Exercise 4.1 — Complete Solutions
Question 1 — Intersecting, Parallel or Coincident?
By comparing the ratios a₁/a₂, b₁/b₂, c₁/c₂, state whether the lines represented by each pair intersect at a point, are parallel, or are coincident.
(a) 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0
a₁ = 5, b₁ = −4, c₁ = 8 | a₂ = 7, b₂ = 6, c₂ = −9
a₁/a₂ = 5/7
b₁/b₂ = −4/6 = −2/3
5/7 ≠ −2/3 ⟹ a₁/a₂ ≠ b₁/b₂
∴ The lines intersect at a point. (Unique solution)
(b) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
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 = 1/2
∴ The lines are coincident. (Infinitely many solutions)
(c) 6x − 3y + 10 = 0 and 2x − y + 9 = 0
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
a₁/a₂ = b₁/b₂ but ≠ c₁/c₂
∴ The lines are parallel. (No solution)
Question 2 — Consistent or Inconsistent? Solve Graphically.
Q2 (a)
3x + 2y = 5 and 2x − 3y = 7
Step 1 — Ratio Check
a₁/a₂ = 3/2, b₁/b₂ = 2/−3 = −2/3
3/2 ≠ −2/3 ⟹ a₁/a₂ ≠ b₁/b₂
Consistent — Lines intersect. Unique solution exists.
| Eq (1): y = (5 − 3x) / 2 | ||
|---|---|---|
| x | y | (x, y) |
| −1 | 4 | (−1, 4) |
| 1 | 1 | (1, 1) |
| 3 | −2 | (3, −2) |
| Eq (2): y = (2x − 7) / 3 | ||
|---|---|---|
| x | y | (x, y) |
| −1 | −3 | (−1, −3) |
| 2 | −1 | (2, −1) |
| 5 | 1 | (5, 1) |
Q2 (b)
2x − 3y = 8 and 4x − 6y = 9
Ratio Check
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = −3/−6 = 1/2, c₁/c₂ = −8/−9 = 8/9
a₁/a₂ = b₁/b₂ but ≠ c₁/c₂ ⟹ Parallel lines
Inconsistent — Lines are parallel. No solution.
| Eq (1): y = (2x − 8) / 3 | ||
|---|---|---|
| x | y | (x, y) |
| 1 | −2 | (1, −2) |
| 4 | 0 | (4, 0) |
| Eq (2): y = (4x − 9) / 6 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | −1.5 | (0, −1.5) |
| 3 | 0.5 | (3, 0.5) |
Both lines have the same slope but different y-intercepts → they never meet.
Q2 (c)
(3/2)x + (5/3)y = 7 and 9x − 10y = 14
Ratio Check (multiply eq1 by 6: 9x + 10y − 42 = 0; eq2 standard: 9x − 10y − 14 = 0)
a₁/a₂ = 9/9 = 1, b₁/b₂ = 10/−10 = −1
1 ≠ −1 ⟹ a₁/a₂ ≠ b₁/b₂
Consistent — Lines intersect. Solution: x ≈ 3.1, y ≈ 1.4
| Eq (1): y = (42 − 9x) / 10 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | 4.2 | (0, 4.2) |
| 2 | 2.4 | (2, 2.4) |
| Eq (2): y = (9x − 14) / 10 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | −1.4 | (0, −1.4) |
| 2 | 0.4 | (2, 0.4) |
Q2 (d)
5x − 3y = 11 and −10x + 6y = −22
Ratio Check
a₁ = 5, b₁ = −3, c₁ = −11 | a₂ = 10, b₂ = −6, c₂ = −22
a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = −11/−22 = 1/2
a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2 ⟹ Coincident lines
Consistent (Dependent) — Infinitely many solutions. Every point on the line is a solution.
| Eq (1): y = (5x − 11) / 3 | ||
|---|---|---|
| x | y | (x, y) |
| 1 | −2 | (1, −2) |
| 4 | 3 | (4, 3) |
| Eq (2): y = (10x − 22) / 6 | ||
|---|---|---|
| x | y | (x, y) |
| −2 | −7 | (−2, −7) |
| 4 | 3 | (4, 3) |
Q2 (e)
(4/3)x + 2y = 8 and 2x + 3y = 12
Ratio Check (eq1 × 3: 4x + 6y − 24 = 0; eq2: 2x + 3y − 12 = 0)
a₁/a₂ = 4/2 = 2, b₁/b₂ = 6/3 = 2, c₁/c₂ = −24/−12 = 2
a₁/a₂ = b₁/b₂ = c₁/c₂ = 2 ⟹ Coincident lines
Consistent (Dependent) — Infinitely many solutions.
