Chapter 4 of 12

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IGCSE Class 9 Mathematics★ 5–7 marks · CBSE Class 9 board

Linear Equations in Two Variables

Every train timetable, every demand-and-supply curve, every flight path — they're all linear equations in two variables. Class 9 chapter on solutions, plotting, slope-intercept ideas, geometric meaning, axis-equations, and real-world applications with 25+ worked exam problems.

⏱ 40 min readEasy difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Express a linear equation in two variables in standard form ax + by + c = 0
2Verify whether an ordered pair is a solution of a given equation
3Find any number of solutions of a linear equation
4Plot the graph of ax + by + c = 0 using ≥ 2 (preferably 3) solutions
5Find x- and y-intercepts and use them to plot quickly
6Write the equation of horizontal/vertical lines and the equations of the axes
7Translate a real-world condition into a linear equation in two variables and graph it
8Find a missing coefficient or constant given a solution of the equation

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Why this chapter matters

A single linear equation in two variables = a straight line. Two equations = an intersection. This chapter teaches you to translate any real-world linear relationship (taxi fare, currency conversion, demand-supply, exam scoring) into algebra and back into geometry — the core literacy of applied math.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

🔗

Linear equations in one variable (Class 8)

Solving 3x + 5 = 14 etc.

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Coordinate Geometry (Class 9, Chapter 3)

Plotting points (x, y) on the plane

Linear Equations in Two Variables — Class 9 (CBSE)

A single linear equation in two variables has infinitely many solutions — and each is a point. Plot all of them and you get a line. That's the whole chapter in one sentence. From this idea spring train schedules, GST calculations, profit maximisation, even GPS triangulation.

1. The story — why two variables matter

In Class 8 you solved linear equations in one variable: $3 x + 5 = 14 \Rightarrow x = 3$ . One equation, one unknown, one answer.

But the real world rarely has one unknown. A street vendor sells two products at different prices and you know the total revenue. A car travels at two different speeds across two stretches. A movie ticket costs different amounts on weekdays vs weekends. To model these, we need TWO variables — and equations relating them.

A linear equation in two variables $x$ and $y$ has the form

$a x + b y + c = 0$

where $a, b, c$ are real numbers with $a$ and $b$ not both zero.

The astonishing fact: such an equation has infinitely many solutions $(x, y)$ — and plotted on the Cartesian plane (from the previous chapter), they form a straight line. Linear equation, straight line — the names match.

2. The big picture — two ideas to take away

Every linear equation $a x + b y + c = 0$ corresponds to a straight line on the Cartesian plane. Each solution $(x_{0}, y_{0})$ is a point on the line.
Solving simultaneous equations geometrically = finding the intersection of two lines. That's the bridge to Class 10's "pair of linear equations" chapter.

3. The standard form and special cases

The standard (general) form is $a x + b y + c = 0$ .

Reading the form. Three coefficients describe a line:

$a$ — coefficient of $x$ .
$b$ — coefficient of $y$ .
$c$ — constant (sign included).

Special cases.

Equation	Description	Shape on the plane
$y = c$	Horizontal line (every point has $y = c$ )	Horizontal, parallel to x-axis
$x = c$	Vertical line (every point has $x = c$ )	Vertical, parallel to y-axis
$y = 0$	The x-axis itself	Coincides with x-axis
$x = 0$	The y-axis itself	Coincides with y-axis
$y = x$	Diagonal through the origin	Bisector of Quadrants I and III
$y = - x$	Other diagonal through the origin	Bisector of Quadrants II and IV

Worked example. Express $3 x = 5 y - 7$ in standard form.

Rearrange: $3 x - 5 y + 7 = 0$ . So $a = 3, b = - 5, c = 7$ .

4. Solutions of a linear equation

A solution of $a x + b y + c = 0$ is an ordered pair $(x_{0}, y_{0})$ that satisfies the equation when substituted.

Worked example. Is $(2, 1)$ a solution of $3 x - 2 y + 1 = 0$ ?

