ML Aggarwal Simultaneous Linear Equations Exe-5.2 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-5.2 Class 9 ICSE Maths Solutions
| Board | ICSE |
| Subject | Maths |
| Class | 9th |
| Chapter-5 | Simultaneous Linear Equations |
| Topics | Solution of Exe-5.2 Questions |
| Academic Session | 2024-2025 |
Solution of Exe-5.2 Questions
ML Aggarwal Simultaneous Linear Equations Exe-5.2 Class 9 ICSE Maths Solutions
Solve the following systems of simultaneous linear equations by the elimination method (1 to 9):
Question 1.
(i) 3x + 4y = 10
2x – 2y = 2
(ii) 2x = 5y + 4
3x – 2y + 16 = 0
Answer :
(i) 3x + 4y = 10 ….. (1)
2x – 2y = 2 ….. (2)
Multiplying equation (1) by 1 and (2) by 2
3x + 4y = 10
4x – 4y = 4
By adding both the equations
7x = 14
By division
x = 14/7 = 2
Substituting the value of x in equation (2)
2 × 2 – 2y = 2
By further calculation
4 – 2y = 2
So we get
2y = 4 – 2 = 2
y = 2/2 = 1
Therefore, x = 2 and y = 1.
(ii) 2x = 5y + 4
3x – 2y + 16 = 0
We can write it as
2x – 5y = 4 …. (1)
3x – 2y = – 16 ….. (2)
Now multiply equation (1) by 3 and (2) by 2
6x – 15y = 12 ….. (3)
6x – 4y = – 32 ….. (4)
By subtracting both the equations
– 11y = 44
y = -44/11 = – 4
Substitute the value of y in equation (1)
2x – 5 (-4) = 4
By further calculation
2x + 20 = 4
So we get
2x = 4 – 20 = – 16
x = – 16/2 = – 8
Therefore, x = – 8 and y = – 4.
Question 2.
(i) ¾ x – 2/3 y = 1
3/8 x – 1/6 y = 1
(ii) 2x – 3y – 3 = 0
2x/3 + 4y + ½ = 0.
Answer :
(i) ¾ x – 2/3 y = 1
3/8 x – 1/6 y = 1
We can write it as
¾ x – 2/3 y = 1
(9x – 8y)/ 12 = 1
By cross multiplication
9x – 8y = 12 ….. (1)
3/8 x – 1/6 y = 1
(9x – 4y)/ 24 = 1
By cross multiplication
9x – 4y = 24 ….. (2)
Subtracting equations (1) and (2)
– 4y = – 12
By division
y = – 12/ – 4 = 3
Substitute the value of y in (1)
9x – 8 × 3 = 12
By further calculation
9x – 24 = 12
9x = 12 + 24 = 36
By division
x = 36/ 9 = 4
Therefore, x = 4 and y = 3.
(ii) 2x – 3y – 3 = 0
2x/3 + 4y + ½ = 0
We can write it as
2x – 3y – 3 = 0
2x – 3y = 3 ….. (1)
2x/3 + 4y + ½ = 0
2x/3 + 4y = – ½
Taking LCM
(2x + 12y)/ 3 = – ½
By cross multiplication
2 (2x + 12y) = – 1 × 3
So we get
4x + 24y = – 3 …. (2)
Multiply equation (1) by 2
4x – 6y = 6
4x + 24y = – 3
By subtracting both the equations
– 30y = 9
So we get
y = -9/30 = – 3/10
Substitute the value of y in equation (1)
2x – 3 (-3/10) = 3
By further calculation
2x + 9/10 = 3
We can write it as
2x = 3 – 9/10
By taking LCM
2x = (30 – 9)/ 10
So we get
2x = 21/10
x = 21/20
Therefore, x = 21/20 and y = – 3/10.
Question 3.
(i) 15x – 14y = 117
14x – 15y = 115
(ii) 41x + 53y = 135
53x + 41y = 147.
Answer :
(i) 15x – 14y = 117 ….. (1)
14x – 15y = 115 ….. (2)
Now multiply equation (1) by 14 and (2) by 15
210x – 196y = 1638 …. (3)
210x – 225y = 1725 ….. (4)
By subtracting both the equations
29y = – 87
So we get
y = -87/29 = – 3
Substitute the value of y in equation (1)
15x – 14 (-3) = 117
By further calculation
15x + 42 = 117
So we get
15x = 117 – 42 = 75
By division
x = 75/15 = 5
Therefore, x = 5 and y = – 3.