| Eq (1): y = (24 − 4x) / 6 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | 4 | (0, 4) |
| 3 | 2 | (3, 2) |
| Eq (2): y = (12 − 2x) / 3 | ||
|---|---|---|
| x | y | (x, y) |
| −3 | 6 | (−3, 6) |
| 6 | 0 | (6, 0) |
Q2 (f)
x + y = 5 and 2x + 2y = 10
Ratio Check
a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = −5/−10 = 1/2
a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2 ⟹ Coincident lines
Consistent (Dependent) — Infinitely many solutions.
| Eq (1): y = 5 − x | ||
|---|---|---|
| x | y | (x, y) |
| 1 | 4 | (1, 4) |
| 3 | 2 | (3, 2) |
| Eq (2): y = (10 − 2x) / 2 | ||
|---|---|---|
| x | y | (x, y) |
| 2 | 3 | (2, 3) |
| 4 | 1 | (4, 1) |
Q2 (g)
x − y = 8 and 3x − 3y = 16
Ratio Check
a₁/a₂ = 1/3, b₁/b₂ = −1/−3 = 1/3, c₁/c₂ = −8/−16 = 1/2
a₁/a₂ = b₁/b₂ but ≠ c₁/c₂ ⟹ Parallel lines
Inconsistent — No solution.
| Eq (1): y = x − 8 | ||
|---|---|---|
| x | y | (x, y) |
| 4 | −4 | (4, −4) |
| 5 | −3 | (5, −3) |
| Eq (2): y = (3x − 16) / 3 | ||
|---|---|---|
| x | y | (x, y) |
| 2 | −3.33 | (2, −3.33) |
| 3 | −2.33 | (3, −2.33) |
Q2 (h)
2x + y − 6 = 0 and 4x − 2y − 4 = 0
Ratio Check
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 1/−2 = −1/2
1/2 ≠ −1/2 ⟹ a₁/a₂ ≠ b₁/b₂
Consistent — Lines intersect. Unique solution.
| Eq (1): y = 6 − 2x | ||
|---|---|---|
| x | y | (x, y) |
| 1 | 4 | (1, 4) |
| 3 | 0 | (3, 0) |
| Eq (2): y = (4x − 4) / 2 | ||
|---|---|---|
| x | y | (x, y) |
| 1 | 0 | (1, 0) |
| 3 | 4 | (3, 4) |
Q2 (i)
2x − 2y − 2 = 0 and 4x − 4y − 5 = 0
Ratio Check
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = −2/−4 = 1/2, c₁/c₂ = −2/−5 = 2/5
a₁/a₂ = b₁/b₂ but ≠ c₁/c₂ ⟹ Parallel lines
Inconsistent — No solution.
| Eq (1): y = (2x − 2) / 2 | ||
|---|---|---|
| x | y | (x, y) |
| 1 | 0 | (1, 0) |
| 3 | 2 | (3, 2) |
| Eq (2): y = (4x − 5) / 4 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | −1.25 | (0, −1.25) |
| 2 | 0.75 | (2, 0.75) |
Question 3 — Pants and Skirts (Word Problem)
Q3
Neha bought some pants and skirts. The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants. Find how many pants and skirts Neha bought.
Solution
Let pants = x, skirts = y
Condition 1: y = 2x − 2 …(1)
Condition 2: y = 4x − 4 …(2)
Equating (1) and (2): 2x − 2 = 4x − 4
⟹ 4 − 2 = 4x − 2x ⟹ 2 = 2x ⟹ x = 1
Substitute x = 1 in (1): y = 2(1) − 2 = 0
∴ Neha bought 1 pant and 0 skirts.
| Eq (1): y = 2x − 2 | ||
|---|---|---|
| x | y | (x, y) |
| 2 | 2 | (2, 2) |
| 3 | 4 | (3, 4) |
| Eq (2): y = 4x − 4 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | −4 | (0, −4) |
| 2 | 4 | (2, 4) |
Lines intersect at (1, 0) → x = 1, y = 0
Question 4 — Mathematics Quiz
Q4
10 students of Class X took part in a Maths quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part.
