Utilizing OP Malhotra Class 9 Solutions Chapter 5 Simultaneous Linear Equations in Two Variables Ex 5(A) as a study aid can enhance exam preparation.
S Chand Class 9 ICSE Maths Solutions Chapter 5 Simultaneous Linear Equations in Two Variables Ex 5(A)
Question 1.
If (5, k) is a solution of the equation 2x + y – 7 = 0 find the value of k.
Solution:
∵ (5, k) is a solution of the eqution 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:
Question 2.
x + y = 5, x – y = 3
Solution:
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
⇒ \(\frac { 2 }{ 2 }\) = 1
∴ x = 3 + y = 3 + 1 = 4
Hence x = 4, y = 1
Question 3.
y = 2x – 6, y = 0
Solution:
y = 2x – 6 … (i)
y = 0 … (ii)
From (i) and (ii)
2x – 6 = 0 ⇒ 2x = 6
⇒ x = \(\frac { 6 }{ 2 }\)
Hence x = 3, y = 0
Question 4.
p = 2q – 1, q = 5 – 3p
Solution:
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 = \(\frac { 8 }{ 7 }\) – 1
From (ii),
P = 2 x \(\frac { 8 }{ 7 }\) – 1
⇒ \(\frac{16}{7}-1=\frac{16-7}{7}=\frac{9}{7}\)
∴ p = \(\frac { 9 }{ 7 }\), q = \(\frac { 8 }{ 7 }\)
Question 5.
9x + 4y = 5, 4x – 5y = 9
Solution:
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 = \(\frac { 61 }{ 61 }\) = 1
From (i),
9 x 1 + 4y = 5 ⇒ 4y = 5 – 9 = – 4
⇒ y = \(\frac { -4 }{ 4 }\) = – 1
∴ x = 1, y = – 1
Question 6.
x + 3y = 5, 3x – y = 5
Solution:
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 = \(\frac { -10 }{ -10 }\) = 1
∴ x = 5 – 3y = 5 – 3 x 1 = 5
Hence x = 2, y = 1
Question 7.
3x – 7y + 10 = 0, y – 2x – 3 = 0
Solution:
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 = \(\frac { 11 }{ -11 }\)
∴ y = 3 + 2x = 3 + 2 x (- 1)
= 3 – 2 = 1
Hence x = – 1, y = 1
Question 8.
20u – 30v = 13, 10v – 10u = – 5
Solution:
20u – 30v = 13 … (i)
10v – 10u =- 5
10u – 10v = 5 … (ii)
Multiplying (i) by 1 and (ii) by 2, we get
From (ii) 10u – 10 x (\(\frac { -3 }{ 10 }\)) = 5
⇒ 10u + 3 = 5 ⇒ 104 = 5 – 3 = 2
∴ u = \(\frac { 2 }{ 10 }\) = \(\frac { 1 }{ 5 }\)
Hence u = \(\frac { 1 }{ 5 }\), v = \(\frac { -3 }{ 10 }\)
Question 9.
2x – 3y = 1.3, y – x = 0.5
Solution:
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
Question 10.
11x + 15y + 23 = 0, 7x – 2y – 20 = 0
Solution:
11x + 15y + 23 = 0 … (i)
7x – 2y – 20 = 0 … (ii)
Multiplying (i) by 2 and (ii) by 15, we get
⇒ 127x = 254
⇒ x = \(\frac { 254 }{ 127 }\) = 2
From (ii)
7 (2) – 2y – 20 = 0 ⇒ 14 – 7y – 20 = 0
⇒ – 2y = 20 – 14 = 6 ⇒ y = \(\frac { 6 }{ – 2 }\) = – 3
∴ x = 2, y = – 3
Question 11.
