Simultaneous Linear Equations in Two Variables Class 9 OP Malhotra Exe-5A ICSE Maths Solutions Ch-5. We Provide Step by Step Solutions / Answer of Questions on Simultaneous Linear Equations in Two Variables OP Malhotra Maths. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.
Simultaneous Linear Equations in Two Variables Class 9 OP Malhotra Exe-5A ICSE Maths Solutions Ch-5
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 9th |
| Chapter-5 | Simultaneous Linear Equations in Two Variables |
| Writer | OP Malhotra |
| Exe-5A | More complex problems |
| Edition | 2025-2026 |
Exercise – 5A
Simultaneous Linear Equations in Two Variables Class 9 OP Malhotra ICSE Maths Solutions Ch-5
Que-1: If (5, k) is a solution of the equation 2x + y – 7 = 0 find the value of k.
Sol: (5, k) is a solution of the equation 2x + y – 7 = 0.
2 x 5 + k – 7 = 0
⇒ 10 + k – 7 = 0
⇒ 3 + k = 0
⇒ k = – 3
∴ k = – 3
Solve the following simultaneous equations:
Que-2: x + y = 5, x – y = 3
Sol: x + y = 5 … (i)
x – y = 3 … (ii)
From (ii) x = 3 + y
Substituting the value of x in (i)
3 + y + y = 5 ⇒ 2y = 5
⇒ 2/2 = 1
∴ x = 3 + y = 3 + 1 = 4
Hence x = 4, y = 1
Que-3: y = 2x – 6, y = 0
Sol: y = 2x – 6 … (i)
y = 0 … (ii)
From (i) and (ii)
2x – 6 = 0
⇒ 2x = 6
⇒ x = 6/2
Hence x = 3, y = 0
Que-4: p = 2q – 1, q = 5 – 3p
Sol: p = 2q – 1 … (i)
q = 5 – 3p … (ii)
Substituting the value of p, in (ii)
q = 5 – 3 (2q – 1)
⇒ q = 5 – 6q + 3
⇒ q + 6q = 8 ⇒ 7q = 8
⇒ q = (8/7) – 1
From (ii),
P = {2 x (8/7)} – 1
⇒ (16/7) − 1
= (16−7)/7 = 9/7
∴ p = 9/7, q = 8/7
Que-5: 9x + 4y = 5, 4x – 5y = 9
Sol: 9x + 4y = 5 … (i)
4x – 5y = 9 … (ii)
Multiplying (i) by 5 and (ii) by 4, we get
45x + 20y = 25
16x – 20y = 36
Adding we get, 61x = 61
⇒ x = 61/61 = 1
From (i),
9 x 1 + 4y = 5 ⇒ 4y = 5 – 9 = – 4
⇒ y = −4/4 = – 1
∴ x = 1, y = – 1
Que-6: x + 3y = 5, 3x – y = 5
Sol: x + 3y = 5 … (i)
3x – y = 5 … (ii)
From (i) x = 5 – 3y
Substituting the value of x in (ii)
3 (5 – 3y) – y = 5 ⇒ 15 – 9y – y = 5
⇒ – 10y = 5 – 15 = – 10
⇒ y = −10/−10 = 1
∴ x = 5 – 3y = 5 – (3 x 1) = 5
Hence x = 2, y = 1
Que-7: 3x – 7y + 10 = 0, y – 2x – 3 = 0
Sol: 3x – 7y + 10 = 0
3x – 7y = – 10 … (i)
y – 2x – 3 = 0 … (ii)
From (ii) y = 3 + 2x
Substituting the value of y in (i)
3x – 7 (3 + 2x) = – 10
⇒ 3x – 21 – 14x = – 10
⇒ – 11x = – 10 + 21 = 11
⇒ x = 11/−11
∴ y = 3 + 2x = 3 + {2 x (- 1)}
= 3 – 2 = 1
Hence x = – 1, y = 1
Que-8: 20u – 30v = 13, 10v – 10u = – 5
Sol: 20u – 30v = 13 … (i)
10v – 10u =- 5
10u – 10v = 5 … (ii)
Multiplying (ii) by 2, we get
20u – 20v = 10 ….. (iii)
