Linear Equations in Two Variables

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CLASS IX Mathematics ~4–6 marks/year Ch 4 of 12
Linear Equations in Two Variables

Class 9 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • A linear equation in two variables has the standard form $ax + by + c = 0$, where $a, b, c$ are real numbers and at least one of $a$ or $b$ is non-zero.
  • Every linear equation in two variables has infinitely many solutions — each solution is an ordered pair $(x, y)$.
  • The graph of a linear equation in two variables is always a straight line. Every point on the line is a solution; every solution is a point on the line.
  • If $c = 0$, the line passes through the origin $(0, 0)$.
  • $x = k$ (where $k$ is a constant) is a line parallel to the $y$-axis (vertical line).
  • $y = k$ (where $k$ is a constant) is a line parallel to the $x$-axis (horizontal line).
  • Board weightage: ~4--6 marks/year — usually one graphing question (2--3 marks) and one equation-from-situation or solutions question (1--2 marks).
Detailed Notes

1. Standard form — $ax + by + c = 0$

Any equation of the form $ax + by + c = 0$ is called a linear equation in two variables, where:

  • $a$, $b$, $c$ are real numbers (can be integers, fractions, decimals, or zero).
  • $a$ and $b$ are called coefficients of $x$ and $y$ respectively.
  • $c$ is the constant term.
  • Crucial restriction: $a$ and $b$ cannot both be zero simultaneously — otherwise the "equation" would collapse to $c = 0$, which has no variables at all.
  • The word "linear" means the highest power of each variable is exactly 1.

Examples of linear equations in two variables:

  • $2x + 3y = 7 \;\Rightarrow\; 2x + 3y - 7 = 0$ (here $a=2,\; b=3,\; c=-7$).
  • $x - 4y = 0$ (here $a=1,\; b=-4,\; c=0$).
  • $5x = 3$ can be written as $5x + 0 \cdot y - 3 = 0$ (linear in two variables, with $b = 0$).
  • $y = 7$ can be written as $0 \cdot x + y - 7 = 0$ (linear in two variables, with $a = 0$).
  • $\sqrt{2}\,x + y = 3$ — valid, since coefficients can be irrational.

Non-examples: $x^2 + y = 4$ (power of $x$ is 2, not linear); $xy = 5$ (product of variables — not linear); $\sqrt{x} + y = 1$ (variable under a root).

NCERT Example 1 — Write in standard form

Write each of the following as an equation in two variables:

(i) $x = -5$: Write as $1 \cdot x + 0 \cdot y + 5 = 0$, i.e. $x + 0\cdot y + 5 = 0$.

(ii) $y = 2$: Write as $0 \cdot x + 1 \cdot y - 2 = 0$.

(iii) $2x = 3$: Write as $2x + 0 \cdot y - 3 = 0$.

(iv) $5y = 2$: Write as $0 \cdot x + 5y - 2 = 0$.

Key idea: even a one-variable equation can be treated as a linear equation in two variables by introducing the other variable with coefficient 0. This perspective is crucial for graphing such equations as lines in the plane.

2. Solutions — ordered pairs, infinitely many

A solution of a linear equation $ax + by + c = 0$ is an ordered pair $(x_0, y_0)$ such that substituting $x = x_0$ and $y = y_0$ satisfies the equation: $ax_0 + by_0 + c = 0$.

Why infinitely many solutions? For any value of $x$ you choose, you can always find a corresponding $y$ (when $b \neq 0$): $y = \dfrac{-(ax + c)}{b}$. Since we can choose $x$ to be any real number and there are infinitely many real numbers, there are infinitely many solutions.

Notation matters: A solution is an ordered pair $(x, y)$ — order is important. $(2, 3)$ and $(3, 2)$ are different ordered pairs and will generally not both satisfy the same equation.

NCERT Example 2 — Checking solutions of $2x + 3y = 12$

Check which of the following are solutions: $(3, 2)$, $(1, 3)$, $(0, 4)$, $(-1, 5)$.

$(3, 2)$: $2(3) + 3(2) = 6 + 6 = 12$. Yes, a solution.

$(1, 3)$: $2(1) + 3(3) = 2 + 9 = 11 \neq 12$. Not a solution.

$(0, 4)$: $2(0) + 3(4) = 0 + 12 = 12$. Yes, a solution.

$(-1, 5)$: $2(-1) + 3(5) = -2 + 15 = 13 \neq 12$. Not a solution.