Substitute: $3 (2) - 2 (1) + 1 = 6 - 2 + 1 = 5 \neq = 0$ . NOT a solution.

Key fact. A linear equation in two variables has infinitely many solutions. Pick any $x$ ; solve for $y$ . You get one solution. Try another $x$ ; another solution. You can keep doing this forever.

How to find solutions methodically

To list a handful of solutions for $2 x + 3 y = 12$ :

Pick $x$	Compute $y$	Solution $(x, y)$
$0$	$3 y = 12 \Rightarrow y = 4$	$(0, 4)$
$3$	$6 + 3 y = 12 \Rightarrow y = 2$	$(3, 2)$
$6$	$12 + 3 y = 12 \Rightarrow y = 0$	$(6, 0)$
$- 3$	$- 6 + 3 y = 12 \Rightarrow y = 6$	$(- 3, 6)$

These four points all lie on the same straight line.

5. Graph of a linear equation — the big result

Theorem. The graph of every linear equation $a x + b y + c = 0$ in two variables is a straight line.

Why? Pick any two solutions. Plot them on the plane. Now pick any third solution. It will lie exactly on the line joining the first two. (Class 11 calculus proves this with vectors; for now, take it as a remarkable fact.)

Method to plot a line.

Find at least two solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ of the equation.
Plot both points on the Cartesian plane.
Join them with a straight line and extend in both directions.

Teacher's tip. Always find three solutions, not two — the third acts as a check. If all three are collinear, your arithmetic is correct.

Worked example. Plot $2 x + y = 4$ .

Find three solutions:

Let $x = 0$ : $y = 4$ . Solution: $(0, 4)$ .
Let $y = 0$ : $2 x = 4 \Rightarrow x = 2$ . Solution: $(2, 0)$ .
Let $x = 1$ : $2 + y = 4 \Rightarrow y = 2$ . Solution: $(1, 2)$ .

Plot the three points. They're collinear → draw the straight line through them and extend.

6. Where does a line cut the axes? — intercepts

The x-intercept is the value of $x$ where the line crosses the x-axis (i.e. where $y = 0$ ). The y-intercept is the value of $y$ where the line crosses the y-axis (i.e. where $x = 0$ ).

Worked example. Find the intercepts of $3 x + 4 y = 12$ .

y-intercept (put $x = 0$ ): $4 y = 12 \Rightarrow y = 3$ . The line cuts the y-axis at $(0, 3)$ .
x-intercept (put $y = 0$ ): $3 x = 12 \Rightarrow x = 4$ . The line cuts the x-axis at $(4, 0)$ .

Why this matters. Intercepts are the fastest way to plot a line — find both, plot, draw. Two intercepts uniquely determine a line (as long as the line isn't parallel to either axis).

7. Equations of the axes and lines parallel to them

The x-axis is the equation $y = 0$ . (Every point on the x-axis has y-coordinate 0.)
The y-axis is the equation $x = 0$ .
A horizontal line through point $(0, k)$ is $y = k$ .
A vertical line through point $(h, 0)$ is $x = h$ .

Worked example. Write the equation of the horizontal line passing through $(7, - 3)$ .

Horizontal lines have constant y-coordinate. The y-coordinate here is $- 3$ . So the equation is $y = - 3$ .

8. Linear equation through the origin

If the constant $c = 0$ , the equation $a x + b y = 0$ passes through the origin. (Substitute $x = 0, y = 0$ — it satisfies trivially.)

Examples. $y = x$ , $y = 2 x$ , $y = - 3 x$ , $3 x + 2 y = 0$ . All pass through $(0, 0)$ .

Worked example. Find any three solutions of $3 x = - 2 y$ .

Rewrite: $3 x + 2 y = 0$ .

$x = 0 \Rightarrow y = 0$ . Solution $(0, 0)$ .
$x = 2 \Rightarrow y = - 3$ . Solution $(2, - 3)$ .
$x = - 2 \Rightarrow y = 3$ . Solution $(- 2, 3)$ .