3 – (x – 5) = y + 2
2(x + y) = 4 – 3y
Solution:
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 = \(\frac { -8 }{ 3 }\)
∴ x = 6 – y = 6 + \(\frac{8}{3}=\frac{18+8}{3}=\frac{26}{3}\)
Hence x = \(\frac { 26 }{ 3 }\), y = \(\frac { -8 }{ 3 }\)
Question 12.
(a) \(\frac{x}{2}+\frac{y}{4}=6, \frac{x}{5}-\frac{y}{2}\)
(b) \(\frac{x}{4}-3=\frac{y}{6}, \frac{1}{2} x-y=-2\)
Solution:
(a) \(\frac { x }{ 2 }\) + \(\frac { y }{ 4 }\) = 6 … (i)
\(\frac { x }{ 5 }\) – \(\frac { y }{ 2 }\) = 0 … (ii)
Multiplying (i) by 1 and (ii) by \(\frac { 1 }{ 2 }\) , we get
\(\frac { x }{ 2 }\) + \(\frac { y }{ 4 }\) = 6
\(\frac { x }{ 10 }\) – \(\frac { y }{ 4 }\) = 0
Adding we get,
\(\frac { x }{ 2 }\) + \(\frac { y }{ 10 }\) = 6
⇒ \(\frac{5 x+x=60}{10}\) ⇒ 6x = 60
⇒ x = \(\frac { 60 }{ 6 }\) = 10
From (ii),
\(\frac{10}{5}-\frac{y}{2}=0 \Rightarrow 2-\frac{y}{2}\) = 0
⇒ \(\frac { y }{ 2 }\) = 2 ⇒ y = 4
∴ x = 10, y = 4
(b) \(\frac { x }{ 4 }\) – 3 = \(\frac { y }{ 6 }\) ⇒ \(\frac { x }{ 4 }\) – \(\frac { y }{ 6 }\) = 3 … (i)
\(\frac { 1 }{ 2 }\)x – y = – 2
Multiplying (i) 1 and (ii) by \(\frac { 1 }{ 6 }\), we get
\(\frac { x }{ 4 }\) – \(\frac { y }{ 6 }\) = 3
\(\frac{1}{12} x-\frac{y}{6}=\frac{-2}{6}=\frac{-1}{3}\)
Subtracting we get,
\(\frac{x}{4}-\frac{x}{12}=3+\frac{1}{3}=\frac{10}{3}\)
\(\frac{3 x-x=40}{12}\) ⇒ 2x = 40
⇒ x = \(\frac { 40 }{ 2 }\) = 20
From (ii)
\(\frac { 20 }{ 2 }\) – y = – 2 ⇒ 10 – y = – 2
⇒ y = 10 + 2 = 12
∴ x = 20, y = 12
Question 13.
(a) \(\frac{7}{x}+\frac{8}{y}=2, \frac{2}{x}+\frac{12}{y}\) = 20
(b) \(\frac{1}{7 x} + \frac{1}{6 y}=3, \frac{1}{2 x}-\frac{1}{3 y}\) = 5
Solution:
(a) \(\frac { 7 }{ x }\) + \(\frac { 8 }{ y }\) = 2 … (i)
\(\frac { 2 }{ x }\) + \(\frac { 12 }{ y }\) = 20 … (ii)
Multiplying (i) by 3 and (ii) by 2, we get
(b) \(\frac { 1 }{ 7x }\) + \(\frac { 1 }{ 6y }\) = 3 … (i)
\(\frac { 1 }{ 2x }\) – \(\frac { 1 }{ 3y }\) = 5 … (ii)
Multiplying (i) by 1 and (ii) by \(\frac { 1 }{ 2 }\), we get
\(\frac { 1 }{ 7x }\) + \(\frac { 1 }{ 6y }\) = 3
\(\frac { 1 }{ 4x }\) – \(\frac { 1 }{ 6y }\) = \(\frac { 5 }{ 2 }\)
Adding we get,
Question 14.
