Solution:

Let number of boys = x, number of girls = y

Equation 1: x + y = 10

Equation 2: y = x + 4

Graphical Solution:

For x + y = 10: (0,10), (5,5), (10,0)

For y = x + 4: (0,4), (2,6), (4,8)

Both lines intersect at (3,7)

∴ Boys = 3, Girls = 7

Solution:

Let cost of 1 pencil = ₹x, cost of 1 pen = ₹y

Equation 1: 5x + 7y = 50

Equation 2: 7x + 5y = 46

Add equation 1 and 2: 12x + 12y = 96 → x + y = 8 ...(3)

Subtract equation 2 from 1: (5x+7y) - (7x+5y) = 50 - 46 → -2x + 2y = 4 → -x + y = 2 ...(4)

From (3) and (4): y = 5, x = 3

∴ Pencil = ₹3, Pen = ₹5

Solution:

Equation 1: a₁ = 5, b₁ = -4, c₁ = 8

Equation 2: a₂ = 7, b₂ = 6, c₂ = -9

a₁/a₂ = 5/7

b₁/b₂ = -4/6 = -2/3

Since 5/7 ≠ -2/3, therefore a₁/a₂ ≠ b₁/b₂

∴ Lines intersect at a point (unique solution)

Solution:

a₁ = 9, b₁ = 3, c₁ = 12

a₂ = 18, b₂ = 6, c₂ = 24

a₁/a₂ = 9/18 = 1/2

b₁/b₂ = 3/6 = 1/2

c₁/c₂ = 12/24 = 1/2

a₁/a₂ = b₁/b₂ = c₁/c₂

∴ Lines are coincident (infinitely many solutions)

Solution:

a₁ = 6, b₁ = -3, c₁ = 10

a₂ = 2, b₂ = -1, c₂ = 9

a₁/a₂ = 6/2 = 3

b₁/b₂ = (-3)/(-1) = 3

c₁/c₂ = 10/9 ≠ 3

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

∴ Lines are parallel (no solution)

Solution:

a₁ = 1, b₁ = 1, c₁ = -5

a₂ = 2, b₂ = 2, c₂ = -10

a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = (-5)/(-10) = 1/2

a₁/a₂ = b₁/b₂ = c₁/c₂

∴ Consistent (coincident lines)

Solution:

a₁ = 1, b₁ = -1, c₁ = -8

a₂ = 3, b₂ = -3, c₂ = -16

a₁/a₂ = 1/3, b₁/b₂ = (-1)/(-3) = 1/3, c₁/c₂ = (-8)/(-16) = 1/2

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

∴ Inconsistent (parallel lines, no solution)

Solution:

a₁ = 2, b₁ = 1, c₁ = -6

a₂ = 4, b₂ = -2, c₂ = -4

a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 1/(-2) = -1/2

a₁/a₂ ≠ b₁/b₂

∴ Consistent (intersecting lines, unique solution)

Solution:

a₁ = 2, b₁ = -2, c₁ = -2

a₂ = 4, b₂ = -4, c₂ = -5

a₁/a₂ = 2/4 = 1/2, b₁/b₂ = (-2)/(-4) = 1/2, c₁/c₂ = (-2)/(-5) = 2/5

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

∴ Inconsistent (parallel lines)

Solution:

Let length = l, width = w

Half perimeter = l + w = 36 ...(1)

Length is 4 m more than width: l = w + 4 ...(2)

Sub (2) in (1): w + 4 + w = 36 → 2w = 32 → w = 16 m

l = 16 + 4 = 20 m

∴ Length = 20 m, Width = 16 m

Solution:

(i) Intersecting lines: a₁/a₂ ≠ b₁/b₂

Take equation: 3x + 2y - 7 = 0 (since 2/3 ≠ 3/2)

(ii) Parallel lines: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Take equation: 4x + 6y - 5 = 0 (since 2/4 = 3/6 = 1/2, but -8/-5 ≠ 1/2)

(iii) Coincident lines: a₁/a₂ = b₁/b₂ = c₁/c₂

Take equation: 6x + 9y - 24 = 0 (multiply original by 3)

Solution:

Equation 1: x - y + 1 = 0 → y = x + 1

Equation 2: 3x + 2y - 12 = 0 → 2y = 12 - 3x → y = (12 - 3x)/2

Finding intersection point:

x + 1 = (12 - 3x)/2 → 2x + 2 = 12 - 3x → 5x = 10 → x = 2, y = 3

Intersection point A = (2, 3)