9. Linear equation in ONE variable — when does it appear?

If you set $b = 0$ in $a x + b y + c = 0$ , you get $a x + c = 0$ , i.e. $x = - c / a$ . That's an equation in only $x$ — a vertical line in 2D.

Similarly setting $a = 0$ gives a horizontal line $y = - c / b$ .

Subtle point. "The equation $x = 4$ " can be interpreted as a one-variable equation (with one solution $x = 4$ ) OR as a two-variable equation (with infinitely many solutions of the form $(4, y)$ for any $y$ ). In Class 9 we adopt the two-variable view by default.

10. Finding the equation of a line from two given points

This is technically Class 10 territory, but you'll see it in HOTS problems. Given two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :

$\frac{y - y _{1}}{x - x _{1}} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} .$

Worked example. Find the equation of the line through $(1, 2)$ and $(3, 8)$ .

$\frac{y - 2}{x - 1} = \frac{8 - 2}{3 - 1} = \frac{6}{2} = 3$ .

Cross-multiply: $y - 2 = 3 (x - 1) \Rightarrow y - 2 = 3 x - 3 \Rightarrow 3 x - y - 1 = 0$ .

Verify with the second point: $3 (3) - 8 - 1 = 0$ ✓.

11. Eight worked exam examples

Example 1 — Standard form (1 mark)

Express $4 x + 5 y = 12$ in standard form. $4 x + 5 y - 12 = 0$ , so $a = 4, b = 5, c = - 12$ .

Example 2 — Check a solution (1 mark)

Is $(3, - 2)$ a solution of $2 x + y = 4$ ? $2 (3) + (- 2) = 6 - 2 = 4$ ✓. Yes.

Example 3 — Find solutions (2 marks)

Find four solutions of $5 x - 2 y = 10$ .

$x = 0 : - 2 y = 10 \Rightarrow y = - 5 \Rightarrow (0, - 5)$ .
$y = 0 : 5 x = 10 \Rightarrow x = 2 \Rightarrow (2, 0)$ .
$x = 4 : 20 - 2 y = 10 \Rightarrow y = 5 \Rightarrow (4, 5)$ .
$y = 10 : 5 x - 20 = 10 \Rightarrow x = 6 \Rightarrow (6, 10)$ .

Example 4 — Plot (3 marks)

Plot the line $x + 2 y = 6$ . Two solutions: $(0, 3)$ and $(6, 0)$ . Plot, join, extend.

Example 5 — Equation from condition (2 marks)

A line is parallel to the x-axis and passes through $(2, - 5)$ . Find its equation. Parallel to x-axis → horizontal → $y =$ constant = the y-coordinate of the point = $- 5$ . Equation: $y = - 5$ .

Example 6 — Intercepts (3 marks)

Find the intercepts of $4 x - 3 y = 12$ and use them to plot the line. y-intercept (x = 0): $- 3 y = 12 \Rightarrow y = - 4$ . Point $(0, - 4)$ . x-intercept (y = 0): $4 x = 12 \Rightarrow x = 3$ . Point $(3, 0)$ . Plot both; draw the line.

Example 7 — Find a constant (3 marks)

If $(2, k)$ is a solution of $5 x + 3 y = 19$ , find $k$ . Substitute: $5 (2) + 3 k = 19 \Rightarrow 10 + 3 k = 19 \Rightarrow 3 k = 9 \Rightarrow k = 3$ .

Example 8 — HOTS (4 marks)

A taxi charges ₹20 as a fixed fee plus ₹10 per km. If $x$ is the distance in km and $y$ is the total fare in rupees, write the linear equation, find three solutions, and plot the line.

Equation: $y = 20 + 10 x$ , or in standard form $10 x - y + 20 = 0$ . Solutions:

$x = 0$ : $y = 20$ → $(0, 20)$ . (Fare for 0 km — just the booking fee.)
$x = 5$ : $y = 70$ → $(5, 70)$ .
$x = 10$ : $y = 120$ → $(10, 120)$ .

