MathsCBSE Class 9 · Two Variables, One Line

Linear Equations in Two Variables: Solutions, Slope and Graph

Updated 17 October 2026 · Fact-checked

A linear equation in two variables has the form $ax + by = c$ with $a$ and $b$ not both zero. Each solution is an ordered pair $(x, y)$, there are infinitely many solutions, and their points make up a straight line. In slope-intercept form $y = mx + d$, $m$ is the slope and $d$ the y-intercept; the x-intercept is $-\frac{d}{m}$. Horizontal lines are $y = k$ and vertical lines are $x = h$.

Understand Linear Equations in Two Variables: Solutions, Slope and Graph

Equation and solutions. An equation $a x + b y = c$ (with $a, b$ not both $0$ ) is a linear equation in two variables. A solution is an ordered pair $(x, y)$ that makes it true. For $3 x - 2 y = 12$ : $(4, 0)$ is a solution since $12 - 0 = 12$ , $(2, - 3)$ since $6 + 6 = 12$ and $(6, 3)$ since $18 - 6 = 12$ , but $(1, 2)$ is not ( $3 - 4 = - 1$ ). Choosing any value of $x$ gives a value of $y$ , so there are infinitely many solutions. For $5 x + 3 y = 7$ the integer solutions come in a family $x = 2 + 3 t$ , $y = - 1 - 5 t$ : $(2, - 1)$ , $(5, - 6)$ , $(- 1, 4)$ and so on, since $5 (2 + 3 t) + 3 (- 1 - 5 t) = 10 - 3 = 7$ .

The graph is a straight line. Plotting the solutions gives points on a straight line; to draw it, find two (better three) solutions, plot them and join them. Choosing the intercepts is convenient: put $x = 0$ for the y-intercept and $y = 0$ for the x-intercept. A point lies on the line exactly when its coordinates satisfy the equation. To find an unknown coefficient, substitute a point on the line: if $(2, - 1)$ lies on $k x + 3 y = 7$ then $2 k - 3 = 7$ and $k = 5$ .

Slope. The slope (steepness) of a line through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ : the change in $y$ for each unit change in $x$ . A positive slope rises left to right, a negative slope falls. For $(- 2, 5)$ and $(4, - 1)$ : $m = \frac{- 6}{6} = - 1$ .

Slope-intercept form. Solving for $y$ gives $y = m x + d$ . For $4 x + 3 y = 12$ : $y = - \frac{4}{3} x + 4$ , so the slope is $- \frac{4}{3}$ , the y-intercept is $4$ and the x-intercept is $3$ . The line through $(0, - 3)$ with slope $2$ is $y = 2 x - 3$ , meeting the x-axis at $(\frac{3}{2}, 0)$ . For $a x + b y = c$ with $b \neq = 0$ : slope $= - \frac{a}{b}$ , y-intercept $= \frac{c}{b}$ , x-intercept $= \frac{c}{a}$ (if $a \neq = 0$ ). Lines with the same slope are parallel: the line through $(1, 1)$ parallel to $y = - x + 3$ is $y = - x + 2$ .

Special lines. $y = k$ is a horizontal line (slope $0$ ), $x = h$ is a vertical line (slope not defined), and $y = m x$ passes through the origin. The line $3 x + 4 y = 12$ cuts the axes at $(4, 0)$ and $(0, 3)$ : the triangle with the axes has legs $4$ and $3$ , hypotenuse $5$ and area $6$ .

When only certain values are allowed. In a situation like 5 rupees per pen and 3 rupees per pencil with exactly 60 rupees spent, $5 x + 3 y = 60$ has only whole-number, non-negative solutions $(0, 20), (3, 15), (6, 10), (9, 5), (12, 0)$ . The points lie on the line, but only these points make sense; the points between them (such as $x = 4$ , $y = \frac{40}{3}$ ) do not. The graph is then a set of separate points on a line.

Exam approach. Verify solutions by substitution; draw graphs with a table of three points and labelled axes; convert to $y = m x + d$ to read slope and intercepts; and restrict to meaningful values in word problems.