(a) 65x – 33p = 97, 33x – 65y = 1
(b) 23x + 31y = 77, 31x + 23y = 85
Solution:
(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 = \(\frac { 4 }{ 2 }\) = 2
and subtracting (iii), form (iv)
2y = 2 ⇒ y = \(\frac { 2 }{ 2 }\) = 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 = \(\frac { 4 }{ 2 }\) = 2
and subtracting (iv) from (iii)
2y = 2 ⇒ y = \(\frac { 2 }{ 2 }\) = 1
∴ x = 2, y = 1
Question 15.
(a) 4x + \(\frac { 6 }{ y }\) = 15, 6x – \(\frac { 8 }{ y }\) = 14; y ≠ 0
Find p, if y = px – 2
(b) \(\frac { 4 }{ x }\) + 5y = 7, \(\frac { 3 }{ x }\) + 4y = 5; x ≠ 0
Solution:
y = px – 2
(a) 4x + \(\frac { 6 }{ y }\) = 15 … (i)
6x – \(\frac { 8 }{ y }\) = 14 … (ii)
Multiplying (i) by 4 and (ii) by 3, we get
16x + \(\frac { 24 }{ y }\) = 60
18x – \(\frac { 24 }{ y }\) = 42
Adding we get,
34x = 102 ⇒ x = \(\frac { 102 }{ 34 }\) = 3
From (i)
4 x 3 + \(\frac { 6 }{ y }\) = 15 ⇒ 12 + \(\frac { 6 }{ y }\) = 17
⇒ \(\frac { 6 }{ y }\) = 15 – 12 = 3
⇒ 3y = 6 ⇒ y = \(\frac { 6 }{ 3 }\) = 2
Hence x = 3, y = 2
Now y = px – 2
⇒ 2 = 3p – 2 ⇒ 3p = 2 + 2 = 4
p = \(\frac { 4 }{ 3 }\)
(b) \(\frac { 4 }{ x }\) + 5y = 7 … (i)
\(\frac { 3 }{ x }\) + 4y = 5 … (ii)
Multiplying (i) by 4 and (ii) by 5, we get
Question 16.
(a) \(\frac{6}{x+y}=\frac{7}{x-y}+3, \frac{1}{2(x+y)}\) = \(\frac{1}{3(x-y)}\), where x + y ≠ 0, x – y ≠ 0.
(b) \(\frac{44}{x+y}+\frac{30}{x-y}=10, \frac{55}{x+y}+\frac{40}{x-y}\) = 13, where x + y ≠ 0, x – y ≠ 0.
Solution:
(a) \(\frac{6}{x+y}=\frac{7}{x-y}\) + 3
Now x + y = \(\frac { -3 }{ 2 }\)
x – y = – 1
Adding we get,
2x = \(\frac { -5 }{ 2 }\) ⇒ x = \(\frac { – 5 }{ 4 }\)
and subtracting,
2y = \(\frac { -1 }{ 2 }\) ⇒ y = \(\frac{-1}{2 \times 2}=\frac{-1}{4}\)
Hence x = \(\frac { – 5 }{ 4 }\), y = \(\frac { – 1 }{ 4 }\)
(b) \(\frac{44}{x-y}+\frac{30}{x-y}\) = 10
\(\frac{55}{x+y}+\frac{40}{x-y}\) = 13
Let x + y = a and x – y = b, then
\(\frac{44}{a}+\frac{30}{b}\) = 10 … (i)
\(\frac{55}{a}+\frac{40}{b}\) = 13 … (ii)
Multiplying (i) by 4 and (ii) by 3, we get
Now x + y = 11
x – y = 5
Adding we get,
2x – 16 ⇒ x = \(\frac { 16 }{ 2 }\) = 8
and subtracting,
2y = 6 ⇒ y = \(\frac { 6 }{ 2 }\) = 3
Hence x = 8, y = 3
Question 17.