Subtracting (iii) and (i) we get,
-10v = 3
v = -3/10
From (ii) 10u – 10 x (−3/10) = 5
⇒ 10u + 3 = 5 ⇒ 104 = 5 – 3 = 2
∴ u = 2/10 = 1/5
Hence u = 1/5, v = −3/10
Que-9: 2x – 3y = 1.3, y – x = 0.5
Sol: 2x – 3y = 1.3 … (i)
y – x = 0.5 … (ii)
From (ii) y = 0.5 + x
Substituting the value of y in (i)
2x – 3 (0.5 + x) = 1.3
⇒ 2x – 1.5 – 3x = 1.3 ⇒ – x = 1.3 + 1.5
⇒ x = 2.8 ⇒ x = – 2.8
∴ y = 0.5 + (- 2.8) = 0.5 – 2.8 = – 2.3
Hence x = – 2.8 and y = – 2.3
Que-10: 11x + 15y + 23 = 0, 7x – 2y – 20 = 0
Sol: 11x + 15y + 23 = 0 … (i)
7x – 2y – 20 = 0 … (ii)
Multiplying (i) by 2 and (ii) by 15, we get
22x + 30y + 46 = 0 ……. (iii)
105x – 30y – 300 0 …… (iv)
Adding (iii) and (iv) we get,
⇒ 127x = 254
⇒ x = 254/127 = 2
From (ii)
7 (2) – 2y – 20 = 0
⇒ 14 – 7y – 20 = 0
⇒ – 2y = 20 – 14 = 6
⇒ y = 6/–2 = – 3
∴ x = 2, y = – 3
Que-11: 3 – (x – 5) = y + 2
2(x + y) = 4 – 3y
Sol: 3 – (x – 5) = y + 2
⇒ 3 – x + 5 = y + 2
⇒ – x – y = 2 – 3 – 5
⇒ – x – y = – 6
⇒ x + y = 6 … (i)
and 2 (x + y) = 4 – 3y ⇒ 2x + 2y = 4 – 3y
⇒ 2x + 2y + 3y = 4 ⇒ 2x + 5y = 4 … (ii)
From (i) x = 6 – y
Substituting the value of x in (ii)
2 (6 – y) + 5y = 4 ⇒ 12 – 2y + 5y = 4
⇒ 3y = 4 – 12 = – 8
⇒ y = −8/3
∴ x = 6 – y = 6 + (8/3)
= (18+8)/3 = 26/3
Hence x = 26/3, y = −8/3
Que-12:
(a) (x/2) + (y/4) = 6, (x/5) − (y/2)
(b) (x/4) − 3 = (y/6), (x/2) − y = −2
Sol: (a) (x/2) + (y/4) = 6 … (i)
(x/5) – (y/2) = 0 … (ii)
Multiplying (i) by 1 and (ii) by (1/2) , we get
(x/2) + (y/4) = 6
(x/10) – (y/4) = 0
Adding we get,
(x/2) + (y/10) = 6
⇒ 5x + x = 60
⇒ 6x = 60
⇒ x = 60/6 = 10
From (ii),
(10/5) − (y/2) = 0
⇒ 2 − (y/2) = 0
⇒ (y/2) = 2 ⇒ y = 4
∴ x = 10, y = 4
(b) (x/4) – 3 = (y/6)
⇒ (x/4) – (y/6) = 3 … (i)
(x/2) – y = – 2
Multiplying (i) 1 and (ii) by 16, we get
(x/4) – (y/6) = 3
(x/12) − (y/6) = −2/6 = −1/3
Subtracting we get,
(x/4) − (x/12) = 3 + (1/3) = 10/3
3x−x = 40
⇒ 2x = 40
⇒ x = 40/2 = 20
From (ii)
(20/2) – y = – 2 ⇒ 10 – y = – 2
⇒ y = 10 + 2 = 12
∴ x = 20, y = 12
Que-13:
(a) (7/x) + (8/y) = 2, (2/x) + (12/y) = 20
(b) (1/7x) + (1/6y) = 3, (1/2x) − (1/3y) = 5
Sol: (a) (7/x) + (8/y) = 2, (2/x) + (12/y) = 20
Let (1/x) = u and (1/y) = v then equation (1) and (2) can be written as
7u + 8v = 2 ….(3)
and 2u + 12v = 20 ….(4)
Multiplying equation (3) by 2 and (4) by 7, we get
14u + 16v = 4 ….(5)
and 14u = 84v = 140 ….(6)
On subtracting equation (6) from (5), we get
-68v = -136
implies v = 2
Substituting v = 2 in equation (3), we get
7u + (8×2) = 2
implies 7u + 16 = 2
implies 7u = – 14 implies u = – 2
when u = -1/2 implies (1/x) = -2 implies x = -1/2.