Linear Equations in Two Variables: Solutions, Slope and Graph: formulas to remember

Standard form: $a x + b y = c, (a, b) \neq = (0, 0)$; Infinitely many solutions.
Slope from two points: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$; Rise over run.
Slope-intercept form: $y = m x + d$; $m$ slope, $d$ y-intercept.
From standard form: $m = - \frac{a}{b}, d = \frac{c}{b}$; For $a x + b y = c$ , $b \neq = 0$ .
Intercepts: $(\frac{c}{a}, 0) and (0, \frac{c}{b})$; Put $y = 0$ and $x = 0$ .
Special lines: $y = k, x = h, y = m x$; Horizontal, vertical, through the origin.
Parallel lines: $m_{1} = m_{2}$; Same slope, different intercepts.

How to solve Linear Equations in Two Variables: Solutions, Slope and Graph questions

Substitute to test, rearrange to read slope, and use intercepts to draw.

1To test a point, substitute $x$ and $y$ into the equation and check both sides are equal.
2To find an unknown coefficient, substitute a known point and solve.
3To read the slope, solve for $y$ to get $y = m x + d$ .
4To graph, find the intercepts (put $x = 0$ , then $y = 0$ ), plot them, and check with a third point.
5For a line through two points, find the slope first, then use one point to find $d$ .
6In word problems, list only the solutions allowed by the situation (whole numbers, non-negative).

Linear Equations in Two Variables: Solutions, Slope and Graph in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Equation with limited solutions

Suppose a juice stall sells small cups for 5 rupees and large cups for 8 rupees, and collected exactly 80 rupees during one rush.

List the possible numbers of small cups $x$ and large cups $y$ sold, as whole numbers.

Try it first, then show the solutionSolution

$5 x + 8 y = 80$ , so $y = \frac{80 - 5 x}{8}$ .
$y$ is a whole number when $5 x$ is a multiple of $8$ , that is when $x$ is a multiple of $8$ (since 5 and 8 have no common factor).
$x = 0, y = 10$ ; $x = 8, y = 5$ ; $x = 16, y = 0$ .

Answer: The possible sales are $(0, 10)$ , $(8, 5)$ and $(16, 0)$ .

Memory hook: Many points on the line, but only some make sense.

Tricks to remember Linear Equations in Two Variables: Solutions, Slope and Graph

Short memory aids. Say each one aloud once, then use it on a question the same day.

Two points make a line
Two solutions fix the line; a third checks it.
Use it when: Graphing.
Zero trick for intercepts
x = 0 gives the y-intercept; y = 0 gives the x-intercept.
Use it when: Quick plotting.
Solve for y
Make y the subject to read slope and intercept at a glance.
Use it when: Slope-intercept form.

Linear Equations in Two Variables: Solutions, Slope and Graph: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Solutions of a linear equation · 3 marks · Short answerBoard level

Which of $(4, 0)$ , $(2, - 3)$ , $(6, 3)$ , $(1, 2)$ are solutions of $3 x - 2 y = 12$ ? Express $y$ in terms of $x$ and find $y$ when $x = - 2$ .

Try it first, then show the solutionSolution

$(4, 0)$ : $12 - 0 = 12$ ✓; $(2, - 3)$ : $6 + 6 = 12$ ✓; $(6, 3)$ : $18 - 6 = 12$ ✓; $(1, 2)$ : $3 - 4 = - 1 \neq = 12$ . (1½)
$y = \frac{3 x - 12}{2}$ . (½)
For $x = - 2$ : $y = \frac{- 6 - 12}{2} = - 9$ . (1)

Answer: The first three are solutions; $y = \frac{3 x - 12}{2}$ and $y = - 9$ at $x = - 2$ .

Example 2: Slope-intercept form · 3 marks · Short answerBoard level

Write the equation of the line through $(0, - 3)$ with slope 2 and find where it meets the x-axis. Also express $4 x + 3 y = 12$ in the form $y = m x + d$ and give both intercepts.

Try it first, then show the solutionSolution

$y = 2 x - 3$ . Setting $y = 0$ : $x = \frac{3}{2}$ , so it meets the x-axis at $(\frac{3}{2}, 0)$ . (1½)
$4 x + 3 y = 12$ gives $y = - \frac{4}{3} x + 4$ : slope $- \frac{4}{3}$ , y-intercept $4$ . (1)
x-intercept: $y = 0$ gives $x = 3$ . (½)

Answer: $y = 2 x - 3$ , meeting the x-axis at $(1.5, 0)$ ; $y = - \frac{4}{3} x + 4$ with intercepts $(3, 0)$ and $(0, 4)$ .

