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 :–
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