ML Aggarwal Solutions Class 9 Math Fifth Chapter Simultaneous Linear Equations Exercise 5.4

ML Aggarwal Solutions Class 9 Math 5th Chapter Simultaneous Linear Equations Exercise 5.4

ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Fifth Chapter Simultaneous Linear Equations Exercise 5.4. APC Solution Class 9 Exercise 5.4.

 

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

2/x – 1/y = 1

Solution:

2/x + 2/3y = 1/6

Let, 1/x = p, 1/y = m

∴ 2p + 2m/3 = 1/6

Or, 6p + 2m = 1/2 —- (i)

2/x – 1/y = 1

Or, 2p – m = 1 — (ii)

Multiplying equation (ii) by 2 then adding them we get,

4p – 2m = 2
6p + 2m = 1/2
____________________
10p = 2 + 1/2

Or, p = 4+1/2×10

Or, p = 5/2×10

Or, p = 1/4

Substituting the value of p  in equation (ii) we get,

2 × 1/4 – m = 1

Or, m = 1 – 1/2

Or, m = 1/2

Substituting the value of p & m into x and y we get,

p = 1/x

Or, 1/x = 1/4

Or, x = 4

m = 1/2

Or, 1/y = 1/2

Or, y = 2

 

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

5/x – 3/y = 1

Solution:

3/2x + 2/3y = 5,

Let, 1/x = m, 1/y = n

∴ 3m/2 + 2n/3 = 5

Or, 9m+4n/6 = 5

Or, 9m + 4n = 30 — (i)

5/x – 3/y = 1

5m – 3n = 1 —- (ii)

Multiplying equation (i) with 4 & (ii) with 3 then adding then we get,

20m – 12n = 4
27m + 12n = 90
_________________
47m = 94

Or, m = 2

Substituting the value of m in equation (i) we get,

9×2 + 4n = 30

Or, 4n = 30 – 18

Or, n = 12/4

Or, n = 3

 

Substituting the value of m & n with 1/x & 1/y we get,

m = 2

Or, 1/x = 2

Or, x = 1/2

n = 3

Or, 1/y = 3

Or, y = 3

 

(2) (i) 7x-2y/xy = 5

8x+7y/xy = 15

Solution:

(i) 7x-2y/xy = 5

Or, 7x/xy – 2y/xy = 5

Or, 7/y – 2/x = 5

8x + 7y/xy = 15

Or, 8x/xy + 7y/xy = 15

Or, 8/y + 7/x = 15

Let, 1/x = m, 1/y = n

∴ 7n – 2m = 5 —- (i) 8n + 7m = 15 —– (ii)

Multiplying equation (i) with 7 & 2 with 2 then adding there we get,

49n – 14m = 35
16n + 14m = 30
___________________
55n = 55

Substituting the value of n in equation (i) we get,

7 × 1 – 2m = 5

Or, -2m = 5 – 7

Or, -m = -2/2

Or, m = 1

Substituting the values of m & n as 1/x & 1/y we get

n = 1

Or, 1/y = 1

Or, y = 1

m = 1

Or, 1/x = 1

Or, x = 1

 

(ii) 99x + 101y = 499xy

101x + 99y = 501xy

Solution:

99x + 101y = 449xy

Or, 99x+101y/xy = 499

Or, 99/y + 101/x = 499

Let, 1/x = m, 1/y = n

∴ 99n + 101m = 499 —- (i)

101x + 99y = 501xy

Or, 101x+99y/xy = 501

Or, 101/y + 99/x = 501

101n + 99m = 501 —- (ii)

Adding equation (i) & (ii) we get,

99n + 101m = 499
101n + 99m = 501
_________________
200n + 200m = 1000

Or, 200 (n + m) = 1000

Or, n + m = 5 —- (iii)

 

Subtracting equation (ii) from equation (i) we get,

101n + 99m = 501
99n + 101m = 499
(-)   (-)       (-)
___________________
2n – 2m = 2

Or, 2 (n – m) = 2

Or, n – m = 1 — (iv)

Adding equation (iii) & (iv) we get,

n + m = 5
n – m = 1
___________
2n = 6

Or, n = 3

Substituting the value of n in equation (iii) we get,

3 + m = 5

Or, m = 2

substituting the values of m & n as 1/x, 1/y we get,

n = 3

Or, 1/y = 3

Or, y = 1/3

m = 2

Or, 1/x = 2

Or, x = 1/2

 

(3) (i) 3x +14y = 5xy

21y – x = 2xy

Solution:

3x + 14y = 5xy

Or, 3x + 14y/xy = 5

Or, 3/y + 14/x = 5

Or, 14/x + 3/y = 5 —- (i)

21y – x = 2xy

Or, 21y-x/xy = 2

Or, 21/x – 1/y = 2 — (ii)

∴ 14m + 3n = 5 —- (iii)

Let, 1/x = m, 1/y = n

21m – n = 2 (iv)

Multiplying equation (iv) with 3 then adding equation (iii) & (iv) we get,

14m + 3n = 5
63m – 3n = 6
________________
77m = 11

Or, m= 1/7

Substituting the value of m in equation (i) we get,

14 × 1/7 + 3n = 5

Or, 3n = 5 – 2

Or, 3n = 3

Or, n = 1

Substituting the value of m & n as 1/y we get,

m = 1/7

Or, 1/x = 1/7

Or, x = 7

n = 1

Or, 1/y = 1

Or, y = 1

 

(ii) 3x + 5y = 4xy

2y – x = xy

Solution:

3x + 5y = 4xy

Or, 3x+5y/xy = 4

Or, 3/y + 5/x = 4

Or, 5/x + 3/y = 4 —- (i)

2y – 2 = xy

Or, 2y-x/xy = 1

Or, 2/x – 1/y = 1 —- (ii)