Intersection with x-axis (y = 0):

For equation 1: x - 0 + 1 = 0 → x = -1 → B = (-1, 0)

For equation 2: 3x + 0 - 12 = 0 → 3x = 12 → x = 4 → C = (4, 0)

Vertices of triangle: A(2,3), B(-1,0), C(4,0)

Solution:

x - y = 4 → x = y + 4

Substitute in x + y = 14:

(y + 4) + y = 14 → 2y + 4 = 14 → 2y = 10 → y = 5

x = 5 + 4 = 9

∴ x = 9, y = 5

Solution:

s - t = 3 → s = t + 3

Substitute in s/3 + t/2 = 6:

(t+3)/3 + t/2 = 6

Multiply by 6: 2(t+3) + 3t = 36

2t + 6 + 3t = 36 → 5t = 30 → t = 6

s = 6 + 3 = 9

∴ s = 9, t = 6

Solution:

3x - y = 3 → y = 3x - 3

Substitute in 9x - 3y = 9:

9x - 3(3x - 3) = 9

9x - 9x + 9 = 9 → 9 = 9

∴ Infinitely many solutions (dependent equations)

Solution:

Multiply by 10: 2x + 3y = 13 ...(1)

4x + 5y = 23 ...(2)

From (1): 2x = 13 - 3y → x = (13 - 3y)/2

Substitute in (2): 4(13-3y)/2 + 5y = 23

2(13-3y) + 5y = 23 → 26 - 6y + 5y = 23

26 - y = 23 → y = 3

x = (13 - 9)/2 = 4/2 = 2

∴ x = 2, y = 3

Solution:

From first: √2x = -√3y → x = (-√3/√2)y

Substitute in second: √3(-√3/√2 y) - √8y = 0

(-3/√2)y - 2√2y = 0

y(-3/√2 - 2√2) = 0

Since (-3/√2 - 2√2) ≠ 0, therefore y = 0

Then x = 0

∴ x = 0, y = 0

Solution:

Multiply first by 6: 9x - 10y = -12 ...(1)

Multiply second by 6: 2x + 3y = 13 ...(2)

From (2): 2x = 13 - 3y → x = (13 - 3y)/2

Sub in (1): 9(13-3y)/2 - 10y = -12

Multiply by 2: 9(13-3y) - 20y = -24

117 - 27y - 20y = -24 → 117 - 47y = -24

-47y = -141 → y = 3

x = (13 - 9)/2 = 4/2 = 2

∴ x = 2, y = 3

Solution:

2x + 3y = 11 ...(1)

2x - 4y = -24 ...(2)

Subtract (2) from (1): (2x+3y) - (2x-4y) = 11 - (-24)

7y = 35 → y = 5

Substitute in (1): 2x + 15 = 11 → 2x = -4 → x = -2

Given y = mx + 3 → 5 = m(-2) + 3

5 = -2m + 3 → -2m = 2 → m = -1

∴ x = -2, y = 5, m = -1

Solution:

Let numbers be x and y (x > y)

x - y = 26 ...(1)

x = 3y ...(2)

Substitute (2) in (1): 3y - y = 26 → 2y = 26 → y = 13

x = 3 × 13 = 39

∴ Numbers are 39 and 13

Solution:

Let smaller angle = x, larger angle = y

Supplementary: x + y = 180 ...(1)

y = x + 18 ...(2)

Substitute (2) in (1): x + x + 18 = 180 → 2x = 162 → x = 81°

y = 81 + 18 = 99°

∴ Angles are 81° and 99°

Solution:

Let cost of bat = ₹x, ball = ₹y

7x + 6y = 3800 ...(1)

3x + 5y = 1750 ...(2)

From (2): 3x = 1750 - 5y → x = (1750 - 5y)/3

Sub in (1): 7(1750-5y)/3 + 6y = 3800

Multiply by 3: 7(1750-5y) + 18y = 11400

12250 - 35y + 18y = 11400 → 12250 - 17y = 11400

-17y = -850 → y = 50

x = (1750 - 250)/3 = 1500/3 = 500

∴ Bat = ₹500, Ball = ₹50

Solution:

Let fixed charge = ₹x, charge per km = ₹y

For 10 km: x + 10y = 105 ...(1)

