How do you form the linear equation for the auto fare problem in Exercise 6.4?

The auto charges ₹15 for the first kilometre and ₹8 for each subsequent kilometre. For x km total, the fare is: y = 15 + 8(x − 1) = 15 + 8x − 8 = 8x + 7. So the equation is y = 8x + 7. From this, if the fare paid is ₹55: 55 = 8x + 7 → x = 6 km. If the distance is 7 km: y = 8×7 + 7 = ₹63.

What is the equation for the Celsius to Fahrenheit conversion and what does the graph show?

The equation is F = (9/5)C + 32, or in graph form y = (9/5)x + 32 where x is Celsius and y is Fahrenheit. From the graph: 30°C = 86°F (since 9/5 × 30 + 32 = 86); 95°F = 35°C (solving 95 = 9x/5 + 32 gives x = 35). At −40°, both scales give the same numerical value because −40 = (9/5)(−40) + 32 holds true — this is the only temperature where Celsius and Fahrenheit are equal.

How do you form a linear equation for ratio problems like hydrogen-oxygen in water?

The molecular weight ratio of hydrogen to oxygen in water is 1:8. If hydrogen = x grams and oxygen = y grams, then x/y = 1/8, which gives y = 8x. This is a y = mx equation passing through the origin. From the graph: if oxygen = 12 g, then 12 = 8x → x = 1.5 g of hydrogen. If hydrogen = 3/2 g, then y = 8 × 1.5 = 12 g of oxygen.

What is the equation for parking charges at Hyderabad Railway Station and how do you read the graph?

The parking charges are ₹50 for the first 2 hours and ₹10 for each subsequent hour. For x hours total: y = 50 + 10(x − 2) = 10x + 30. The equation y = 10x + 30 is graphed with time on x-axis and charge on y-axis. Reading from the graph: 3 hours → ₹60; 6 hours → ₹90; if Rekha paid ₹80, then 80 = 10x + 30 → x = 5 hours.

Exercise 6.4 — Applications | Class 9 Maths | Linear Equations in Two Variables

Applications of Linear Equations — Graphical Solutions

Exercise 6.4 brings linear equations to life through real-world contexts — elections, ages, auto fares, library charges, parking fees, speed, chemical ratios, and temperature conversion. In each problem, you form the equation from the given situation, draw its graph, and then read specific answers directly from the graph. This is the power of graphical method: once the line is drawn, infinitely many answers are available by inspection.

General approach for every application problem in Exercise 6.4:
1. Identify the two unknown quantities and name them x and y.
2. Translate the given relationship into a linear equation in x and y.
3. Find two or three solution points by substituting convenient values.
4. Plot the points and draw the line.
5. Read the required values directly from the graph using dashed reference lines.

Question 1 — Election Voting (60% turnout)

Problem: 60% of total voters cast their votes. Form an equation and draw the graph.

Let total voters = y, votes cast = x

Condition: x = 60% of y = (60/100) × y = (3/5)y

Rearranging: y = (5/3)x

Equation: y = (5x)/3

x (votes cast)	y = 5x/3 (total voters)	(x, y)
300	500	(300, 500)
600	1000	(600, 1000)

(i) If 1200 votes cast → total voters?

y = 5×1200/3 = 2000
Total voters = 2000

(ii) If total voters = 800 → votes cast?

800 = 5x/3 → x = 480
Votes cast = 480

Question 2 — Father's Age (Rupa's Birth)

Problem: When Rupa was born, her father was 25 years old.

Let Rupa's age = x years, Father's age = y years

At Rupa's birth: Father was 25. At any time: Father = Rupa + 25

Equation: y = x + 25

x (Rupa's age)	y = x + 25 (Father's age)	(x, y)
10	35	(10, 35)
20	45	(20, 45)

(i) Rupa is 25 → Father's age?

y = 25 + 25 = 50
Father's age = 50 years

(ii) Father is 40 → Rupa's age?

40 = x + 25 → x = 15
Rupa's age = 15 years

Question 3 — Auto Fare (₹15 first km, ₹8 each subsequent km)

Problem: Auto charges ₹15 for 1st km, ₹8 per km thereafter.

Let distance = x km, total fare = ₹y

Fare = ₹15 + ₹8 × (x − 1) = 15 + 8x − 8 = 8x + 7

Equation: y = 8x + 7

x (km)	y = 8x + 7 (₹)	(x, y)
4	39	(4, 39)
8	71	(8, 71)

Fare paid = ₹55 → distance?

55 = 8x + 7 → x = 6
Distance = 6 km

Distance = 7 km → fare?

y = 8×7 + 7 = 63
Fare = ₹63

Question 4 — Lending Library (Fixed + Per-day Charges)

Problem: Fixed charges for 3 days + ₹y per extra day. John paid ₹27 for 7 days.