Solution
Let boys = x, girls = y
x + y = 10 …(1)
y = x + 4 …(2)
Substitute (2) in (1): x + (x + 4) = 10 ⟹ 2x = 6 ⟹ x = 3
y = 3 + 4 = 7
∴ Number of boys = 3, Number of girls = 7
| Eq (1): y = 10 − x | ||
|---|---|---|
| x | y | (x, y) |
| 4 | 6 | (4, 6) |
| 5 | 5 | (5, 5) |
| Eq (2): y = x + 4 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | 4 | (0, 4) |
| 2 | 6 | (2, 6) |
Lines intersect at (3, 7) → x = 3 (boys), y = 7 (girls)
Question 5 — Cost of Pencils and Pens
Q5
5 pencils and 7 pens together cost ₹50. 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and one pen.
Solution
Let cost of one pencil = ₹x, one pen = ₹y
5x + 7y = 50 …(1)
7x + 5y = 46 …(2)
Adding (1) + (2): 12x + 12y = 96 ⟹ x + y = 8 …(3)
Subtracting (2) − (1): 2x − 2y = −4 ⟹ x − y = −2 …(4)
From (3) + (4): 2x = 6 ⟹ x = 3
From (3): y = 8 − 3 = 5
∴ Cost of one pencil = ₹3, Cost of one pen = ₹5
| Eq (1): y = (50 − 5x) / 7 | ||
|---|---|---|
| x | y | (x, y) |
| −4 | 10 | (−4, 10) |
| 10 | 0 | (10, 0) |
| Eq (2): y = (46 − 7x) / 5 | ||
|---|---|---|
| x | y | (x, y) |
| −2 | 12 | (−2, 12) |
| 8 | −2 | (8, −2) |
Lines intersect at (3, 5) → pencil = ₹3, pen = ₹5
Question 6 — Rectangular Garden
Q6
Half the perimeter of a rectangular garden is 36 m. If the length is 4 m more than its width, find the dimensions of the garden.
Solution
Let length = x m, breadth = y m
Half perimeter: x + y = 36 …(1)
Length condition: x − y = 4 …(2)
Adding (1) + (2): 2x = 40 ⟹ x = 20
From (1): y = 36 − 20 = 16
∴ Length = 20 m, Breadth = 16 m
| Eq (1): y = 36 − x | ||
|---|---|---|
| x | y | (x, y) |
| 12 | 24 | (12, 24) |
| 16 | 20 | (16, 20) |
| Eq (2): y = x − 4 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | −4 | (0, −4) |
| 8 | 4 | (8, 4) |
Lines intersect at (20, 16) → length = 20 m, breadth = 16 m
Question 7 — Writing New Equations
Q7
We have a linear equation 2x + 3y − 8 = 0. Write another linear equation in x and y such that the pair forms: (i) intersecting lines, (ii) parallel lines. Also write one more equation so that all three together include a coincident pair.
Solution
Given: 2x + 3y − 8 = 0 here a₁ = 2, b₁ = 3, c₁ = −8
(i) Intersecting line (a₁/a₂ ≠ b₁/b₂): 4x + 9y − 7 = 0
Check: a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/9 = 1/3 ⟹ 1/2 ≠ 1/3 ✓
(ii) Parallel line (a₁/a₂ = b₁/b₂ ≠ c₁/c₂): 8x + 12y + 13 = 0
Check: a₁/a₂ = 2/8 = 1/4, b₁/b₂ = 3/12 = 1/4, c₁/c₂ = −8/13 ⟹ Equal a,b ratios, different c ✓
(iii) Coincident line (a₁/a₂ = b₁/b₂ = c₁/c₂): 6x + 9y − 24 = 0
Check: 2/6 = 3/9 = −8/−24 = 1/3 ✓
Many answers are valid — the ratio conditions are the deciding test.
Question 8 — Rectangle Area Problem
Q8
The area of a rectangle gets reduced by 80 sq units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area increases by 50 sq units. Find the length and breadth of the rectangle.
