ML Aggarwal Simultaneous Linear Equations Exe-5.4 Class 9 ICSE Maths Solutions

ML Aggarwal Simultaneous Linear Equations Exe-5.4 Class 9 ICSE Maths Solutions. Step by step solutions  of  Simultaneous Linear Equations problems as council prescribe guideline. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Simultaneous Linear Equations Exe-5.4 Class 9 ICSE Maths Solutions

Board ICSE
Subject Maths
Class 9th
Chapter-5 Simultaneous Linear Equations
Topics Solution of Exe-5.4 Questions
Academic Session 2024-2025

Solution of Exe-5.4 Questions

ML Aggarwal Simultaneous Linear Equations Exe-5.4 Class 9 ICSE Maths Solutions

Solve the following pairs of linear equations (1 to 5):

Question 1.

(i) 2/x + 2/3y = 1/6

2/x – 1/y = 1

(ii) 3/2x + 2/3y = 5

5/x – 3/y = 1.

Answer :

(i) 2/x + 2/3y = 1/6 ….. (1)

2/x – 1/y = 1 ….. (2)

By subtracting both the equations

5/3y = -5/6

By cross multiplication

– 15y = 30

By division

y = 30/ -15 = – 2

Substitute the value of y in equation (1)

2/x + 2/ (3 × (-2)) = 1/6

By further calculation

2/x – 1/3 = 1/6

So we get

2/x = 1/6 + 1/3

Taking LCM

2/x = (1 + 2)/ 6 = 3/6

By cross multiplication

x = (2 × 6)/3 = 12/3 = 4

Therefore, x = 4 and y = – 2.

(ii) 3/2x + 2/3y = 5 ….. (1)

5/x – 3/y = 1 ….. (2)

Multiply equation (1) by 1 and (2) by 2/9

3/2x + 2/3y = 5

10/9x – 2/3y = 2/9

By adding both the equations

(3/2 + 10/9)1/x = 5 + 2/9

Taking LCM

(27 + 20)/ 18 × 1/x = (45 + 2)/ 9

By further calculation

47/18x = 47/9

By cross multiplication

x = (47 × 9)/ (47 × 18) = ½

Substitute the value of x in equation (2)

5/ ½ – 3/y = 1

By further calculation

10 – 3/y = 1

3/y = 10 – 1 = 9

So we get

y = 3/9 = 1/3

Therefore, x = ½ and y = 1/3.

Question 2.

(i) (7x – 2y)/ xy = 5

(8x + 7y)/ xy = 15

(ii) 99x + 101y = 499xy

101x + 99y = 501xy.

Answer :

(i) (7x – 2y)/ xy = 5

(8x + 7y)/ xy = 15

We can write it as

7x/xy – 2y/xy = 5

8x/xy + 7y/xy = 15

By further simplification

7/y – 2/x = 5 …. (1)

8/y + 7/x = 15 ….. (2)

Now multiply equation (1) by 7 and (2) by 2

49/y – 14/x = 35

16/y + 14/x = 30

By adding both the equations

65/y = 65

So we get

y = 65/65 = 1

Substitute the value of y in equation (1)

7/1 – 2/x = 5

By further calculation

2/x = 7 – 5 = 2

So we get

x = 2/2 = 1

Therefore, x = 1 and y = 1.

(ii) 99x + 101y = 499xy

101x + 99y = 501xy

Now divide each term by xy

99x/xy + 101y/xy = 499xy/xy

101y/xy + 99x/xy = 501xy/xy

By further calculation

99/y + 101/x = 499 ….. (1)

101/y + 99/x = 501 ….. (2)

By adding both the equations

200/y + 200/x = 1000

Divide by 200

1/y + 1/x = 5 …… (3)

Subtracting both the equations

-2/y + 2/x = – 2

Divide by 2

-1/y + 1/x = – 1 …. (4)

By adding equation (3) and (4)

2/x = 4

So we get

x = 2/4 = ½

By subtracting equation (3) and (4)

2/y = 6

So we get

y = 2/6 = 1/3

Therefore, x = ½ and y = 1/3 if x ≠ 0, y ≠ 0.

Question 3.

(i) 3x + 14y = 5xy

21y – x = 2xy

(ii) 3x + 5y = 4xy

2y – x = xy.

Answer :

(i) 3x + 14y = 5xy

21y – x = 2xy

Now dividing each equation by xy of x ≠ 0, y ≠ 0

3x/xy + 14y/xy = 5xy/xy

By further calculation

3/y = 14/x = 5 ….. (1)

(ii) 3x + 5y = 4xy

2y – x = xy

We can write it as

3x + 5y = 4xy

– x + 2y = xy

Divide each equation by xy if x≠ 0 and y ≠ 0

3x/xy + 5y/xy = 4xy/xy

So we get

3/y + 5/x = 4 ….. (1)

-x/xy + 2y/xy = xy/xy

So we get

-1/y + 2/x = 1 ….. (2)

Now multiply equation (1) by 1 and (2) by 3

3/y + 5/x = 4

-3/y + 6/x = 3

By adding both the equations

11/x = 7

So we get

x = 11/7

Substitute the value of x in equation (2)

-1/y + 2/11/7 = 1

By further calculation

-1/y + (2 × 7)/ 11 = 1

-1/y + 14/11 = 1

We can write it as

-1/y = 1 – 14/11

Taking LCM

-1/y = (11 – 14)/ 11

So we get

-1/y = -3/11

By cross multiplication

-3y = – 11

y = – 11/-3 = 11/3

Therefore, x = 11/7 and y = 11/3.