Fixed charge = ₹x (for first 3 days), extra charge per day = ₹y

John's 7 days = 3 fixed + 4 extra: x + 4y = 27

Equation: x + 4y = 27 ⟹ y = (27 − x) / 4

x (fixed ₹)	y = (27−x)/4	(x, y)
3	6	(3, 6)
15	3	(15, 3)

If extra charge = ₹4/day → fixed charge?

x + 4×4 = 27 → x = 11
Fixed charge = ₹11

If fixed charge = ₹7 → extra/day?

7 + 4y = 27 → y = 5
Extra charge = ₹5/day

Question 5 — Parking Charges (₹50 first 2 hrs, ₹10/hr after)

Problem: Hyderabad Railway Station parking — ₹50 for first 2 hours, ₹10 each subsequent hour.

Let time parked = x hours, charge = ₹y

Charge = 50 + 10(x − 2) = 50 + 10x − 20 = 10x + 30

Equation: y = 10x + 30

x (hours)	y = 10x + 30 (₹)	(x, y)
4	70	(4, 70)
7	100	(7, 100)

(i) Charge for 3 hours?

y = 10×3 + 30 = 60
₹60

(ii) Charge for 6 hours?

y = 10×6 + 30 = 90
₹90

(iii) Rekha paid ₹80 → time parked?

80 = 10x + 30 → x = 5
5 hours

Question 6 — Distance-Time Graph (Speed = 60 kmph)

Problem: Sameera drives at uniform speed 60 kmph. Draw distance-time graph.

Distance = x km, Time = y hours

Time = Distance/Speed: y = x/60

Equation: y = x/60

x (km)	y = x/60 (hours)	(x, y)
180	3	(180, 3)
300	5	(300, 5)

(i) Distance in 1½ hours?

x = 60 × 1.5 = 90
90 km

(ii) Distance in 2 hours?

x = 60 × 2 = 120
120 km

(iii) Distance in 3½ hours?

x = 60 × 3.5 = 210
210 km

Question 7 — Hydrogen and Oxygen Ratio in Water (1:8)

Problem: Molecular weight ratio of H : O in water = 1 : 8.

Let Hydrogen = x grams, Oxygen = y grams

x : y = 1 : 8 ⟹ 8x = y

Equation: y = 8x

x (H in grams)	y = 8x (O in grams)	(x, y)
2	16	(2, 16)
4	32	(4, 32)

(i) Oxygen = 12 g → Hydrogen?

12 = 8x → x = 1.5
Hydrogen = 1.5 grams

(ii) Hydrogen = 3/2 g → Oxygen?

y = 8 × 1.5 = 12
Oxygen = 12 grams

Question 8 — Milk-Water Mixture (Ratio 5:2, 28 litres total)

Problem: Mixture:Milk ratio = 7:5 (since milk:water = 5:2, total parts = 7).

Let mixture = x litres, milk = y litres

x : y = 7 : 5 ⟹ 7y = 5x ⟹ y = (5/7)x

Equation: y = (5/7)x

x (mixture litres)	y = 5x/7 (milk litres)	(x, y)
7	5	(7, 5)
14	10	(14, 10)

From the graph: When mixture = 28 litres → milk = 5×28/7 = 20 litres.

Question 9 — Celsius to Fahrenheit Conversion

Problem: F = (9/5)C + 32. Draw graph (Celsius on x-axis, Fahrenheit on y-axis).

Let Celsius temperature = x, Fahrenheit = y

Equation: y = (9/5)x + 32

x (°C)	y = (9/5)x + 32 (°F)	(x, y)
20	68	(20, 68)
40	104	(40, 104)

(ii) 30°C in Fahrenheit?

y = (9/5)×30 + 32 = 54 + 32 = 86
30°C = 86°F

(iii) 95°F in Celsius?

95 = (9/5)x + 32 → x = 35
95°F = 35°C

(iv) Same value in °C and °F?

x = (9/5)x + 32 → −(4/5)x = 32 → x = −40
−40° is same in both scales

Summary Table — All Exercise 6.4 Equations

Q	Context	Equation	Type
1	Election (60% turnout)	y = (5/3)x	y = mx through origin
2	Father-daughter age	y = x + 25	y = x + c (parallel to y=x)
3	Auto fare	y = 8x + 7	y = mx + c
4	Library charges	x + 4y = 27	ax + by = c
5	Parking charges	y = 10x + 30	y = mx + c
6	Distance-time	y = x/60	y = mx through origin
7	H:O ratio in water	y = 8x	y = mx through origin
8	Milk-mixture ratio	y = (5/7)x	y = mx through origin
9	°C to °F conversion	y = (9/5)x + 32	y = mx + c

Problems of the form y = mx (no constant) always produce lines through the origin — election ratios, speed, chemical ratios all follow this pattern.
Problems with a fixed charge plus a variable rate give y = mx + c — auto fare, parking, library charges all fit this model.
The Celsius-Fahrenheit formula F = (9/5)C + 32 is a standard linear equation. At −40°, both scales give the same numerical reading — a famous mathematical curiosity.
Once the graph is drawn correctly, you can find any value by drawing reference lines — no algebra needed for individual answers.
Always verify graph-read answers algebraically: substitute the value back into the equation to confirm LHS = RHS.