(ii) 41x + 53y = 135 ….. (1)
53x + 41y = 147 ….. (2)
Now multiply equation (1) by 53 and (2) by 41
2173x + 2809y = 7155 ….. (3)
2173x + 1681y = 6027 ….. (4)
By subtracting both the equations
1128y = 1128
So we get
y = 1128/1128 = 1
Substitute the value of y in equation (1)
41x + 53 × 1 = 135
By further calculation
41x + 53 = 135
So we get
41x = 135 – 53 = 82
By division
x = 82/41 = 2
Therefore, x = 2 and y = 1.
Question 4.
(i) x/6 = y – 6
3x/4 = 1 + y
(ii) x – 2/3 y = 8/3
2x/5 – y = 7/5.
Answer :
(i) x/6 = y – 6
3x/4 = 1 + y
We can write it as
x = 6 (y – 6)
x = 6y – 36
x – 6y = – 36 ….. (1)
3x/4 = 1 + y
By cross multiplication
3x = 4 (1 + y)
So we get
3x = 4 + 4y
3x – 4y = 4 …. (2)
Multiply equation (1) by 3
3x – 18y = – 108
3x – 4y = 4
Subtracting both the equations
– 14y = – 112
So we get
y = – 112/- 14 = 8
Substitute the value of y in equation (1)
x – 6 × 8 = – 36
By further calculation
x – 48 = – 36
x = – 36 + 48
x = 12
Therefore, x = 12 and y = 8.
(ii) x – 2/3 y = 8/3
2x/5 – y = 7/5
We can write it as
x – 2/3 y = 8/3
Taking LCM
(3x – 2y)/ 3 = 8/3
By cross multiplication
3x – 2y = 8/3 × 3 = 8
3x – 2y = 8 ….. (1)
2x/5 – y = 7/5
Taking LCM
(2x – 5y)/ 5 = 7/5
By cross multiplication
2x – 5y = 7/5 × 5 = 7
2x – 5y = 7 …… (2)
Multiply equation (1) by 2 and (2) by 3
6x – 4y = 16 ….. (3)
6x – 15y = 21 …… (4)
Subtracting both the equations
11y = – 5
y = – 5/11
Substitute the value of y in equation (1)
3x – 2 (-5/11) = 8
By further calculation
3x + 10/11 = 8
We can write it as
3x = 8 – 10/11
Taking LCM
3x = (88 – 10)/ 11 = 78/11
By cross multiplication
x = 78/ (11 × 3) = 26/11
Therefore, x = 26/11 and y = – 5/11.
Question 5.
(i) 9 – (x – 4) = y + 7
2 (x + y) = 4 – 3y
(ii) 2x + (x – y)/ 6 = 2
x – (2x + y)/3 = 1.
Answer :
(i) 9 – (x – 4) = y + 7
2 (x + y) = 4 – 3y
We can write it as
9 – (x – 4) = y + 7
9 – x + 4 = y + 7
By further calculation
13 – x = y + 7
– x – y = 7 – 13 = – 6
x + y = 6 ….. (1)
2 (x + y) = 4 – 3y
2x + 2y = 4 – 3y
By further calculation
2x + 2y + 3y = 4
So we get
2x + 5y = 4 ….. (2)
Now multiply equation (1) by 5 and (2) by 1
5x + 5y = 30
2x + 5y = 4
By subtracting both the equations
3x = 26
So we get
x = 26/3
Substitute the value of x in (1)
26/3 + y = 6
We can write it as
y = 6 – 26/3
Taking LCM
y = (18 – 26)/3
So we get
y = – 8/3
Therefore, x = 26/3 and y = – 8/3.
(ii) 2x + (x – y)/ 6 = 2
x – (2x + y)/3 = 1
2x + (x – y)/ 6 = 2
Multiply by 6
12x + x – y = 12
By further calculation
13x – y = 12 …… (2)
x – (2x + y)/3 = 1
Multiply by 3
3x – 2x – y = 3
By further calculation
x – y = 3 …… (2)
So we get
x = 3 + y ….. (3)
Substitute the value of x in (1)
13 (3 + y) – y = 12
By further calculation
39 + 13y – y = 12
So we get
12y = 12 – 39 = – 27
By division
y = – 27/12 = – 9/4
Substitute the value of y in (3)
x = 3 + y
x = 3 + (-9)/4
By further calculation
x = 3 – 9/4
Taking LCM
x = (12 – 9)/ 4
x = ¾
Therefore, x = ¾ and y = – 9/4.
