Substituting y = 3 in equation (i),

x = 2y

∴ x = 2(3) = 6

∴ (6, 3) is the solution of the given equations.

v. 2x + 3y + 4 = 0 …(i)

x – 5y = 11

∴x = 11 + 5y …(ii)

Substituting x = 11 + 5y in equation (i),

2(11 +5y) + 3y + 4 = 0

∴ 22 + 10y + 3y + 4 = 0

∴ 13y + 26 = 0

∴ 13y = -26

∴y = \(\frac { -26 }{ 13 }\)

∴ y = -2

Substituting y = -2 in equation (ii),

x = 11 + 5y

∴ x = 11 + 5(-2)

∴ x = 11 – 10 = 1

∴ (1, -2) is the solution of the given equations.

vi. 2x – 7y = 7 …(i)

3x + y = 22

∴ y = 22 – 3x ……(ii)

Substituting y = 22 – 3x in equation (i),

2x – 7(22 – 3x) = 7

∴ 2x – 154 + 21x = 7

∴ 23x = 7 + 154

∴ 23x = 161

∴ x = \(\frac { 161 }{ 23 }\)

∴ x = 7

Substituting x = 7 in equation (ii),

y = 22 – 3x

∴ y = 22 – 3(7)

∴ 7 = 22 -21= 1

∴ (7, 1) is the solution of the given equations.

Question 1.

Solve the following equations. (Textbook pg. no. 80)

i. m + 3 = 5

ii. 3y + 8 = 22

iii. \(\frac { x }{ 3 }\) = 2

iv. 2p = p + \(\frac { 4 }{ 9 }\)

Solution:

i. m + 3 = 5

m = 5 – 3

∴m = 2

ii. 3y + 8 = 22

∴ 3y = 22 – 8

∴ 3y = 14

∴ y = \(\frac { 14 }{ 9 }\)

iii. \(\frac { x }{ 3 }\) = 2

∴ x = 2 × 3

∴ x = 6

iv. 2p = p + \(\frac { 4 }{ 9 }\)

∴ 2p – p = \(\frac { 4 }{ 9 }\)

∴ p = \(\frac { 4 }{ 9 }\)

Question 2.

Which number should be added to 5 to obtain 14? (Textbook pg. no. 80)

Solution:

x + 5 = 14

∴ x = 14 – 5

x = 9

∴ 9 + 5 = 14

Question 3.

Which number should be subtracted from 8 to obtain 2? (Textbook pg. no. 80)

Solution:

8 – y = 2

∴ y = 8 – 2

∴ y = 6

∴ 8 – 6 = 2

Question 4.

x + y = 5 and 2x + 2y= 10 are two equations in two variables. Find live different solutions of x + y = 5, verify whether same solutions satisfy the equation 2x + 2y = 10 also. Observe both equations. Find the condition where two equations in two variables have all solutions in common. (Textbook pg. no. 82)

Solution:

Five solutions of x + y = 5 are given below:

(1,4), (2, 3), (3, 2), (4,1), (0, 5)

The above solutions also satisfy the equation 2x + 2y = 10.

∴ x + y = 5 …[Dividing both sides by 2]

∴ If the two equations are the same, then the two equations in two variables have all solutions common.

Question 5.

3x – 4y – 15 = 0 and y + x + 2 = 0. Can these equations be solved by eliminating x ? Is the solution same? (Textbook pg. no. 84)

Solution:

3x – 4y – 15 = 0

∴ 3x – 4y = 15 …(i)

y + x + 2 = 0

∴ x + y = -2 ……(ii)

Multiplying equation (ii) by 3,

3x + 3y = -6 …(iii)

Subtracting equation (iii) from (i),

∴ y = -3

Substituting y = -3 in equation (ii),

∴ x – 3 = -2

∴ x = – 2 + 3

∴ x = 1

∴ (x, y) = ( 1, -3)

Yes, the given equations can be solved by eliminating x. Also, the solution will remain the same.