Chapter 3: Pair of Linear Equations in Two Variables

Key Conditions for a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0:
• Unique solution: a₁/a₂ ≠ b₁/b₂
• Infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂
• No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
• Cross multiplication: x/(b₁c₂-b₂c₁) = y/(c₁a₂-c₂a₁) = 1/(a₁b₂-a₂b₁)

Exercise 3.1

Q1. Aftab tells his daughter: 7 years ago I was 7 times as old as you. 3 years from now I shall be 3 times as old. Represent algebraically.

Solution

Let father's age = x, daughter's age = y.
7 years ago: x-7 = 7(y-7) → x-7y = -42
3 years from now: x+3 = 3(y+3) → x-3y = 6
∴ Equations: x-7y+42=0 and x-3y-6=0

Q2. 7 bats and 6 balls cost Rs 3800. 3 bats and 5 balls cost Rs 1750. Find cost of each.

Solution

7x+6y = 3800 ...(i)
3x+5y = 1750 ...(ii)
From (ii): x = (1750-5y)/3
Substituting: 7(1750-5y)/3+6y = 3800
12250-35y+18y = 11400
-17y = -850 → y = 50
x = (1750-250)/3 = 500
∴ Bat = Rs 500, Ball = Rs 50

Q5. Half the perimeter of a rectangular garden is 36m, length is 4m more than width. Find dimensions.

Solution

x = y+4 and (x+y)/2 = 36 → x+y = 72
(y+4)+y = 72 → 2y = 68 → y = 34
x = 38
∴ Length = 38m, Width = 34m

Exercise 3.2

Q1. 10 students in math quiz. Girls are 4 more than boys. Find boys and girls.

Solution

x+y=10, y=x+4
x+(x+4)=10 → 2x=6 → x=3, y=7
∴ Boys = 3, Girls = 7

Q2. 5 pencils and 7 pens cost Rs 50. 7 pencils and 5 pens cost Rs 46. Find cost of each.

Solution

5x+7y=50, 7x+5y=46
Multiply (i) by 5 and (ii) by 7:
25x+35y=250, 49x+35y=322
Subtracting: -24x=-72 → x=3
5(3)+7y=50 → 7y=35 → y=5
∴ Pencil = Rs 3, Pen = Rs 5

Q3. Determine consistent/inconsistent: (i) x+y=5; 2x+2y=10 (ii) x-y=8; 3x-3y=16

Solution

(i) 1/2=1/2=5/10 → all equal → Consistent, infinitely many solutions
(ii) 1/3=(-1)/(-3)=1/3 ≠ 8/16=1/2 → Inconsistent (no solution)

Q5. Half perimeter of rectangle is 36m, length 4m more than width. Find dimensions.

Solution

Same as Exercise 3.1 Q5: Length=38m, Width=34m

Exercise 3.3

Q1. Solve by substitution: (i) x+y=14; x-y=4 (ii) s-t=3; s/3+t/2=6

Solution

(i)

x=14-y
(14-y)-y=4 → 14-2y=4 → y=5
x=14-5=9
∴ x=9, y=5

(ii)

2s+3t=36 (multiplying by 6)
s=t+3
2(t+3)+3t=36 → 5t+6=36 → t=6
s=6+3=9
∴ s=9, t=6

Q2. Solve 2x+3y=11 and 2x-4y=-2, find m if y=mx+3.

Solution

Subtracting: (2x+3y)-(2x-4y)=11-(-2)
7y=13 → y=13/7
2x=11-3(13/7)=38/7 → x=19/7
For y=mx+3: 13/7=m(19/7)+3
13/7-21/7=19m/7
-8/7=19m/7 → m=-8/19

Q4. Solve by elimination: (i) x+y=5; 2x-3y=4 (ii) 3x+4y=10; 2x-2y=2

Solution

(i)

Multiply (i) by 3: 3x+3y=15
Add to (ii): 5x=19 → x=19/5
y=5-19/5=6/5
∴ x=19/5, y=6/5

(ii)

Multiply (ii) by 2: 4x-4y=4
Add to (i): 7x=14 → x=2
2(2)-2y=2 → y=1
∴ x=2, y=1

Q5. A library charges fixed + per day. Sheela paid Rs 27 for 7 days, Sunita paid Rs 21 for 5 days. Find charges.

Solution

Fixed charge = x, extra per day = y
x+4y = 27 (3 days fixed + 4 extra)
x+2y = 21 (3 days fixed + 2 extra)
Subtracting: 2y=6 → y=3
x=21-6=15
∴ Fixed = Rs 15, Extra = Rs 3/day

Exercise 3.4

Q1. Solve by cross multiplication: (i) x-3y-3=0; 3x-9y-2=0 (ii) 2x+y-5=0; 3x+2y-8=0

Solution

(i) 1/3=(-3)/(-9)=1/3 ≠ -3/(-2)=3/2 → No solution

(ii) x/(-8+10) = y/(-15+16) = 1/(4-3)
x/2 = y/1 = 1/1
∴ x=2, y=1

Exercise 3.5

Q2. Find a and b for infinitely many solutions: 2x+3y=7; (a-b)x+(a+b)y=3a+b-2

Solution

2/(a-b) = 3/(a+b) = 7/(3a+b-2)
From first two: 2(a+b)=3(a-b)
2a+2b=3a-3b → a=5b ...(i)
From first and third: 2(3a+b-2)=7(a-b)
6a+2b-4=7a-7b → a=9b-4 ...(ii)
From (i) and (ii): 5b=9b-4 → 4b=4 → b=1
a=5(1)=5
∴ a=5, b=1

Q3. For what k does 3x+y=1; (2k-1)x+(k-1)y=2k+1 have no solution?

Solution

3/(2k-1) = 1/(k-1) ≠ 1/(2k+1)
3(k-1) = 2k-1
3k-3 = 2k-1 → k=2
Check: 3/(3)=1/(1)=1 ≠ 1/(5) ✓
∴ k=2

Exercise 3.6

Q1. Solve: 1/x+1/y=10; 1/x-1/y=2

Solution

Let p=1/x, q=1/y
p+q=10, p-q=2
Adding: 2p=12 → p=6, q=4
∴ x=1/6, y=1/4

Q2. Solve: 6/x+3/y=6; 4/x+9/y=5

Solution

Let p=1/x, q=1/y
6p+3q=6 → 2p+q=2 ...(i)
4p+9q=5 ...(ii)
From (i): q=2-2p
4p+9(2-2p)=5
4p+18-18p=5
-14p=-13 → p=13/14 → x=14/13
q=2-26/14=2/14=1/7 → y=7
∴ x=14/13, y=7

Exercise 3.7 (Optional)

Q4. A boat goes 30km upstream and 44km downstream in 10h. It goes 40km upstream and 55km downstream in 13h. Find speed of boat and stream.

Solution

Let boat speed=x, stream speed=y
30/(x-y)+44/(x+y)=10 ...(i)
40/(x-y)+55/(x+y)=13 ...(ii)
Let p=1/(x-y), q=1/(x+y)
30p+44q=10, 40p+55q=13
Multiply first by 4, second by 3:
120p+176q=40, 120p+165q=39
11q=1 → q=1/11 → x+y=11
30p=10-44/11=6 → p=1/5 → x-y=5
x=(11+5)/2=8, y=(11-5)/2=3
∴ Boat=8 km/h, Stream=3 km/h

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