For 15 km: x + 15y = 155 ...(2)

Subtract (1) from (2): 5y = 50 → y = 10 (per km)

Sub in (1): x + 100 = 105 → x = 5 (fixed)

For 25 km: 5 + 25 × 10 = 5 + 250 = ₹255

∴ Fixed = ₹5, per km = ₹10, for 25 km = ₹255

Solution:

Let fraction = x/y

(x+2)/(y+2) = 9/11 → 11x + 22 = 9y + 18 → 11x - 9y = -4 ...(1)

(x+3)/(y+3) = 5/6 → 6x + 18 = 5y + 15 → 6x - 5y = -3 ...(2)

From (1): 11x = 9y - 4 → x = (9y - 4)/11

Sub in (2): 6(9y-4)/11 - 5y = -3

Multiply by 11: 6(9y-4) - 55y = -33

54y - 24 - 55y = -33 → -y - 24 = -33 → -y = -9 → y = 9

x = (81 - 4)/11 = 77/11 = 7

∴ Fraction = 7/9

Solution:

Let Jacob = x, son = y

Five years hence: x + 5 = 3(y + 5) → x + 5 = 3y + 15 → x - 3y = 10 ...(1)

Five years ago: x - 5 = 7(y - 5) → x - 5 = 7y - 35 → x - 7y = -30 ...(2)

Subtract (2) from (1): (x-3y) - (x-7y) = 10 - (-30)

4y = 40 → y = 10

x = 3(10) + 10 = 40

∴ Jacob = 40 years, Son = 10 years

🔷 METHOD 1: Elimination Method

Given equations:
(1) x + y = 5
(2) 2x - 3y = 4

Step 1: Eliminate x. Multiply equation (1) by 2:

2(x + y) = 2 × 5 → 2x + 2y = 10 ...(3)

Step 2: Subtract equation (2) from equation (3):

(2x + 2y) - (2x - 3y) = 10 - 4

2x + 2y - 2x + 3y = 6

5y = 6

y = 6/5 = 1.2

Step 3: Substitute y = 6/5 in equation (1):

x + 6/5 = 5

x = 5 - 6/5 = (25 - 6)/5 = 19/5

x = 19/5 = 3.8

✅ Answer using Elimination: x = 19/5, y = 6/5

🔷 METHOD 2: Substitution Method

From equation (1): x = 5 - y

Substitute in equation (2): 2(5 - y) - 3y = 4

10 - 2y - 3y = 4

10 - 5y = 4

-5y = 4 - 10 = -6

y = 6/5

x = 5 - 6/5 = 19/5

✅ Answer using Substitution: x = 19/5, y = 6/5

🔷 METHOD 1: Elimination Method

Given equations:
(1) 3x + 4y = 10
(2) 2x - 2y = 2

Step 1: Simplify equation (2) by dividing by 2:

x - y = 1 ...(3)

Step 2: Eliminate y. Multiply equation (3) by 4:

4x - 4y = 4 ...(4)

Step 3: Add equation (1) and (4):

(3x + 4y) + (4x - 4y) = 10 + 4

7x = 14

x = 2

Step 4: Substitute x = 2 in equation (3):

2 - y = 1 → -y = 1 - 2 = -1 → y = 1

✅ Answer: x = 2, y = 1

🔷 METHOD 2: Substitution Method

From equation (3): x = y + 1

Substitute in (1): 3(y + 1) + 4y = 10

3y + 3 + 4y = 10 → 7y = 7 → y = 1

x = 1 + 1 = 2

Step 1: Write equations in standard form:

(1) 3x - 5y = 4

(2) 9x - 2y = 7

Step 2: Eliminate x. Multiply equation (1) by 3:

9x - 15y = 12 ...(3)

Step 3: Subtract equation (2) from equation (3):

(9x - 15y) - (9x - 2y) = 12 - 7

9x - 15y - 9x + 2y = 5

-13y = 5

y = -5/13

Step 4: Substitute y = -5/13 in equation (1):

3x - 5(-5/13) = 4

3x + 25/13 = 4

3x = 4 - 25/13 = (52 - 25)/13 = 27/13

x = 9/13

✅ Answer: x = 9/13, y = -5/13

Step 1: Eliminate fractions by multiplying:

Equation (1): x/2 + 2y/3 = -1 → Multiply by 6: 3x + 4y = -6 ...(1)

