- 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).
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).
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.
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.
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:
- 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.
- Plot the points on a coordinate plane with labelled axes.
- 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.
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.
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 $\Leftrightarrow$ $c = 0$.
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).
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.
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:
| Equation | Type | Parallel to | Cuts 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:
- Identify the two unknown quantities and assign variable names ($x$ and $y$).
- Translate the given condition into an algebraic equation using $x$ and $y$.
- Simplify and write in standard form $ax + by + c = 0$.
"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. $(10, 5)$.
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.
(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.
- $x^2 + y = 5$
- $x + y^2 = 3$
- $2x + 3y = 7$
- $xy = 6$
- Exactly one
- Exactly two
- Exactly three
- Infinitely many
- $(1,\,0)$
- $(0,\,1)$
- The origin $(0,\,0)$
- $(3,\,-2)$ only
- A horizontal line 3 units above the $x$-axis
- A vertical line 3 units to the right of the $y$-axis
- A vertical line 3 units to the left of the $y$-axis
- A line through the origin
- $(0,\,4)$
- $(3,\,2)$
- $(1,\,2)$
- $(-1,\,6)$
- $x = 0$
- $y = 0$
- $x = y$
- $x + y = 0$
- $-1$
- $1$
- $2$
- $-2$
- Parallel to the $y$-axis at distance 5
- Parallel to the $x$-axis at distance 5
- Passing through the origin
- Making a $45^\circ$ angle with the $x$-axis
- $a \neq 0$ and $b \neq 0$
- $c \neq 0$
- $a$ and $b$ are not both zero
- $a = b$
- $2x + 3y = 6$
- $x - y = 1$
- $4x + 5y = 0$
- $3x - 2y + 1 = 0$
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.
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 $(0, 6)$.
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 $x + 4 = 0$.
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.
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: $(4, 0)$.
Book a free demo class