The line slopes upward (positive direction) and has y-intercept 20 (the fixed fee). The real-world domain is $x \geq 0$ — you can't travel a negative distance.

12. Common pitfalls

Counting only one solution. A linear equation in one variable has one solution; in two variables it has infinitely many. Don't confuse the two.
Plotting a line with only one point. You need at least two points to determine a line. Always plot three (third is a check).
Swapping intercepts. y-intercept = put $x = 0$ (not $y = 0$ ). Easy to confuse under exam stress.
Forgetting the equation of the axes. $y = 0$ is the x-axis; $x = 0$ is the y-axis. Examiners love testing this.
Mixing up vertical and horizontal. $y = 5$ is HORIZONTAL (y is fixed, x varies); $x = 5$ is VERTICAL.
Negative-sign slips in standard form. $3 x = 4 y - 7$ rearranges to $3 x - 4 y + 7 = 0$ , NOT $3 x + 4 y + 7 = 0$ .
Misreading the sign of $c$ . In standard form $a x + b y + c = 0$ , $c$ includes its sign. For $2 x + 3 y - 5 = 0$ , $c = - 5$ , not $+ 5$ .

13. Beyond NCERT — stretch problems

Stretch 1 — Express in two equivalent forms

Show that $2 x + 3 y = 12$ and $4 x + 6 y = 24$ represent the same line.

Divide the second by 2: $2 x + 3 y = 12$ . Same equation, so same line. Any scalar multiple of a linear equation represents the same line.

Stretch 2 — Olympiad — locus

Find all points equidistant from the x-axis and the y-axis.

Distance from x-axis is $∣ y ∣$ . Distance from y-axis is $∣ x ∣$ . Equidistant means $∣ x ∣ = ∣ y ∣$ , i.e. $y = x$ or $y = - x$ . Two lines: the diagonal bisectors of the quadrants.

Stretch 3 — Real-world (PISA)

A mobile phone plan charges ₹100 base + ₹2 per minute of talk-time. Another plan charges ₹0 base + ₹3 per minute. For how many minutes are the two plans equally expensive?

Plan A: $y = 100 + 2 x$ . Plan B: $y = 3 x$ . Equate: $100 + 2 x = 3 x \Rightarrow x = 100$ minutes. At 100 minutes both plans cost ₹300. Below 100 minutes, Plan B is cheaper; above, Plan A is.

14. Real-world linear equations

Cost-revenue analysis. Fixed cost + per-unit cost = total cost. Always linear.
Currency conversion. 1 USD = 83 INR → $y = 83 x$ — a line through the origin.
Temperature conversion. $F = (9/5) C + 32$ — a linear equation relating Celsius to Fahrenheit.
Demand and supply curves. Both are typically modelled as linear at the introductory level. Their intersection is the market equilibrium.
GPS trilateration. Each satellite gives a linear (well, hyperbolic in 3D, linear in 2D approximation) constraint. Multiple intersections pin down your position.
Train timetables. A train moving at constant speed traces a linear position-time graph; the slope is the speed.
Tax calculations. A flat tax + a per-unit tax is a linear equation. India's GST has linear components.

15. CBSE exam blueprint

Type	Marks	Typical question	Time
VSA	1	Standard form; identify $a, b, c$ ; check if a point is a solution	30 sec
SA-I	2	Find four solutions; equation of the axes	2 min
SA-II	3	Plot a line; find intercepts; find a missing constant	4–5 min
LA	4	Real-world modeling; plot from a verbal description	6–8 min

Total marks: 5–7 / 80 in Class 9 finals. The graph-based questions earn diagram marks easily — always label axes, scale, and points clearly.

Three exam-day strategies:

Always make a small solution table (3 rows minimum) before plotting. It catches arithmetic errors.
Use intercepts when both exist — fastest path to drawing a line.
Label everything on your graph: axes, scale, the equation of the line, and each plotted point.