Example 3: Unknown coefficient and integer solutions · 5 marks · Long answerHarder board level

The point $(2, - 1)$ lies on the line $k x + 3 y = 7$ . Find $k$ , the slope of the line, and three more solutions with integer coordinates. Show that $x = 2 + 3 t$ , $y = - 1 - 5 t$ gives a solution for every integer $t$ .

Try it first, then show the solutionSolution

Substitute: $2 k + 3 (- 1) = 7$ , so $2 k = 10$ and $k = 5$ . (1)
The line is $5 x + 3 y = 7$ , i.e. $y = - \frac{5}{3} x + \frac{7}{3}$ , with slope $- \frac{5}{3}$ . (1)
Three more solutions: $(- 1, 4)$ since $- 5 + 12 = 7$ ; $(5, - 6)$ since $25 - 18 = 7$ ; $(8, - 11)$ since $40 - 33 = 7$ . (1½)
Check the family: $5 (2 + 3 t) + 3 (- 1 - 5 t) = 10 + 15 t - 3 - 15 t = 7$ for every $t$ . (1½)

Answer: $k = 5$ ; slope $- \frac{5}{3}$ ; solutions include $(- 1, 4)$ , $(5, - 6)$ , $(8, - 11)$ .

Example 4: Line through two points · 3 marks · Short answerHarder board level

Find the equation of the line through $(- 2, 5)$ and $(4, - 1)$ , check that $(3, 0)$ lies on it, and write the line through $(1, 1)$ parallel to it.

Try it first, then show the solutionSolution

Slope $m = \frac{- 1 - 5}{4 - ( - 2 )} = - 1$ ; $5 = - 1 (- 2) + d$ gives $d = 3$ , so $y = - x + 3$ . (1)
At $(3, 0)$ : $- 3 + 3 = 0$ , so it lies on the line. (1)
A parallel line has slope $- 1$ : $y = - x + q$ with $1 = - 1 + q$ , so $q = 2$ and $y = - x + 2$ . (1)

Answer: $y = - x + 3$ ; $(3, 0)$ lies on it; the parallel line is $y = - x + 2$ .

Example 5: Pens and pencils · 4 marks · Case-basedHOTS / competency

Case: a shopkeeper sells pens at 5 rupees and pencils at 3 rupees. A customer spends exactly 60 rupees. (i) Write the equation and list all solutions in non-negative whole numbers. (ii) What is the slope of the line? (iii) Is $x = 4$ pens possible? (iv) What is the maximum number of pens?

Try it first, then show the solutionSolution

(i) $5 x + 3 y = 60$ ; $y = \frac{60 - 5 x}{3}$ is a whole number when $x$ is a multiple of $3$ : $(0, 20), (3, 15), (6, 10), (9, 5), (12, 0)$ . (1½)
(ii) Slope $= - \frac{5}{3}$ . (½)
(iii) $x = 4$ gives $y = \frac{40}{3}$ , not a whole number, so it is not possible (the point lies on the line but is not a valid purchase). (1)
(iv) The largest $x$ is $12$ pens (with no pencils). (1)

Answer: (i) as listed; (ii) $- \frac{5}{3}$ ; (iii) no; (iv) 12 pens.

Where students lose marks in Linear Equations in Two Variables: Solutions, Slope and Graph

The slips examiners see most often, and what to write instead.

Mistake: Wrong sign in the slope from standard form
Taking slope $= \frac{a}{b}$ .
Wrong working
$3 x + 4 y = 12$ has slope $\frac{3}{4}$ .
Right working
Solve for $y$ : $y = - \frac{3}{4} x + 3$ , slope $- \frac{3}{4}$ .
Do this: Rearrange to $y = m x + d$ .
Usually costs: 1 mark.
Mistake: Intercept confusion
Setting the wrong variable to zero.
Wrong working
x-intercept found by putting $x = 0$ .
Right working
x-intercept: put $y = 0$ .
Do this: The x-intercept lies on the x-axis, where $y = 0$ .
Usually costs: 1 to 2 marks.
Mistake: Plotting every point of a restricted situation
Joining points that are not allowed.
Wrong working
Joining the solutions of $5 x + 3 y = 60$ into a continuous line.
Right working
Only whole-number points are valid; draw separate points.
Do this: Read the situation for allowed values.
Usually costs: 1 mark.