Takeaway: not every pair you guess is a solution — you must always substitute and verify. But there are still infinitely many pairs that do satisfy the equation.

NCERT Example 3 — Finding four solutions of $2x + y = 7$

Rearrange: $y = 7 - 2x$. Substitute different values of $x$:

$x$$y = 7 - 2x$Solution
$0$$7$$(0,\,7)$
$1$$5$$(1,\,5)$
$2$$3$$(2,\,3)$
$-1$$9$$(-1,\,9)$

All four are valid solutions. We could freely choose any value of $x$ and find a matching $y$, demonstrating the infinite nature of the solution set.

3. Graphing a linear equation — table of values, plot, join

Since every solution $(x, y)$ corresponds to a point in the coordinate plane and all solutions lie on one straight line, we graph the equation by a simple three-step process:

  1. Make a table of values: choose at least 3 values of $x$ (convenient integers make arithmetic easier), compute the corresponding $y$ to get ordered pairs.
  2. Plot the points on a coordinate plane with labelled axes.
  3. Join the points with a straight line, and extend it in both directions with arrowheads to show it continues indefinitely.

Why 3 points? Two points determine a unique line. The third point acts as a check — if all three are collinear, your arithmetic is likely correct. If they are not collinear, re-check your table.

Choosing easy values: pick $x = 0$ to get the $y$-intercept, pick $y = 0$ to get the $x$-intercept. These two axis intercepts are the most natural plotting points.

NCERT Example 4 — Graph of $2x + y = 6$

Rearrange: $y = 6 - 2x$. Build a table:

$x$$y = 6 - 2x$Point
$0$$6$$(0,\,6)$
$3$$0$$(3,\,0)$
$1$$4$$(1,\,4)$

Plot $(0, 6),\; (3, 0),\; (1, 4)$ and join. The three points are collinear — our graph is correct.

Reading the graph: pick any point on this line and its coordinates satisfy $2x + y = 6$. For instance, the point $(2, 2)$: $2(2) + 2 = 6$. Confirmed!

From the graph: as $x$ increases by 1, $y$ decreases by 2. The line has a downward slope.

NCERT Example 5 — Graph of $3x + 4y = 12$

Rearrange: $y = \dfrac{12 - 3x}{4}$. Choose values that make $12 - 3x$ divisible by 4:

$x$$y$Point
$0$$3$$(0,\,3)$
$4$$0$$(4,\,0)$
$-4$$6$$(-4,\,6)$

Plot and join. The line cuts the $y$-axis at $(0, 3)$ and the $x$-axis at $(4, 0)$.

Tip — intercept method: to find the $x$-intercept, put $y = 0$ and solve; to find the $y$-intercept, put $x = 0$ and solve. These two intercept points alone are enough to draw the entire line, with the third point as a check.

Fundamental facts about the graph of $ax + by + c = 0$:

  • Always a straight line — never a curve.
  • Every point on the line is a solution; every solution is a point on the line.
  • The line extends infinitely in both directions — there are infinitely many solutions.
  • Draw arrowheads at both ends of the line segment to indicate this.

4. Graph passes through the origin when $c = 0$

If the constant term $c = 0$, the equation becomes $ax + by = 0$. Substituting $x = 0,\; y = 0$: $a(0) + b(0) = 0$. This is always true, so $(0, 0)$ is always a solution. Therefore, the graph passes through the origin.

Conversely, if $c \neq 0$, then $a(0) + b(0) + c = c \neq 0$, so the origin is not on the line.

Rule: Graph of $ax + by + c = 0$ passes through the origin $\Leftrightarrow$ $c = 0$.

NCERT Example 6 — Graph of $x - 2y = 0$ (i.e. $c = 0$)

Since $c = 0$, the line passes through the origin. We only need one more point:

$x$$y = x/2$Point
$0$$0$$(0,\,0)$ — origin
$2$$1$$(2,\,1)$
$4$$2$$(4,\,2)$

Join the points. The line passes through the origin and is entirely determined by the origin plus one other point.

Observation: For this equation, every solution has $y = x/2$, so the $y$-value is always half the $x$-value.

Memory trick: "No constant — through the origin." If you see no constant term in the equation (or the constant is zero after rearranging), the line goes through $(0, 0)$.