Equation (2): x - y/3 = 3 → Multiply by 3: 3x - y = 9 ...(2)

Step 2: Subtract equation (2) from equation (1):

(3x + 4y) - (3x - y) = -6 - 9

3x + 4y - 3x + y = -15

5y = -15

y = -3

Step 3: Substitute y = -3 in equation (2):

3x - (-3) = 9 → 3x + 3 = 9 → 3x = 6 → x = 2

✅ Answer: x = 2, y = -3

Step 1: Let fraction = x/y (x = numerator, y = denominator)

Step 2: First condition: If we add 1 to numerator and subtract 1 from denominator → becomes 1

(x + 1)/(y - 1) = 1

x + 1 = y - 1

x - y = -2 ...(1)

Step 3: Second condition: If only 1 is added to denominator → becomes 1/2

x/(y + 1) = 1/2

2x = y + 1

2x - y = 1 ...(2)

Step 4: Subtract equation (1) from equation (2):

(2x - y) - (x - y) = 1 - (-2)

x = 3

Step 5: Substitute x = 3 in equation (1):

3 - y = -2 → -y = -5 → y = 5

✅ Answer: Fraction = 3/5

Check: (3+1)/(5-1) = 4/4 = 1 ✓; 3/(5+1) = 3/6 = 1/2 ✓

Step 1: Let Nuri's present age = x, Sonu's present age = y

Step 2: Five years ago: Nuri's age = x - 5, Sonu's age = y - 5

(x - 5) = 3(y - 5)

x - 5 = 3y - 15

x - 3y = -10 ...(1)

Step 3: Ten years later: Nuri's age = x + 10, Sonu's age = y + 10

(x + 10) = 2(y + 10)

x + 10 = 2y + 20

x - 2y = 10 ...(2)

Step 4: Subtract equation (1) from equation (2):

(x - 2y) - (x - 3y) = 10 - (-10)

y = 20

Step 5: Substitute y = 20 in equation (2):

x - 2(20) = 10 → x - 40 = 10 → x = 50

✅ Answer: Nuri = 50 years, Sonu = 20 years

Check: 5 years ago: 45 = 3×15 ✓; 10 years later: 60 = 2×30 ✓

Step 1: Let tens digit = x, units digit = y

Number = 10x + y, Reversed number = 10y + x

Step 2: First condition: Sum of digits = 9

x + y = 9 ...(1)

Step 3: Second condition: 9 times number = 2 times reversed number

9(10x + y) = 2(10y + x)

90x + 9y = 20y + 2x

90x - 2x + 9y - 20y = 0

88x - 11y = 0

Divide by 11: 8x - y = 0 → y = 8x ...(2)

Step 4: Substitute y = 8x in equation (1):

x + 8x = 9 → 9x = 9 → x = 1

y = 8 × 1 = 8

Step 5: Number = 10x + y = 10(1) + 8 = 18

✅ Answer: Number = 18

Check: Sum = 1+8=9 ✓; 9×18=162, 2×81=162 ✓

Step 1: Let number of ₹50 notes = x, number of ₹100 notes = y

Step 2: Total notes:

x + y = 25 ...(1)

Step 3: Total amount:

50x + 100y = 2000

Divide by 50: x + 2y = 40 ...(2)

Step 4: Subtract equation (1) from equation (2):

(x + 2y) - (x + y) = 40 - 25

y = 15

Step 5: x = 25 - 15 = 10

✅ Answer: ₹50 notes = 10, ₹100 notes = 15

Check: Total notes = 10+15=25 ✓; Amount = 500+1500=2000 ✓

Step 1: Let fixed charge for first 3 days = ₹x

Additional charge per day = ₹y

Step 2: For Saritha (7 days): Fixed for 3 days + extra for 4 days

x + 4y = 27 ...(1)

Step 3: For Susy (5 days): Fixed for 3 days + extra for 2 days

x + 2y = 21 ...(2)

Step 4: Subtract equation (2) from equation (1):

(x + 4y) - (x + 2y) = 27 - 21

2y = 6 → y = 3

Step 5: Substitute y = 3 in equation (2):

x + 2(3) = 21 → x + 6 = 21 → x = 15

✅ Answer: Fixed charge for first 3 days = ₹15, Additional charge per day = ₹3

Check: Saritha: 15 + 4×3 = 15+12=27 ✓; Susy: 15 + 2×3 = 15+6=21 ✓