Exam tips

Write the substitution line when testing a point.
Label axes, scale and intercepts on every graph.
Check your line with a third point.
In context questions, state the meaning of slope and intercept.

Practice questions on Linear Equations in Two Variables: Solutions, Slope and Graph

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The equation of the line parallel to the x-axis passing through $(3, - 4)$ is
(a) $x = 3$
(b) $y = - 4$
(c) $x = - 4$
(d) $y = 3$
Hint
A line parallel to the x-axis has a constant y-value.
Answer
(b) $y = - 4$ .
Q2 · 3 marks · Short answerBoard level
Write $2 x + 5 y = 10$ in the form $y = m x + d$ and state the slope and both intercepts.
Hint
Solve for $y$ .
Answer
$y = - \frac{2}{5} x + 2$ . Slope $- \frac{2}{5}$ . y-intercept $2$ . x-intercept: $y = 0$ gives $x = 5$ .
Q3 · 3 marks · Short answerHarder board level
(i) Find $c$ if the line $x + 2 y = c$ passes through $(3, 4)$ . (ii) Find the equation of the line through $(- 1, - 1)$ parallel to $3 x - y = 2$ .
Hint
Substitute the point; use the same slope for parallel lines.
Answer
(i) $3 + 8 = c$ , so $c = 11$ . (ii) $3 x - y = 2$ gives $y = 3 x - 2$ with slope $3$ . The parallel line through $(- 1, - 1)$ : $- 1 = 3 (- 1) + d$ , $d = 2$ , so $y = 3 x + 2$ .
Q4 · 5 marks · Long answerHarder board level
The line $3 x + 4 y = 12$ meets the axes at $A$ and $B$ . Find $A$ , $B$ , the area of triangle $O A B$ ( $O$ the origin), the length $A B$ , the perimeter and the perpendicular distance from $O$ to the line.
Hint
Area of a triangle can be written two ways.
Answer
$A = (4, 0)$ , $B = (0, 3)$ . Area $= \frac{1}{2} \times 4 \times 3 = 6$ . $A B = 16 + 9 = 5$ . Perimeter $= 4 + 3 + 5 = 12$ . Using area $= \frac{1}{2} \times A B \times h$ : $6 = \frac{1}{2} \times 5 \times h$ , so $h = \frac{12}{5} = 2.4$ , the distance from $O$ to the line.
Q5 · 4 marks · Case-basedHOTS / competency
Case: a phone battery is charged to 100% and drains steadily, so that the charge $y$ (in per cent) after $x$ hours is $y = 100 - 4 x$ . (i) State the slope and its meaning. (ii) State both intercepts and their meaning. (iii) Find the charge after 4.5 hours. (iv) For what values of $x$ is the equation meaningful?
Hint
Read the slope and the intercepts from the equation.
Answer
(i) Slope $- 4$ : the battery loses 4 percentage points per hour. (ii) y-intercept $100$ (full charge at the start); x-intercept: $0 = 100 - 4 x$ gives $x = 25$ , the battery is empty after 25 hours. (iii) $100 - 4 (4.5) = 82%$ . (iv) $0 \leq x \leq 25$ , since the charge cannot be negative.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Number of solutions of a linear equation in two variables?Show answer
Infinitely many.
2.Slope-intercept form?Show answer
$y = m x + d$ .
3.Slope of $a x + b y = c$ ?Show answer
$- \frac{a}{b}$ .
4.Equation of the x-axis?Show answer
$y = 0$ .
5.Do parallel lines have equal slopes?Show answer
Yes.

Linear Equations in Two Variables: Solutions, Slope and Graph: frequently asked questions

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

Each choice of one variable gives a value of the other, and there are infinitely many choices.

What is the slope of a vertical line?

It is not defined, since the change in x is zero.

How many points do I need to draw the line?

Two points are enough; a third point is used to check.