∴ 5m + 3n = 4 —- (iii)

2m – n = 1 — (iv)

Multiplying equation (iv) with 3 then adding equation (iii) & (iv) we get,

5m + 3n = 4
6m – 3n = 3
________________
11m = 7

Or, m = 7/11

Substituting the value of m in equation (iii) we get

2 × 7/11 – n = 1

Or, n = 14/11 – 1

Or, n = 3/11

Substituting the value of m & n as 1/x & 1/y we get,

m = 7/11

Or, 9 1/x = 7/11

Or, x = 11/7

n = 3/11

Or, 1/4 = 3/11

Or, y = 11/3

 

(4) (i) 20/x+1 + 4/y-1 = 5

10/x+1 – 4/y-1 = 1 

Solution:

20/x+1 + 4/y-1 = 5

10/x+1 – 4/y-1 = 1

Let, 1/x+1 = m, 1/y-1 = n

∴ 20m + 4n = 5 —- (i)

10m – 4n = 1 —- (ii)

Adding equation (i) & (ii) we get,

20m + 4n = 5
10m – 4n = 1
________________
30m = 6

Or, m = 1/5

Substituting the value of m in equation (ii) we get,

10 × 1/8 – 4n = 1

Or, 4n = 2 – 1

Or, n = 1/4

Substituting the value of m & n as 1/x+1, 1/y-1, we get

m = 1/5

Or, 1/x+1 = 1/5

Or, x+1 = 5

Or, x = 4

n = 1/4

Or, 1/y-1 = 1/4

Or, y – 1 = 4

Or, y = 5

 

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

2/x+y + 3/x-y = 11/3

Solution:

3/x+y + 2/x-y = 3, 2/x+y + 3/x-y = 11/3

Let, 1/x+y = m, 1/x-y = n

∴3m + 2n = 3 —- (i), 2m + 3n = 11/3 —– (ii)

Multiplying equation (i) with 2 & (ii) with 3 we get, then subtract equation (ii) from (i)

6m + 4n = 6
6m + 9n = 11
(-)    (-)    (-)
_____________
-5n = -5

Or, n = 1

Substituting the value of m in equation (i) we get,

3m + 2 × 1 = 3

Or m = 3-2/3

Or, m = 1/3

Putting the values of m & n

n = 1

Or, 1/x-y = 1

Or, x – y = 1 —– (iii)

m = 1/3

Or, 1/x+y = 1/3

Or, x + y = 3 — (iv)

Adding equation (iii) & (iv) we get,

x-b = 1
x+y = 3
__________
2x = 4

Or, x = 2

Substituting the value of x in equation (iv) we get,

2 + y = 3

Or, y = 1

 

(5) (i) 1/2(2x+3y) + 12/7(3x-2y) = 1/2

7/2x + 3y + 4/3x – 2y = 2

Solution:

1/2(2x+3y) + 12/7(3x-2y) = 1/2, 7/2x+3y + 4/3x-2y = 2

Let, 1/2x+3y = m, 1/3x-2y = n

∴ m/2 + 12n/7 = 1/2, 7m + 4n = 2 —- (ii)

Or, 7m+24n/14 = 1/2

Or, 7m + 24n = 7 —- (i)

Subtracting equation (ii) from equation (i) we get,

7m + 24n = 7
7m + 4n = 2
(-)  (-)    (-)
__________________
20n = 5

Or, n = 1/4

Substituting the value of m in equation (i) we get,

7m + 24 × 1/4 = 7

Or, m = 1/7

Putting the values of m & n as 1/2x+3y & 1/3x-2y we get

n = 1/4

Or, 1/3x – 24 = 4 —– (iii)

m= 1/7

Or, 1/2x + 3y = 1/7 —- (iv)

Multiplying equation (iii) with 3 & (iv) with 2 then adding these we get,

9x – 6y = 12
4x + 6y = 14
____________
13x = 26

Or, x = 2

Substituting the values of x from equation (iii) we get

3 × 2 – 2y = 4

Or, -24 = 4 – 6

Or, – 4 = -2/2

Or, y = 1

 

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

5/4(x+2y) – 3/5(3x-2y) = 61/60

Solution:

1/2(x+2y) + 5/3(3x-2y) = -3/2

Let, 1/x+2y = m, 1/3x-2y = n

∴ m/2 + 5n/3 = -3/2

Or, 3m+10n/6 = -3/2

Or, 3m + 10n = -9 — (i)

5/4(x+2y) – 3/5(3x-2y) = 61/60

5m/4 – 3n/5 = 61/60

Or, 25m-12n/20 = 61/60

Or, 25m – 12n = 61/3 —- (ii)

Multiplying equation (i) with 6 & (ii) with 5 then adding them we get,

18m + 60n = – 54
125m – 60n = 305/3
_________________________
143m = 305/3 – 54

Or, 143m = 305-162/3

Or, 143m = 143/3

Or, m = 1/3

Substituting the values of m in equation (i) we get,

3 × 1/3 + 10n = -9

Or, n = -9-1/10

Or, n = -1

Putting the values of m & n as 1/x+2y & 1/3x-2y we get,

m = 1/3

Or, 1/x+2y = 1/3

Or, x + 2y = 3 — (iii)

n = -1

Or, 1/3x-2y = -1

Or, 3x – 2y = -1 —– (iv)

Adding equation (iii) & (iv) we get,

x + xy = 3
3x – xy = -1
_____________
4x = 2

Or, x = 1/2

Substituting the value of x in equation (ii) we get,

1/2 + 24 = 3

Or, 2y = 3 – 1/2

Or, y = 5/2×2

Or, y = 5/4

Updated: June 18, 2022 — 2:30 pm

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