and when v = 2 implies (1/y) = 2 implies y = 1/2.
(b) (1/7x) + (1/6y) = 3 ……(1)
(1/2x) – (1/3y) = 5 …….(2)
Multiplying (2) by 1/2 we get
(1/4x) – (1/6y) = 5/2 ….(3)
Solving (1) and (3), we get
(1/7x) + (1/6y) = 3
Adding the equations we get,
⇒ (4+7)/28x = (6+5)/2
⇒ 11/28x = 11/2
⇒ x = (11×2)/(28×11) = 11/4
when x = 1/14 we get
[1/{7(1/14)}] + (1/6y) = 3 using (1)
⇒ 2 + (1/6y) = 3
⇒ (1/6y) = 3-2 = 1
⇒ y = 1/6
Thus, the solution of given equation is x = 1/14 and y = 1/6.
Que-14:
(a) 65x – 33p = 97, 33x – 65y = 1
(b) 23x + 31y = 77, 31x + 23y = 85
Sol: (a) 65x – 33y = 97 … (i)
33x – 65y = 1 … (ii)
Adding, we get
98x – 98y = 98
x – y – 1 … (iii)
(Dividing by 98)
Subtracting (ii) from (i)
32x + 32y = 96
x + y = 3 … (iv)
(Dividing by 32)
Adding (iii) and (iv),
2x = 4 ⇒ x = 4/2 = 2
and subtracting (iii), form (iv)
2y = 2 ⇒ y = 2/2 = 1
∴ x = 2, y = 1
(b) 23x + 31y = 77 … (i)
31x + 23y = 85 … (ii)
Adding, we get,
54x + 54y = 162
x + y = 3 … (iii)
(Dividing by 54)
and subtracting (/) from (it),
8x – 8y = 8
x – y = 1 … (iv)
(Dividing by 8)
Now adding (iii) and (iv), we get
2x = 4 ⇒ x = 4/2 = 2
and subtracting (iv) from (iii)
2y = 2 ⇒ y = 2/2 = 1
∴ x = 2, y = 1
Que-15:
(a) 4x + (6/y) = 15, 6x – (8/y) = 14; y ≠ 0
Find p, if y = px – 2
(b) (4/x) + 5y = 7, (3/x) + 4y = 5; x ≠ 0
Sol: y = px – 2
(a) 4x + (6/y) = 15 … (i)
6x – (8/y) = 14 … (ii)
Multiplying (i) by 4 and (ii) by 3, we get
16x + (24/y) = 60
18x – (24/y) = 42
Adding we get,
34x = 102 ⇒ x = 102/34 = 3
From (i)
4 x 3 + (6/y) = 15 ⇒ 12 + (6/y) = 17
⇒ 6y = 15 – 12 = 3
⇒ 3y = 6 ⇒ y = (6/3) = 2
Hence x = 3, y = 2
Now y = px – 2
⇒ 2 = 3p – 2
⇒ 3p = 2 + 2 = 4
p = 4/3
(b) (4/x) + 5y = 7 … (i)
(3/x) + 4y = 5 … (ii)
Multiplying (i) by 4 and (ii) by 5, we get
(16/x) + 20y = 28 …. (iii)
(15/x) + 20y = 25 … (iv)
Subtract (iii) and (iv) we get
1/x = 3
x = 1/3
From (i), {4/(1/3)} + 5y = 7
12 + 5y = 7
5y = 7 – 12
y = -5/5 = -1.
x = 1/3 and y = -1.