(i) 2 (3u – v) = 5uv, 2 (w + 3v) = 5uv
(ii) 3 (a + 3b) = 11ab, 3 (2a + b) = 7ab
Solution:
(i) 2 (3u – v) = 5uv
⇒ 6u – 2v = 5uv
Dividing by uv,
\(\frac{6}{v}-\frac{2}{u}\) = 5 … (i)
and 2 (u + 3v) = 5uv
⇒ 2u + 6v = 5uv
Dividing by uv
\(\frac{2}{v}+\frac{6}{u}\) = 5 … (ii)
Multiplying (i) by 3 and (ii) by 1,
\(\frac{18}{v}-\frac{6}{u}\) = 5
\(\frac{2}{v}+\frac{6}{u}\) = 5
Adding we get,
\(\frac { 20 }{ v }\) = 20 ⇒ v = \(\frac { 20 }{ 20 }\) = 1
From (i) \(\frac { 6 }{ 1 }\) – \(\frac { 2 }{ u }\) = 5
⇒ \(\frac { – 2 }{ u }\) = 5 – 6 = – 1
⇒ u = 2
Hence u = 2, v = 1
(ii) 3 (a + 3b) = 11ab
⇒ 3a + 9b = 11ab
Dividing by ab,
\(\frac{3}{b}+\frac{9}{a}\) = 11 … (i)
and 3 (2a + b) = 7ab
⇒ 6a + 3b = 7ab
Dividing by ab,
\(\frac{6}{b}+\frac{3}{a}\) = 7 … (ii)
Multiply (i) by 1 and (ii) by 3
Question 18.
\(\frac{3 x-6}{4}+3-\frac{5 y-4}{2}=\frac{5 y}{2}\)
\(\frac{y-x}{4}+\frac{x}{8}-\frac{7 x-5 y}{3}=y-2 x\)
Solution:
Question 19.
x – y = 0.9, \(\frac{11}{2(x+y)}\) = 1
Solution:
x – y = 0.9 = \(\frac { 9 }{ 10 }\) … (i)
\(\frac{11}{2(x+y)}\) = 1 ⇒ x + y = \(\frac { 11 }{ 2 }\) … (ii)
Adding we get,
2x = \(\frac{9}{10}+\frac{11}{2}=\frac{9 \times 55}{10}=\frac{64}{10}\)
x = \(\frac{64}{10 \times 2}=\frac{32}{10}\) = 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
Question 20.
\(\frac{x}{2}+y=0.8, \frac{7}{x+\frac{y}{2}}\) = 10
Solution:
\(\frac{x}{2}+y=0.8 \Rightarrow \frac{x}{2}+y=\frac{8}{10}\)
⇒ 5x + 10y = 8 … (i)
From (i)
5 x 0.4 + 10y = 8
2.0 + 10y = 8 ⇒ 10y = 8 – 2 = 6
⇒ y = \(\frac { 6 }{ 10 }\) = \(\frac { 3 }{ 5 }\)
Hence x = \(\frac { 2 }{ 5 }\), y = \(\frac { 3 }{ 5 }\)
Question 21.
If 2x +y = 35, 3x + 4y = 65, find the value of \(\frac { x }{ y }\).
Solution:
2x + y = 35 … (i)
3x + 4y = 65 … (ii)
Multiplying (i) by 4 and (ii) by 1
⇒ x = \(\frac { 75 }{ 5 }\) = 15
From (i) y = 35 – 2x = 35 – 2 x 15 = 35 – 30 = 5
∴ \(\frac{x}{y}=\frac{15}{5}\) = 3
P.Q. Solve:
(a) x – y – a – b, ax – by = a² x b²
(b) ax + by = a – b, bx – ay = a + b
Solution:
(a) x + y = a – b
ax – by = a² + b²
Multiplying (i) by b and (ii) by 1,
(a + b) x = a(a + b)
⇒ x = \(\frac{a(a+b)}{(a+b)}\) = a
From (i)
a + y = a – b
y = a – b – a = – b
Hence x = a, y = – b
(b) ax + by = a – b … (i)
bx – ay = a + b … (ii)
Multiplying (i) by a and (ii) by b,
x = \(\frac{a^2+b^2}{a^2+b^2}\)
From (i),
a x 1 + by = a – b ⇒ a + by = a – b
⇒ by = a – b – a = – b
⇒ y = \(\frac { -b }{ b }\) = – 1
Hence x = 1, y = – 1
Question 22.