Solution
Let length = x, breadth = y, original area = xy
Condition 1: (x − 5)(y + 2) = xy − 80
⟹ xy + 2x − 5y − 10 = xy − 80
⟹ 2x − 5y + 70 = 0 …(1)
Condition 2: (x + 10)(y − 5) = xy + 50
⟹ xy − 5x + 10y − 50 = xy + 50
⟹ −5x + 10y − 100 = 0 ⟹ x − 2y + 20 = 0 …(2)
From (2): x = 2y − 20. Substitute in (1): 2(2y − 20) − 5y + 70 = 0
⟹ 4y − 40 − 5y + 70 = 0 ⟹ −y + 30 = 0 ⟹ y = 30
x = 2(30) − 20 = 40
∴ Length = 40 units, Breadth = 30 units
| Eq (1): y = (2x + 70) / 5 | ||
|---|---|---|
| x | y | (x, y) |
| 15 | 20 | (15, 20) |
| −35 | 0 | (−35, 0) |
| Eq (2): y = (x + 20) / 2 | ||
|---|---|---|
| x | y | (x, y) |
| 20 | 20 | (20, 20) |
| 0 | 10 | (0, 10) |
Lines intersect at (40, 30) → length = 40, breadth = 30
Question 9 — Benches and Students
Q9
In a class, if 3 students sit on each bench, one student will be left over. If 4 students sit on each bench, one bench will be left empty. Find the number of students and benches in the class.
Solution
Let benches = x, students = y
Condition 1: y = 3x + 1 …(1)
Condition 2: y = 4(x − 1) = 4x − 4 …(2)
Equating: 3x + 1 = 4x − 4 ⟹ 5 = x ⟹ x = 5
y = 3(5) + 1 = 16
∴ Number of benches = 5, Number of students = 16
| Eq (1): y = 3x + 1 | ||
|---|---|---|
| x | y | (x, y) |
| 3 | 10 | (3, 10) |
| 7 | 22 | (7, 22) |
| Eq (2): y = 4x − 4 | ||
|---|---|---|
| x | y | (x, y) |
| 1 | 0 | (1, 0) |
| 4 | 12 | (4, 12) |
Lines intersect at (5, 16) → benches = 5, students = 16
Quick Summary — Exercise 4.1 All Results
| Q | Equations | Type of Lines | Solution |
|---|---|---|---|
| 1a | 5x−4y+8=0, 7x+6y−9=0 | Intersecting | Unique solution |
| 1b | 9x+3y+12=0, 18x+6y+24=0 | Coincident | Infinite solutions |
| 1c | 6x−3y+10=0, 2x−y+9=0 | Parallel | No solution |
| 2a | 3x+2y=5, 2x−3y=7 | Intersecting | x ≈ 2.2, y ≈ 0.8 |
| 2b | 2x−3y=8, 4x−6y=9 | Parallel | No solution |
| 2c | (3/2)x+(5/3)y=7, 9x−10y=14 | Intersecting | x ≈ 3.1, y ≈ 1.4 |
| 2d | 5x−3y=11, −10x+6y=−22 | Coincident | Infinite solutions |
| 2e | (4/3)x+2y=8, 2x+3y=12 | Coincident | Infinite solutions |
| 2f | x+y=5, 2x+2y=10 | Coincident | Infinite solutions |
| 2g | x−y=8, 3x−3y=16 | Parallel | No solution |
| 2h | 2x+y−6=0, 4x−2y−4=0 | Intersecting | x = 2, y = 2 |
| 2i | 2x−2y−2=0, 4x−4y−5=0 | Parallel | No solution |
| 3 | y = 2x−2, y = 4x−4 | Intersecting | Pants = 1, Skirts = 0 |
| 4 | x+y=10, y=x+4 | Intersecting | Boys = 3, Girls = 7 |
| 5 | 5x+7y=50, 7x+5y=46 | Intersecting | Pencil = ₹3, Pen = ₹5 |
| 6 | x+y=36, x−y=4 | Intersecting | L = 20 m, B = 16 m |
| 8 | 2x−5y+70=0, x−2y+20=0 | Intersecting | L = 40, B = 30 |
| 9 | y=3x+1, y=4x−4 | Intersecting | Benches = 5, Students = 16 |