Que-16: (a) {6/(x+y)} = {7/(x−y)} + 3, {1/2(x+y)} = {1/3(x−y)}, where x + y ≠ 0, x – y ≠ 0.
(b) {44/(x+y)} + {30/(x−y)} = 10, {55/(x+y)} + {40/(x−y)} = 13, where x + y ≠ 0, x – y ≠ 0.
Sol: (a) Let 1/(x+y) = u and 1/(x-y) = v Then, the given system of equation becomes
6u = 7v + 3
=> 6u – 7v = 3 …..(i)
And u/2 = v/3
=> 3u = 2v
=> 3u – 2v = 0 ……(ii)
Multiplying equation (ii) by 2, and equation (i) by 1, we get
6u – 7v = 3 ….(iii)
6u – 4v = 0 ……(iv)
Subtracting equation (iv) from equation (iii), we get
-7 + 4v = 3
=> -3v = 3
=> v = -1
Putting v = -1 in equation (ii), we get
3u – {2×(-1)} = 0
=> 3u + 2 = 0
=> 3u = -2
⇒ u = -2/3
Now u = -2/3
=> 1/(x+2) = -2/3
⇒ x+y = -3/2 …(v)
And v = -1
⇒ 1/(x-y) = -1
=> x – y = -1 …(vi)
Adding equation (v) and equation (vi), we get
2x = (-3/2) – 1
⇒ 2x = (-3-2)/2
⇒ 2x = -5/2
⇒ x = -5/4
Putting x = (-5)/4 in equation (vi), we get
(-5/4) – y = -1
⇒ (-5/4) + 1 = y
⇒ (-5+4)/4 = y
⇒ -1/4 = y
⇒ y = -1/4
Hence, solution of the system of equation is x = -5/4 and y = -1/4.
(b) The given equations are
{44/(x+y)} + {30/(x−y)} = 10 ……(i)
{55/(x+y)} – {40/(x−y)} = 13 ……(ii)
Putting 1/(x+y) = u and 1/(x−y) = v, we get:
44u + 30v = 10 ……..(iii)
55u + 40v = 13 …….(iv)
On multiplying (iii) by 4 and (iv) by 3, we get:
176u + 120v = 40 …..(v)
165u + 120v = 39 …..(vi)
On subtracting (vi) and (v), we get:
11u = 1
⇒ u = 1/11
⇒ 1/(x+y) = 1/11
⇒ x+y = 11 …….(vii)
On substituting u = 1/11 in (iii), we get:
4 + 30v = 10
⇒ 30v = 6
⇒ v = 6/30 = 1/5
⇒ 1/(x−y) = 1/5
⇒ x–y = 5 …….(viii)
On adding (vii) and (viii), we get
2x = 16
⇒ x = 8
On substituting x =8 in (vii), we get:
8 + y = 11
⇒ y = 11 – 8 = 3
Hence, the required solution is x = 8 and y = 3.