3 – 2(3x + 4y) = x, \(\frac{x-3}{4}-\frac{y-4}{5}=2 \frac{1}{10}\)
Solution:
3 – 2(3x + 4y) = x
3 – 6x – 8y = x ⇒ – 6x – 8y – x = – 3
⇒ – 7x – 8y = – 3
7x + 8y = 3 … (i)
From (i),
7 x 5 + 8y = 3 ⇒ 35 + 8y = 3
⇒ 8y = 3 – 35 = – 32 ⇒ y = \(\frac { -32 }{ 8 }\) = 4
∴ x = 5, y = – 4
Question 23.
Can the following system of equations hold simultaneously?
\(\frac { 3 }{ x }\) + 4y = 7, \(\frac { -2 }{ x }\) + 7y = 5, 5x + \(\frac { 4 }{ y }\) = 9. if yes, find x and y.
Solution:
\(\frac { 3 }{ x }\) + 4y = 7 … (i)
\(\frac { -2 }{ x }\) + 7y = 5 … (ii)
5x + \(\frac { 4 }{ y }\) = 9 … (iii)
Multiplying (i) by 2 and (ii) by 3,
\(\frac { 6 }{ x }\) + 8y = 14
\(\frac { -6 }{ x }\) + 21y = 15
Adding we get,
29y = 29 ⇒ y = \(\frac { 29 }{ 29 }\) = 1
From (i)
\(\frac { 3 }{ x }\) + 4 x 1 ⇒ \(\frac { 3 }{ x }\) + 4 = 1
⇒ \(\frac { 3 }{ x }\) = 7 – 4 = 3 ⇒ x = \(\frac { 3 }{ 3 }\) = 1
∴ x = 1, y = 1
Now substituting the value of x and y in (iii)
5 x 1 + \(\frac { 4 }{ 1 }\) = 9 ⇒ 5 + 4 = 9
⇒ 9 = 9 which is true
Yes, the system of equations are simultaneously and x = 1, y = 1
Question 24.
If the following three equations hold simultaneously for x and y, find p.
3x – 2y = 6, \(\frac{x}{3}-\frac{y}{6}=\frac{1}{2}\), x – py = 6
Solution:
3x – 2y = 6 … (i)
\(\frac{x}{3}-\frac{y}{6}=\frac{1}{2}\) ⇒ 2x – y = 3 … (ii)
(Multiplying by 6)
Multiply (i) by 1 and (ii) by 2
From (i)
3 x 0 – 2y = 6 ⇒ – 2y = 6
∴ The equations are simultaneously
∴ x = 0, y = – 3 will satisfy the third equaiton
∴ x – py = 6 ⇒ 0 – p x (- 3) = 6
= 3p = 6 ⇒ p = \(\frac { 6 }{ 3 }\) = 2
Hence p = 2
Question 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.
Solution:
∵ 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 (i) by (i) and (ii) by 2,
⇒ x = \(\frac{-1}{-2}=\frac{1}{2}\)
From (i)
2x + 6y = 5
⇒ 2 x \(\frac { 1 }{ 2 }\) + 6y = 5
⇒ 1 + 6y = 5 ⇒ 6y = 5 – 1 = 4
⇒ y = \(\frac { 4 }{ 6 }\) = \(\frac { 2 }{ 3 }\)
Now 6x + 3y = 6 x \(\frac { 1 }{ 2 }\) + 3 x \(\frac { 2 }{ 3 }\)
= 3 + 2 = 5
∴ Each side of the triangle = 5 units