Question 6. x – 3y = 3x – 1 = 2x – y.
Answer :
It is given that
x – 3y = 3x – 1 = 2x – y
Here
x- 3y = 3x – 1
x – 3x – 3y = – 1
By further calculation
– 2x – 3y = – 1
2x + 3y = 1 ….. (1)
3x – 1 = 2x – y
3x – 2x + y = 1
By further simplification
x + y = 1 ….. (2)
Multiply equation (2) by 2 and subtract from equation (1)
2x + 3y = 1
2x + 2y = 2
So we get
y = – 1
Substitute the value of y in equation (1)
2x + 3 (-1) = 1
So we get
2x – 3 = 1
2x = 1 + 3 = 4
By division
x = 4/2 = 2
Therefore, x = 2 and y = – 1.
Question 7.
(i) 4x + (x – y)/ 8 = 17
2y + x – (5y + 2)/3 = 2
(ii) (x + 1)/2 + (y – 1)/3 = 8
(x – 1)/3 + (y + 1)/ 2 = 9.
Answer :
(i) 4x + (x – y)/ 8 = 17
2y + x – (5y + 2)/3 = 2
We can write it as
4x + (x – y)/ 8 = 17
(32 + x – y)/ 8 = 17
By further calculation
(33x – y)/ 8 = 17
By cross multiplication
33x – y = 136 ….. (1)
2y + x – (5y + 2)/3 = 2
Taking LCM
[3 (2y + x) – 5 (5y + 2)]/ 3 = 2
By further calculation
6y + 3x – 5y – 2 = 2 × 3
So we get
y + 3x – 2 = 6
3x + y = 6 + 2
3x + y = 8 ….. (2)
By adding both the equations
36x = 144
By division
x = 144/36 = 4
Substitute the value of x in equation (1)
33 × 4 – y = 136
By further calculation
132 – y = 136
– y = 136 – 132
So we get
– y = 4
y = – 4
Therefore, x = 4 and y = – 4.
(ii) (x + 1)/2 + (y – 1)/3 = 8
(x – 1)/3 + (y + 1)/ 2 = 9
We can write it as
(x + 1)/2 + (y – 1)/3 = 8
Taking LCM
(3x + 3 + 2y – 2)/6 = 8
By further calculation
3x + 2y + 1 = 48
So we get
3x + 2y = 47 ….. (1)
(x – 1)/3 + (y + 1)/ 2 = 9
Taking LCM
(2x – 2 + 3y + 3)/6 = 9
By further calculation
2x + 3y + 1 = 54
So we get
2x + 3y = 53 ….. (2)
By adding equation (1) and (2)
5x + 5y = 100
Dividing by 5
x + y = 20 …… (3)
By subtracting equation (1) and (2)
x – y = – 6 ….. (4)
Now add equation (3) and (4)
2x = 14
x = 14/2 = 7
Subtracting equation (4) and (3)
2y = 26
y = 26/2 = 13
Therefore, x = 7 and y = 13.
Question 8.
(i) 3/x + 4y = 7
5/x + 6y = 13
(ii) 5x – 9 = 1/y
x + 1/y = 3.
Answer :
(i) 3/x + 4y = 7 ….. (1)
5/x + 6y = 13 ….. (2)
Substitute 1/x = a in equation (1) and )@)
3a + 4y = 7 …. (3)
5a + 6y = 13 ….. (4)
Multiply equation (3) by 5 and (4) by 3
15a + 20y = 35
15a + 18y = 39
Subtracting both the equations
2y = -4
So we get
y = – 4/2 = – 2
Substitute the value of y in equation (3)
3a + 4 (-2) = 7
By further calculation
3a – 8 = 7
3a = 7 + 8 = 15
So we get
3a = 15
a = 15/3 = 5
Here x = 1/a = 1/5
Therefore, x = 1/5 and y = – 2.