5. Equations of lines parallel to the $y$-axis: $x = k$

The equation $x = k$ (where $k$ is a real constant) means that the $x$-coordinate is always equal to $k$, regardless of the value of $y$. In standard form: $1\cdot x + 0 \cdot y - k = 0$, so $a = 1,\; b = 0,\; c = -k$.

Solutions: $(k, 0),\; (k, 1),\; (k, -1),\; (k, 5),\ldots$ — all pairs where the first coordinate equals $k$.

Graph: All these points have the same $x$-coordinate, so they lie on a vertical line passing through the point $(k, 0)$ on the $x$-axis. This line is parallel to the $y$-axis.

  • $k = 0$: the equation $x = 0$ is the $y$-axis itself.
  • $k > 0$: vertical line to the right of the $y$-axis.
  • $k < 0$: vertical line to the left of the $y$-axis.

All lines of the form $x = k$ are parallel to each other and parallel to the $y$-axis (and hence perpendicular to the $x$-axis).

NCERT Example 7 — Graph of $x + 4 = 0$, i.e. $x = -4$

Rewrite: $x = -4$. Some solutions: $(-4,\,0),\; (-4,\,3),\; (-4,\,-2)$.

Graph: a vertical line 4 units to the left of the $y$-axis, cutting the $x$-axis at $(-4, 0)$.

Important distinction: In one variable (on the number line), $x = -4$ means just the single point $-4$. In two variables (on the coordinate plane), $x = -4$ means the entire infinite vertical line at $x = -4$. Same equation, different geometric meaning depending on context.

6. Equations of lines parallel to the $x$-axis: $y = k$

The equation $y = k$ means the $y$-coordinate is always $k$, and $x$ can be anything. In standard form: $0 \cdot x + 1 \cdot y - k = 0$, so $a = 0,\; b = 1,\; c = -k$.

Solutions: $(0,\,k),\; (1,\,k),\; (-3,\,k),\ldots$ — all pairs where the second coordinate equals $k$.

Graph: All points have the same $y$-coordinate — a horizontal line at height $k$, parallel to the $x$-axis.

  • $k = 0$: the equation $y = 0$ is the $x$-axis itself.
  • $k > 0$: horizontal line above the $x$-axis.
  • $k < 0$: horizontal line below the $x$-axis.
NCERT Example 8 — Graph of $y - 3 = 0$, i.e. $y = 3$

Rewrite: $y = 3$. Solutions: $(0,\,3),\; (2,\,3),\; (-5,\,3)$, etc.

Graph: a horizontal line 3 units above the $x$-axis, cutting the $y$-axis at $(0, 3)$.

Contrast with $x = k$: $x = k$ is a vertical line (parallel to $y$-axis); $y = k$ is a horizontal line (parallel to $x$-axis). A memory trick: "$x$ is horizontal position, so fixing $x$ gives a vertical cut; $y$ is vertical position, so fixing $y$ gives a horizontal cut."

Summary table — special lines in the plane:

EquationTypeParallel toCuts axis at
$x = k\; (k \neq 0)$Vertical$y$-axis$(k,\,0)$ on $x$-axis
$y = k\; (k \neq 0)$Horizontal$x$-axis$(0,\,k)$ on $y$-axis
$x = 0$Vertical— (it IS the $y$-axis)Origin
$y = 0$Horizontal— (it IS the $x$-axis)Origin

7. Representing real-life situations as linear equations

Many everyday relationships between two quantities can be expressed as linear equations in two variables. The process:

  1. Identify the two unknown quantities and assign variable names ($x$ and $y$).
  2. Translate the given condition into an algebraic equation using $x$ and $y$.
  3. Simplify and write in standard form $ax + by + c = 0$.
NCERT Example 9 — Cost of notebook and pen

"The cost of a notebook is twice the cost of a pen." Let $x$ = cost of one notebook (Rs) and $y$ = cost of one pen (Rs). The condition "twice" gives: $x = 2y$, i.e. $x - 2y = 0$ (standard form with $c = 0$).

Since $c = 0$, the graph passes through the origin. One solution: if a pen costs Rs 5, a notebook costs Rs 10, i.e. $(10, 5)$.

NCERT Example 10 — Fahrenheit and Celsius

The relationship between Fahrenheit ($F$) and Celsius ($C$) is: $F = \dfrac{9}{5}C + 32$.

Rearranging: $5F = 9C + 160$, i.e. $9C - 5F + 160 = 0$ (standard form with $a=9, b=-5, c=160$).