Question 4.

(i) 20/ (x + 1) + 4/ (y – 1) = 5

10/ (x + 1) – 4/ (y – 1) = 1

(ii) 3/ (x + y) + 2/ (x – y) = 3

2/ (x + y) + 3/ (x – y) = 11/3.

Answer :

(i) 20/ (x + 1) + 4/ (y – 1) = 5 ….. (1)

10/ (x + 1) – 4/ (y – 1) = 1 ….. (2)

Add equation (1) and (2)

30/ (x + 1) = 6

By cross multiplication

30 = 6 (x + 1)

By further calculation

30/6 = x + 1

5 = x + 1

So we get

x = 5 – 1 = 4

Substitute the value of x in equation (1)

20/ (x + 1) + 4/ (y – 1) = 5

20/ (4 + 1) + 4/ (y – 1) = 5

By further calculation

20/5 + 4/ (y – 1) = 5

4 + 4/ (y – 1) = 5

We can write it as

4/ (y – 1) = 5 – 4 = 1

4/ (y – 1) = 1

By cross multiplication

4 = 1 (y – 1)

So we get

4 = y – 1

y = 4 + 1 = 5

Therefore, x = 4 and y = 5.

(ii) 3/ (x + y) + 2/ (x – y) = 3 …. (1)

2/ (x + y) + 3/ (x – y) = 11/3 ….. (2)

Multiply equation (1) by 3 and (2) by 2

9/ (x + y) + 6/ (x – y) = 9 ….. (3)

4/ (x + y) + 6/ (x – y) = 22/3 ….. (4)

Subtracting both the equations

5/ (x + y) = 9 – 22/3

Taking LCM

5/ (x + y) = 5/3

By cross multiplication

5 × 3 = 5 (x + y)

By further calculation

(5 × 3)/ 5 = x + y

x + y = (3 × 1)/ 3

x + y = 3 …… (5)

Substitute equation (5) in (1)

3/3 + 2/ (x – y) = 3

By further calculation

1 + 2/ (x – y) = 3

2/ (x – y) = 3 – 1 = 2

So we get

2/2 = x – y

Here

1 = x – y ….. (6)

We can write it as

x – y = 1

x + y = 3

By adding both the equations

2x = 4

x = 4/2 = 2

Substitute x = 2 in equation (5)

2 + y = 3

y = 3 – 2 = 1

Therefore, x = 2 and y = 1.

Question 5.

(i) 1/ 2(2x + 3y) + 12/ 7(3x – 2y) = ½

7/ (2x + 3y) + 4/ (3x – 2y) = 2

(ii) 1/ 2(x + 2y) + 5/ 3(3x – 2y) = – 3/2

5/ 4(x + 2y) – 3/ 5 (3x – 2y) = 61/60.

Answer :

(i) 1/ 2(2x + 3y) + 12/ 7(3x – 2y) = ½

7/ (2x + 3y) + 4/ (3x – 2y) = 2

Consider 2x + 3y = a and 3x – 2y = b

We can write it as

1/2a + 12/7b = ½

7/a + 4/b = 2

Now multiply equation (1) by 7 and (2) by ½

7/2a + 12/b = 7/2

7/2a + 2/b = 1

Subtracting both the equations

10/b = 5/2

So we get

b = (10 × 2)/ 5 = 4

Substitute the value of b in equation (2)

7/a + 4/4 = 2

7/a + 1 = 2

So we get

7/a = 2 – 1 = 1

a = 7

Here

2x + 3y = 7 ….. (3)

3x – 2y = 4 ….. (4)

Multiply equation (3) by 2 and (4) by 3

4x + 6y = 14

9x – 6y = 12

So we get

13x = 26

x = 26/13 = 2

Substitute the value of x in (3)

2 × 2 + 3y = 7

By further calculation

4 + 3y = 7

So we get

3y = 7 – 4 = 3

y = 3/3 = 1

Therefore, x = 2 and y = 1.

(ii) 1/ 2(x + 2y) + 5/ 3(3x – 2y) = – 3/2

5/ 4(x + 2y) – 3/ 5 (3x – 2y) = 61/60

Consider x + 2y = a and 3x – 2y = b

1/2a + 5/3b = – 3/2 …. (1)

5/4a – 3/5b = 61/60 ….. (2)

Now multiply equation (1) by 5/2 and (2) by (1)

5/4a + 25/6b = – 15/4

5/4a – 3/5b = 61/60

Subtracting both the equations

25/6b + 3/5b = – 15/4 – 61/60

Taking LCM

(125 + 18)/ 30b = (-225 – 61)/ 60

By further calculation

143/30b = – 286/60

By cross multiplication

30b × (-286) = 60 × 143

So we get

b = (60 × 143)/ (30 × -286) = – 1

Substitute the value of b in equation (1)

1/2a + 5/ (3 × – 1) = -3/2

By further calculation

1/2a – 5/4 = – 3/2

We can write it as

1/2a = – 3/2 + 5/3

Taking LCM

1/2a = (-9+ 10)/ 6 = 1/6

So we get

a = 6/2 = 3

Here

x + 2y = 3 ….. (3)

3x – 2y = – 1 ….. (4)

Adding both the equations

4x = 2

x = 2/4 = ½

Substitute the value of x in equation (3)

½ + 2y = 3

By further calculation

2y = 3 – ½

Taking LCM

2y = 5/2

y = 5/ (2 × 2) = 5/4

Therefore, x = ½ and y = 5/4.

—  : End of ML Aggarwal Simultaneous Linear Equations Exe-5.4 Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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