(ii) 5x – 9 = 1/y ….. (1)
x + 1/y = 3 ….. (2)
Substitute 1/y = b in (1) and (2)
5x – 9 = b
5x – b = 9 ….. (3)
x + b = 3 ….. (4)
By adding equation (3) and (4)
5x – b = 9 ….. (3)
x + b = 3 ….. (4)
So we get
6x = 12
By division
x = 12/6 = 2
Substitute the value of x in equation (4)
2 + b = 3
b = 3 – 2
b = 1
Here 1/y = 1
b = 1/y
y = 1
Therefore, x = 2 and y = 1.
Question 9.
(i) px + qy = p – q
qx – py = p + q
(ii) x/a – y/b = 0
ax + by = a2 + b2.
Answer :
(i) px + qy = p – q …. (1)
qx – py = p + q ….. (2)
Now multiply equation (1) by p and (2) by q
p2x + pqy = p2 – pq
q2x – pqy = pq + q2
By adding both the equations
(p2 + q2) x = p2 + q2
By further calculation
x = (p2 + q2)/ (p2 + q2) = 1
From equation (1)
p × 1 + qy = p – q
By further calculation
p – qy = p – q
So we get
qy = p – q – p = – q
Here
y = -q/q = – 1
Therefore, x = 1 and y = – 1.
(ii) x/a – y/b = 0
ax + by = a2 + b2
We can write it as
x/a – y/b = 0
Taking LCM
(bx – ay)/ab = 0
By cross multiplication
bx – ay = 0 …… (1)
ax + by = a2 + b2 ….. (2)
Multiply equation (1) by b and equation (2) by a
b2x – aby = 0
a2x + aby = a2 + ab2
By adding both the equations
(a2 + b2)x = a2+ ab2 = a (a2 + b2)
So we get
x = a (a2 + b2)/ a2 + b2 = a
From equation (2)
b × a – ay = 0
By further calculation
ab – ay = 0
ay = ab
So we get
y = ab/a = b
Therefore, x = a and y = b.
Question 10. Solve 2x + y = 23, 4x – y = 19. Hence, find the values of x – 3y and 5y – 2x.
Answer :
It is given that
2x + y = 23 …. (1)
4x – y = 19 ….. (2)
Adding both the equations
6x = 42
x = 42/6 = 7
Substitute the value of x in equation (1)
2 × 7 + y = 23
By further calculation
14 + y = 23
So we get
y = 23 – 14 = 9
Therefore, x = 7 and y = 9.
x – 3y = 7 – 3 × 9 = 7 – 27 = – 20
5y – 2x = 5 × 9 – 2 × 7 = 45 – 14 = 31
Question 11. The expression ax + by has value 7 when x = 2, y = 1. When x = – 1, y = 1, it has value 1, find a and b.
Answer :
It is given that
ax + by = 7 when x = 2 and y = 1
Substituting the values
a (2) + b (1) = 7
2a + b = 7 ….. (1)
Here
ax + by = 1 when x = – 1 and y = 1
Substituting the values
a (-1) + b (1) = 1
– a + b = 1 ….. (2)
By subtracting both the equations
– 3a = – 6
So we get
a = – 6/ – 3 = 2
Substituting the value of a in equation (1)
2 × 2 + b = 7
By further calculation
4 + b = 7
b = 7 – 4 = 3
Therefore, a = 2 and b = 3.
Question 12. Can the following equations hold simultaneously?
3x – 7y = 7
11x + 5y = 87
5x + 4y = 43.
If so, find x and y.
Answer :
3x – 7y = 7 …… (1)
11x + 5y = 87 ….. (2)
5x + 4y = 43 ….. (3)
Now multiply equation (1) by 5 and (2) by 7
15x – 35y = 35
77x + 35y = 609
By adding both the equations
92x = 644
By division
x = 644/92 = 7
Substitute the value of x in equation (1)
3 × 7 – 7y = 7
By further calculation
21 – 7y = 7
So we get
– 7y = 7 – 21 = – 14
y = – 14/ – 7 = 2
Therefore, x = 7 and y = 2.
If x = 7 and y – 2 satisfy the equation (3) then we can say that the equations hold simultaneously
Substitute the value of x and y in equation (3)
5x + 4y = 43
By further calculation
5 × 7 + 4 × 2 = 43
So we get
35 + 8 = 43
43 = 43 which is true.
Therefore, the equations hold simultaneously.
— : End of ML Aggarwal Simultaneous Linear Equations Exe-5.2 Class 9 ICSE Maths Solutions :–
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