Table of values:

$C$$F = \dfrac{9}{5}C + 32$
$0$$32$
$-40$$-40$
$100$$212$

The solution $(-40, -40)$ is remarkable: both Celsius and Fahrenheit read $-40$ at this temperature. Since $c = 160 \neq 0$, the graph does not pass through the origin.

More situation-to-equation examples

(i) "Sum of two numbers is 25": $x + y - 25 = 0$.

(ii) "A number is 5 more than twice another": $x - 2y - 5 = 0$.

(iii) "Perimeter of a rectangle is 40 cm" (length $x$, breadth $y$): $2(x + y) = 40$, i.e. $x + y - 20 = 0$.

(iv) Taxi fare: Rs 8 for first km, Rs 5 per km after that. Total fare $y$ Rs for $x$ km (where $x \geq 1$): $y = 8 + 5(x - 1) = 5x + 3$, i.e. $5x - y + 3 = 0$.

Important note: when an equation models a real situation, not all mathematical solutions are physically meaningful. For example, negative values of $x$ or $y$ representing lengths or counts are rejected. We restrict solutions to those satisfying the real-life constraints.

8. Common mistakes to avoid

  • Reversing the ordered pair: $(2, 3)$ and $(3, 2)$ are different. Always write $(x, y)$ — $x$ first, $y$ second.
  • Forgetting arrowheads: the line extends infinitely — always draw arrowheads at both ends on the graph.
  • Drawing a curve: linear equation always gives a straight line — never draw a curve.
  • $x = k$ is parallel to the $x$-axis? No! $x = k$ is vertical, hence parallel to the $y$-axis.
  • Saying "one solution" or "two solutions": a linear equation in two variables always has infinitely many solutions.
  • Missing $c = 0$ check: only check whether the graph passes through the origin by checking whether the constant term is zero.
  • Checking only two points: always use a third point to verify your table before drawing the line.
  • Sign errors: when rearranging to standard form, change signs carefully when moving terms across the equals sign.

9. Quick revision checklist

  • Standard form: $ax + by + c = 0$, with $a$ and $b$ not both zero.
  • Every linear equation in two variables has infinitely many solutions.
  • A solution is an ordered pair $(x, y)$ satisfying the equation — substitute both values to verify.
  • Graph = straight line; build a table of at least 3 values, plot, then join with arrowheads.
  • Use $y = 0$ to find the $x$-intercept; use $x = 0$ to find the $y$-intercept.
  • $c = 0$ means the line passes through the origin $(0, 0)$.
  • $x = k$ is a vertical line, parallel to the $y$-axis; $x = 0$ is the $y$-axis itself.
  • $y = k$ is a horizontal line, parallel to the $x$-axis; $y = 0$ is the $x$-axis itself.
  • Real-life situations: identify two unknowns, write one equation, graph it.
Practice MCQs
1. Which of the following is a linear equation in two variables?
  1. $x^2 + y = 5$
  2. $x + y^2 = 3$
  3. $2x + 3y = 7$
  4. $xy = 6$
Answer: (C) $2x + 3y = 7$ — both variables appear to the first power and there is no product of variables.
2. How many solutions does the equation $3x + 2y = 12$ have?
  1. Exactly one
  2. Exactly two
  3. Exactly three
  4. Infinitely many
Answer: (D) Infinitely many — a linear equation in two variables always has infinitely many solutions.
3. The graph of $2x + 3y = 0$ passes through:
  1. $(1,\,0)$
  2. $(0,\,1)$
  3. The origin $(0,\,0)$
  4. $(3,\,-2)$ only
Answer: (C) The origin — since $c = 0$, the line passes through $(0, 0)$. Verify: $2(0) + 3(0) = 0$.
4. The graph of $x = -3$ is:
  1. A horizontal line 3 units above the $x$-axis
  2. A vertical line 3 units to the right of the $y$-axis
  3. A vertical line 3 units to the left of the $y$-axis
  4. A line through the origin
Answer: (C) $x = -3$ is a vertical line (parallel to the $y$-axis) 3 units to the left of the $y$-axis.
5. Which point lies on the line $2x - y = 4$?
  1. $(0,\,4)$
  2. $(3,\,2)$
  3. $(1,\,2)$
  4. $(-1,\,6)$
Answer: (B) $(3, 2)$: $2(3) - 2 = 6 - 2 = 4$. Check others — (A): $0 - 4 = -4 \neq 4$; (C): $2-2=0\neq 4$; (D): $-2-6=-8\neq 4$.
6. The equation of the $x$-axis is:
  1. $x = 0$
  2. $y = 0$
  3. $x = y$
  4. $x + y = 0$
Answer: (B) $y = 0$ — every point on the $x$-axis has $y$-coordinate equal to zero.
7. If $(2,\,k)$ is a solution of $3x - 2y = 8$, then $k =$
  1. $-1$
  2. $1$
  3. $2$
  4. $-2$
Answer: (A) Substitute: $3(2) - 2k = 8 \Rightarrow 6 - 2k = 8 \Rightarrow -2k = 2 \Rightarrow k = -1.$
8. The graph of $y = 5$ is:
  1. Parallel to the $y$-axis at distance 5
  2. Parallel to the $x$-axis at distance 5
  3. Passing through the origin
  4. Making a $45^\circ$ angle with the $x$-axis
Answer: (B) $y = 5$ is a horizontal line parallel to the $x$-axis, 5 units above it.
9. In $ax + by + c = 0$, the condition for this to be a linear equation in two variables is:
  1. $a \neq 0$ and $b \neq 0$
  2. $c \neq 0$
  3. $a$ and $b$ are not both zero
  4. $a = b$
Answer: (C) At least one of $a$ or $b$ must be non-zero. Both can be non-zero, or exactly one can be zero — but both cannot be zero simultaneously.
10. Which equation represents a line passing through the origin?
  1. $2x + 3y = 6$
  2. $x - y = 1$
  3. $4x + 5y = 0$
  4. $3x - 2y + 1 = 0$
Answer: (C) $4x + 5y = 0$ — the constant term is zero, so $(0,0)$ satisfies it: $4(0) + 5(0) = 0$. The other options all have non-zero constant terms.
Previous-year Questions (PYQs)
PYQ 1. Find two solutions of the linear equation $3x - 2y = 6$ and verify them. (CBSE, 2 marks)