16. NCERT exercise walkthrough

Exercise 4.1: 2 questions — write a given description as a linear equation; express in standard form.
Exercise 4.2: 4 questions — check solutions, find solutions of a linear equation.
Exercise 4.3: 8 questions — plot lines, find values of a constant from given conditions, real-world word problems.

The chapter has been trimmed in the 2023+ NCERT; the graph section (formerly Exercise 4.4 and 4.5) is now a single combined exercise.

17. Connections — what's next

Class 10, Pair of Linear Equations. Solve TWO simultaneous equations — geometrically, find the intersection of two lines.
Class 11, Straight Lines. Slope, angle, intercept form, normal form, distance from a point to a line.
Class 11, Linear Programming. Maximise/minimise a linear cost function subject to linear constraints. Modern operations research starts here.
Class 12, Vectors. Lines and planes in 3D, vector form of a line.

18. 60-second recap

Standard form: $a x + b y + c = 0$ .
Solution: ordered pair $(x_{0}, y_{0})$ satisfying the equation. Infinitely many solutions.
Graph: every linear equation in two variables = a straight line.
Plot method: find ≥ 2 solutions (better: 3), plot, join.
x-intercept: put $y = 0$ . y-intercept: put $x = 0$ .
x-axis: $y = 0$ . y-axis: $x = 0$ .
Horizontal line through $(h, k)$ : $y = k$ . Vertical line: $x = h$ .
Through origin: $c = 0$ .

Take the practice quiz and the flashcard deck. Next: Introduction to Euclid's Geometry.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Standard form

ax + by + c = 0,  (a, b) ≠ (0, 0)

a, b, c are real numbers; a and b can't both be zero.

Intercept form

x/p + y/q = 1

p = x-intercept, q = y-intercept.

Solution

(x₀, y₀) satisfies the equation

Infinitely many for a single linear equation in two vars.

Equation of x-axis

y = 0

Every point on x-axis has y = 0.

Equation of y-axis

x = 0

Horizontal line

y = k (constant)

Through (h, k) for any h.

Vertical line

x = h (constant)

Through (h, k) for any k.

Line through origin

ax + by = 0

Constant term c = 0.

Line through 2 points

(y − y₁)/(x − x₁) = (y₂ − y₁)/(x₂ − x₁)

Class 10/11 anticipated.

x-intercept

Set y = 0; solve for x

Where the line cuts the x-axis.

y-intercept

Set x = 0; solve for y

Where the line cuts the y-axis.

⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Saying a single 2-variable linear equation has one solution

✓ It has INFINITELY many. (Two simultaneous equations may have one — that's Class 10.)

WATCH OUT

✗ Plotting only one point and drawing a line

✓ You need at least two. Always find three so the third checks your arithmetic.

WATCH OUT

✗ Confusing x and y intercepts

✓ x-intercept → set y = 0 (cuts the x-axis). y-intercept → set x = 0 (cuts the y-axis).

WATCH OUT

✗ Calling y = 5 a vertical line

✓ y = 5 is HORIZONTAL (y is fixed). x = 5 is VERTICAL.

WATCH OUT

✗ Wrong sign of c in standard form

✓ In ax + by + c = 0, c includes its sign. For 2x + 3y = 5, c = −5 (after moving the 5 to the LHS), not +5.

WATCH OUT

✗ Forgetting to label axes in graphs

✓ Always label x-axis, y-axis, the origin, and at least one scale tick. Worth half a mark.

WATCH OUT

✗ Negative-y as 'below the line'

✓ On a graph, negative y is BELOW the x-axis (not below the line). Don't confuse axes with the line being plotted.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 4.1

Exercise 4.1

Express a description as a linear equation; convert to standard form

Questions

Ex 4.2

Exercise 4.2

Check / find solutions; understand 'infinitely many solutions'

Questions

Ex 4.3

Exercise 4.3

Plot lines, find missing constants, real-world word problems

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Standard form

Express 5y = 3x − 7 in standard form ax + by + c = 0 and write a, b, c.