Que-17:
(i) 2 (3u – v) = 5uv, 2 (w + 3v) = 5uv
(ii) 3 (a + 3b) = 11ab, 3 (2a + b) = 7ab
Sol: (i) 2 (3u – v) = 5uv
⇒ 6u – 2v = 5uv
Dividing by uv,
(6/v) − (2/u) = 5 … (i)
and 2 (u + 3v) = 5uv
⇒ 2u + 6v = 5uv
Dividing by uv
(2/v) + (6/u) = 5 … (ii)
Multiplying (i) by 3 and (ii) by 1,
(18/v) − (6/u) = 5
(2/v) + (6/u) = 5
Adding we get,
20/v = 20
⇒ v = 20/20 = 1
From (i) (6/1) – (2/u) = 5
⇒ –2/u = 5 – 6 = – 1
⇒ u = 2
Hence u = 2, v = 1
(ii) Given,
3 (a + 3b) = 11ab
⇒ 3a + 9b = 11ab …(i)
3 (2a + b) = 7ab
⇒ 6a + 3b = 7ab …(ii)
Now, dividing both sides of (i) and (ii) by ab, where both a and b are non-zero, we get
3/b + 9/a = 11
⇒ 9/a + 3/b = 11 …(iii)
6/b + 3/a = 7
⇒ 3/a + 6/b = 7 …(iv)
Let us take, 1/a = x and 1/b = y. Then, (iii) and (iv) become
9x + 3y = 11
⇒ 3x + y = 11/3 …(v)
3x + 6y = 7 …(iv)
From (iv) and (v), on subtraction, we get
5y = 7 – 11/3
⇒ 5y = (21 – 11)/3
⇒ 5y = 10/3
⇒ y = 2/3
⇒ 1/b = 2/3 [∵ y = 1/b]
⇒ b = 3/2
Putting b = 3/2 in (iii), we get
9/a + 3/(3/2) = 11
⇒ 9/a + 2 = 11
⇒ 9/a = 11 – 2
⇒ 9/a = 9
⇒ 1/a = 1
⇒ a = 1
∴ the required solution be
a = 1 and b = 3/2.
Que-18: {(3x−6)/4} + 3 − {(5y−4)/2} = 5y/2
{(y−x)/4} + (x/8) − {(7x−5y)/3} = y−2x
Sol: {(3x−6)/4} + 3 − {(5y−4)/2} = 5y/2
2(3x-6+12-10y+8) = 5y×4
2(3x-10y+14) = 20y
6x-20y+28 = 20y
6x-20y-20y = -28
6x-40y = -28
Now divide by 2
3x-20y = -14
{(y−x)/4} + (x/8) − {(7x−5y)/3} = y−2x
6(y-x)+3x-8(7x-5y) = y-2x
6y – 6x+3x-56x + 40y = 24(y-2x)
46y-62x +3x = 24y – 48x
46y – 59x = 24y – 48x
-11x = – 22y
– 11x + 22y = 0
divide by 11
– x + 2y = 0_eq2
3x -20y= – 14_eq1
– x + 2y = 0_eq2×3
Now subtract eq1 from eq2
3x – 20y = 14
-3x + 6y= 0
– 14y = -14
y = 1
Now from eq 2
-x + 2y = 0
– x +2×1 =0
– x+2=0
– x = -2
x= 2
y = 1.
Que-19: x – y = 0.9, {11/2(x+y)} = 1.
Sol: x – y = 0.9 = 9/10 … (i)
{11/2(x+y)} = 1
⇒ x + y = 11/2 … (ii)
Adding we get,
2x = (9/10) + (11/2) = {(9×55)/10} = 64/10
x = 64/(10×2) = 32/10 = 3.2
From (i),
x – y = 0.9
3.2 – y = 0.9
⇒ – y = 0.9 – 3.2
⇒ – y = – 2.3
⇒ y = 2.3
Hence x = 3.2, y = 2.3
Que-20: (x/2) + y = 0.8, 7/{x+(y/2)} = 10
Sol: The given pair of equation are:
(x/2) + y = 0.8
⇒ x + 2y = 1.6 …(a)
7/{x+(y/2)} = 10
⇒ 7 = 10 (x+(y/2))
⇒ 7 = 10x+5y
Lets multiply LHS and RHS of equation (a) by 10 for easy calculation
So, we finally get
10x + 20y = 16 ……………..(i) And,
10x + 5y = 7……………..(ii)
Now, Subtracting two equations we get,
⇒ (i) – (ii)
15y = 9
⇒ y = 3/5
Next, putting the value of y in the equation (i) we get,
x = [16-20(3/5)]/10
⇒ (16-12)/10 = 4/10
x = 2/5
Thus, the value of x and y obtained are 2/5 and 3/5 respectively
Que-21: If 2x +y = 35, 3x + 4y = 65, find the value of x/y.