Rearrange: $y = \dfrac{3x - 6}{2}$.

Put $x = 0$: $y = \dfrac{-6}{2} = -3$. Solution: $(0, -3)$. Verify: $3(0) - 2(-3) = 0 + 6 = 6$. Correct.

Put $x = 2$: $y = \dfrac{6 - 6}{2} = 0$. Solution: $(2, 0)$. Verify: $3(2) - 2(0) = 6 - 0 = 6$. Correct.

PYQ 2. Draw the graph of $2x + y = 6$ and find where it cuts the coordinate axes. (CBSE, 3 marks)

Rearrange: $y = 6 - 2x$. Table: $(0, 6),\; (3, 0),\; (1, 4)$. Plot all three and join.

$x$-intercept: put $y = 0 \Rightarrow 2x = 6 \Rightarrow x = 3$. Point: $(3, 0)$.

$y$-intercept: put $x = 0 \Rightarrow y = 6$. Point: $(0, 6)$.

The graph cuts the $x$-axis at $(3, 0)$ and the $y$-axis at $(0, 6)$.

PYQ 3. Write the equation of the line parallel to the $y$-axis and passing through $(-4, 0)$. (CBSE, 1 mark)

A line parallel to the $y$-axis has the form $x = k$. Since it passes through $(-4, 0)$, we have $k = -4$.

Equation: $x = -4$, or equivalently $x + 4 = 0$.

PYQ 4. The taxi fare in a city is Rs. 8 for the first km and Rs. 5/km for subsequent distance. Let $x$ km be the distance and Rs. $y$ the total fare. Write a linear equation for this and find the fare for 15 km. (CBSE, 3 marks)

For $x \geq 1$ km: $y = 8 + 5(x - 1) = 5x + 3$. Linear equation: $5x - y + 3 = 0$.

For $x = 15$: $y = 5(15) + 3 = 75 + 3 = 78$.

Fare for 15 km = Rs. 78.

PYQ 5. If the point $(2, -3)$ lies on the graph of $3x + ky = 12$, find the value of $k$. Write one more solution of this equation. (CBSE, 2 marks)

Substitute $(2, -3)$: $3(2) + k(-3) = 12 \Rightarrow 6 - 3k = 12 \Rightarrow -3k = 6 \Rightarrow k = -2.$

Equation becomes $3x - 2y = 12$, i.e. $y = \dfrac{3x - 12}{2}$.

Put $x = 4$: $y = \dfrac{12 - 12}{2} = 0$. Another solution: $(4, 0)$.

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