▶ Show solution

Step 1 — Rearrange to bring all terms to one side.
  5y = 3x − 7
  ⇒ 3x − 5y − 7 = 0 (move 5y to RHS, then everything to LHS).

Step 2 — Read off coefficients.
  Compare with ax + by + c = 0.
  a = 3, b = −5, c = −7.

✦ Answer: 3x − 5y − 7 = 0; a = 3, b = −5, c = −7.

Q2EASY· Solution check

Is (2, −1) a solution of 3x + 2y = 4?

▶ Show solution

Step 1 — Substitute x = 2, y = −1 into the LHS.
  LHS = 3(2) + 2(−1) = 6 − 2 = 4.

Step 2 — Compare with RHS.
  RHS = 4. LHS = RHS.

✦ Answer: YES, (2, −1) is a solution.

Q3EASY· Solution check

Is (1, 3) a solution of 2x − y + 1 = 0?

▶ Show solution

Step 1 — Substitute x = 1, y = 3.
  2(1) − 3 + 1 = 2 − 3 + 1 = 0.

Step 2 — Compare to RHS = 0.
  Both equal 0.

✦ Answer: YES, (1, 3) is a solution.

Q4EASY· Axis equation

Write the equation of the x-axis.

▶ Show solution

Step 1 — Property of the x-axis.
  Every point on the x-axis has y-coordinate 0 (regardless of x).

Step 2 — Express as an equation.
  y = 0 (or equivalently 0·x + 1·y + 0 = 0).

✦ Answer: y = 0.

Q5EASY· Horizontal line

Equation of the horizontal line through (4, −2)?

▶ Show solution

Step 1 — Horizontal lines have constant y-value.
  Through (4, −2), the y-coordinate is −2.

Step 2 — Write the equation.
  y = −2.

✦ Answer: y = −2.

Q6MEDIUM· Find solutions

Find four solutions of 2x + 3y = 12.

▶ Show solution

Step 1 — Pick convenient x-values and solve for y.
  x = 0: 3y = 12 → y = 4. Solution (0, 4).
  x = 3: 6 + 3y = 12 → y = 2. Solution (3, 2).
  x = 6: 12 + 3y = 12 → y = 0. Solution (6, 0).
  x = −3: −6 + 3y = 12 → y = 6. Solution (−3, 6).

Step 2 — Verify by substitution (sample check).
  (3, 2): 2(3) + 3(2) = 6 + 6 = 12 ✓.

✦ Answer: Four solutions are (0, 4), (3, 2), (6, 0), (−3, 6). Infinitely many more exist.

Q7MEDIUM· Find constant

If (3, k) is a solution of 4x + 3y = 18, find k.

▶ Show solution

Step 1 — Substitute x = 3, y = k.
  4(3) + 3k = 18.
  12 + 3k = 18.

Step 2 — Solve.
  3k = 6 → k = 2.

✦ Answer: k = 2.

Verification: (3, 2) → 4(3) + 3(2) = 12 + 6 = 18. ✓

Q8MEDIUM· Intercepts

Find the x- and y-intercepts of 3x − 4y = 12 and sketch the line.

▶ Show solution

Step 1 — y-intercept: set x = 0.
  3(0) − 4y = 12 → −4y = 12 → y = −3.
  Point: (0, −3).

Step 2 — x-intercept: set y = 0.
  3x − 4(0) = 12 → 3x = 12 → x = 4.
  Point: (4, 0).

Step 3 — Sketch.
  Plot (0, −3) and (4, 0) on the Cartesian plane.
  Draw the straight line through them, extended in both directions.
  Verify with a third point e.g. x = 8: 3(8) − 4y = 12 → y = 3. Point (8, 3). Should lie on the line.

✦ Answer: x-intercept = 4, y-intercept = −3. Line passes through (4, 0) and (0, −3).

Q9MEDIUM· Equation from condition

A line passes through (0, 5) and (5, 0). Find its equation.