Sol: 2x + y = 35 … (i)
3x + 4y = 65 … (ii)
Multiplying (i) by 4 and (ii) by 1
8x + 4y = 140 …. (iii)
3x + 4y = 65 ….. (iv)
Subtracting (iii) and (iv) we get,
5x = 75
x = 75/5 = 15
From (i) y = 35 – 2x
= 35 – 2 x 15
= 35 – 30
= 5
x/y = 15/5 = 3..
Que-22: 3 – 2(3x + 4y) = x, {(x-3)/4} – {(y-4)/5} = 2*(1/10)
Sol: 3 – 2(3x + 4y) = x
3 – 6x – 8y = x
⇒ – 6x – 8y – x = – 3
⇒ – 7x – 8y = – 3
7x + 8y = 3 … (i)
{(x-3)/4} – {(y-4)/5} = 21/10
(5x-15-4y+16)/20 = 42/20
5x-4y = 42+15-16
5x-4y = 41 …. (ii)
Multiply (ii) by 2
10x – 8y = 82 ….. (iii)
Adding (i) and (iii) we get,
17x = 85
x = 5
From (i),
7 x 5 + 8y = 3
⇒ 35 + 8y = 3
⇒ 8y = 3 – 35 = – 32
⇒ y = −32/8 = 4
∴ x = 5, y = – 4
Que-23: Can the following system of equations hold simultaneously?
(3/x) + 4y = 7, (−2/x) + 7y = 5, 5x + (4/y) = 9. if yes, find x and y.
Sol: (3/x) + 4y = 7 … (i)
(−2/x) + 7y = 5 … (ii)
5x + (4/y) = 9 … (iii)
Multiplying (i) by 2 and (ii) by 3,
(6/x) + 8y = 14
(−6/x) + 21y = 15
Adding we get,
29y = 29
⇒ y = 29/29 = 1
From (i)
(3/x) + 4 x 1
⇒ (3/x) + 4 = 1
⇒ (3/x) = 7 – 4 = 3
⇒ x = 3/3 = 1
∴ x = 1, y = 1
Now substituting the value of x and y in (iii)
5 x 1 + (4/1) = 9
⇒ 5 + 4 = 9
⇒ 9 = 9 which is true
Yes, the system of equations are simultaneously and x = 1, y = 1
Que-24: If the following three equations hold simultaneously for x and y, find p.
3x – 2y = 6, (x/3) − (y/6) = 1/2, x – py = 6
Sol: 3x – 2y = 6 …….. (i)
(x/3) − (y/6) = 1/2
⇒ 2x – y = 3 … (ii) (Multiplying by 6)
Multiply (ii) by 2
4x – 2y = 6 ……. (iii)
Subtracting (i) and (iii) we get,
-x = 0
x = 0.
From (i)
3 x 0 – 2y = 6 ⇒ – 2y = 6
∴ The equations are simultaneously
∴ x = 0, y = – 3 will satisfy the third equation
∴ x – py = 6
⇒ 0 – p x (- 3) = 6
= 3p = 6
⇒ p = 6/3 = 2
Hence p = 2
Que-25: The sides of an equilateral triangle are (6x + 33y) cm, (8x + 9y – 5) cm and (10x + 12y – 8) respectively. Find the length of each side.
Sol: The triangle is an equilateral triangle
∴ Its all the sides are equal
Now 6x + 3y = 8x + 9y – 5
⇒ 8x + 9y – 6x – 3y = 5
⇒ 2x + 6y = 5 … (i)
and 8x + 9y – 5 = 10x + 12y – 8
⇒ 10x + 12y – 8x – 9y = – 5 + 8
⇒ 2x + 3y = 3 … (ii)
Multiplying (ii) by 2,
4x + 6y = 6 …… (iii)
Subtracting (i) and (iii) we get,
-2x = -1
x = 1/2
From (i)
2x + 6y = 5
⇒ 2 x (1/2) + 6y = 5
⇒ 1 + 6y = 5
⇒ 6y = 5 – 1 = 4
⇒ y = 4/6 = 2/3
Now 6x + 3y = 6 x (1/2) + 3 x (2/3)
= 3 + 2 = 5
∴ Each side of the triangle = 5 units.
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