▶ Show solution

Step 1 — Recognise the two points as intercepts.
  (5, 0) is the x-intercept → x-intercept p = 5.
  (0, 5) is the y-intercept → y-intercept q = 5.

Step 2 — Use the intercept form x/p + y/q = 1.
  x/5 + y/5 = 1.

Step 3 — Multiply through by 5 to clear fractions.
  x + y = 5.

Step 4 — Write in standard form.
  x + y − 5 = 0.

✦ Answer: x + y = 5 (or equivalently x + y − 5 = 0).

Q10MEDIUM· Through origin

Find three solutions of 2x + y = 0 and identify what's special.

▶ Show solution

Step 1 — Compute solutions.
  x = 0: y = 0. Point (0, 0).
  x = 1: y = −2. Point (1, −2).
  x = −2: y = 4. Point (−2, 4).

Step 2 — Notice (0, 0) is the origin.
  The constant term in 2x + y = 0 is c = 0, so the line passes through the origin.

✦ Answer: (0, 0), (1, −2), (−2, 4). The line passes through the origin (because c = 0).

Q11MEDIUM· Two unknowns

If (4, 2) is a solution of px + 3y = 14, find p.

▶ Show solution

Step 1 — Substitute x = 4, y = 2.
  p(4) + 3(2) = 14.
  4p + 6 = 14.

Step 2 — Solve.
  4p = 8 → p = 2.

✦ Answer: p = 2.

Verify: 2(4) + 3(2) = 8 + 6 = 14 ✓.

Q12HARD· HOTS — modeling

A taxi charges ₹50 booking fee plus ₹15 per km. Let x = distance in km and y = total fare. Write the linear equation, find three solutions, and identify the y-intercept and what it represents in this context.

▶ Show solution

Step 1 — Translate the verbal description.
  Total fare = booking + per-km × distance.
  y = 50 + 15x.

Step 2 — Rewrite in standard form.
  15x − y + 50 = 0.

Step 3 — Find three solutions.
  x = 0: y = 50 → (0, 50). [Fare for 0 km = just the booking fee.]
  x = 4: y = 50 + 60 = 110 → (4, 110).
  x = 10: y = 50 + 150 = 200 → (10, 200).

Step 4 — Read the y-intercept.
  y-intercept = 50 (when x = 0).
  In context: ₹50 is the BOOKING FEE — the cost when you haven't moved an inch.

Step 5 — Read the slope (gradient).
  For every 1 km, fare increases by ₹15. This is the per-km rate.

✦ Answer: Equation y = 50 + 15x; solutions (0, 50), (4, 110), (10, 200); y-intercept = 50 = booking fee.

Real-world note: x must be ≥ 0 (no negative distance) and is typically discrete (km is continuous but the meter often rounds to 100m chunks).

Q13HARD· HOTS — geometry

Show that the line passing through (1, 1), (2, 2) and (3, 3) is y = x, and find the point on this line whose x-coordinate is 5.

▶ Show solution

Step 1 — Spot the pattern.
  For each point, y = x. Hypothesise the line y = x.

Step 2 — Verify all three points satisfy y = x.
  (1, 1): 1 = 1 ✓
  (2, 2): 2 = 2 ✓
  (3, 3): 3 = 3 ✓
  All three points lie on y = x.

Step 3 — Argue uniqueness.
  Three collinear points uniquely determine a straight line. The three given are collinear (all satisfy y = x), so the line is exactly y = x.

Step 4 — Find the point with x = 5.
  y = x = 5. Point: (5, 5).

✦ Answer: The line is y = x; the point with x = 5 is (5, 5).

Q14HARD· HOTS — equilibrium

Plan A: ₹100 fixed + ₹2/min talktime. Plan B: ₹0 fixed + ₹3/min. For what talktime are the two plans equally expensive?

▶ Show solution

Step 1 — Translate each plan to a linear equation.
  Let x = talktime in minutes, y = total cost in rupees.
  Plan A: y = 100 + 2x.
  Plan B: y = 3x.

Step 2 — Set the costs equal.
  100 + 2x = 3x.

Step 3 — Solve for x.
  100 = 3x − 2x → x = 100.

Step 4 — Find the common cost.
  Substitute x = 100 into either plan.
  Plan A: y = 100 + 200 = 300.
  Plan B: y = 300. ✓ (Both agree as expected.)

Step 5 — Interpret.
  At 100 minutes both plans cost ₹300.
  Below 100 min: Plan B is cheaper (no booking fee).
  Above 100 min: Plan A is cheaper (lower per-min rate amortises the fixed fee).

✦ Answer: At 100 minutes both plans cost ₹300. This is the BREAK-EVEN POINT — exactly the intersection of the two lines.

Geometric interpretation: y = 100 + 2x and y = 3x are two lines on the Cartesian plane; they intersect at (100, 300).

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Standard form ax + by + c = 0; not both a and b are zero.
•Linear equation in TWO variables has INFINITELY many solutions.
•Graph of every linear equation = a straight line.
•Plot using ≥ 2 (better 3) solutions.
•x-intercept: set y = 0. y-intercept: set x = 0.
•x-axis: y = 0. y-axis: x = 0.
•Horizontal line y = k; vertical line x = h.
•Through origin ⇔ c = 0.
•Two-point form: (y − y₁)/(x − x₁) = (y₂ − y₁)/(x₂ − x₁).

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

One equation can only constrain one degree of freedom. You have 2 unknowns and 1 equation, so 1 degree of freedom remains — meaning a whole 1-dimensional set of solutions, i.e. a line.

Yes — for example y = x passes through Q I (positive x → positive y) and Q III (negative x → negative y) but not Q II or Q IV. The pattern depends on the slope and intercept.

Yes — just different forms. Rearrange y = 2x + 3 to get 2x − y + 3 = 0. The graph is the same line.

Two distinct points uniquely determine a straight line in the plane. One point gives a 'pencil of lines' (infinitely many through it) — you can't pick which one.

If a = 0 (and b ≠ 0): equation becomes by + c = 0 ⇒ y = −c/b. Horizontal line. If b = 0 (and a ≠ 0): vertical line x = −c/a. If both are zero, you don't have a meaningful equation.

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Linear Equations in Two Variables

By the end of this chapter you'll be able to…

Before you start — revise these

Linear Equations in Two Variables — Class 9 (CBSE)

1. The story — why two variables matter

2. The big picture — two ideas to take away

3. The standard form and special cases

4. Solutions of a linear equation

How to find solutions methodically

5. Graph of a linear equation — the big result

6. Where does a line cut the axes? — intercepts

7. Equations of the axes and lines parallel to them

8. Linear equation through the origin

9. Linear equation in ONE variable — when does it appear?

10. Finding the equation of a line from two given points

11. Eight worked exam examples

Example 1 — Standard form (1 mark)

Example 2 — Check a solution (1 mark)

Example 3 — Find solutions (2 marks)

Example 4 — Plot (3 marks)

Example 5 — Equation from condition (2 marks)

Example 6 — Intercepts (3 marks)

Example 7 — Find a constant (3 marks)

Example 8 — HOTS (4 marks)

12. Common pitfalls

13. Beyond NCERT — stretch problems

Stretch 1 — Express in two equivalent forms

Stretch 2 — Olympiad — locus

Stretch 3 — Real-world (PISA)

14. Real-world linear equations

15. CBSE exam blueprint

16. NCERT exercise walkthrough

17. Connections — what's next

18. 60-second recap

Key formulas & results

Common mistakes & fixes

NCERT exercises (with solutions)

Practice problems

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Questions students ask

Why does a single linear equation in two variables have infinitely many solutions?

Can a linear equation pass through both quadrants I and III?

Is y = 2x + 3 the same kind of equation as 2x − y + 3 = 0?

Why do we always need at least 2 points to plot a line?

What if a = 0 OR b = 0 in standard form?

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Number Systems

Polynomials

Coordinate Geometry

Introduction to Euclid's Geometry

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