MathsCBSE Class 10 · Pair of Linear Equations in Two Variables

Graphical Method of Solving a Pair of Linear Equations

Updated 14 October 2026 · Fact-checked

The graphical method solves a pair of linear equations by drawing both straight lines on the same axes. If they cross, the crossing point is the unique solution. Parallel lines mean no solution, and coincident lines mean infinitely many. Boards ask you to draw the lines, read the solution, and find vertices or area of a triangle with an axis.

The red dot where lines cross is the one solution; parallel lines never meet, and coincident lines share every point.

Illustration drawn for QuestionBuddy.

Understand Graphical Method of Solving Linear Equations

A linear equation in two variables, such as $x + y = 5$ , has infinitely many solutions, and every one of them is a point on a single straight line. So each equation of a pair is a line, and solving the pair means finding a point that lies on both lines at once.

That is the whole idea of the graphical method. Draw both lines on the same axes with the same scale. Where they meet, the coordinates of the meeting point satisfy both equations, because the point sits on both lines. Read off $x$ and $y$ and you have the solution.

To draw a line you need only two points, but always find a third as a check. If the three points do not lie in a straight line, you have made an arithmetic slip. Putting $x = 0$ and then $y = 0$ gives the intercepts, which are usually the quickest points to plot. Choose other values so that both coordinates are whole numbers.

Two lines in a plane can only do three things. They can intersect at one point (unique solution, consistent pair), be parallel (no common point, no solution, inconsistent pair), or coincide (every point common, infinitely many solutions, consistent and dependent pair). So the picture tells you not just the answer but how many answers exist. The ratio conditions on $\frac{a _{1}}{a _{2}}$ , $\frac{b _{1}}{b _{2}}$ , $\frac{c _{1}}{c _{2}}$ give the same verdict without drawing, and you can use them to check your graph.

The method works best when the solution has integer coordinates, since a hand-drawn graph cannot read $x = \frac{17}{11}$ accurately. That is why the algebraic methods (substitution and elimination) come next in the chapter. In board questions with graphs, the numbers are chosen to give clean points, and the question often goes further: find the vertices of the triangle formed by the two lines and the $x$ -axis or $y$ -axis, and its area. The vertices are the intersection point and the two points where the lines cut that axis.

Graphical Method of Solving Linear Equations: formulas to remember

Intercepts of a line: $a x + b y = c : x -intercept (\frac{c}{a}, 0), y -intercept (0, \frac{c}{b})$; Valid when $a \neq = 0$ and $b \neq = 0$ . Fast plotting points and the base vertices of a triangle formed with an axis.
Intersecting lines: $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} \Rightarrow unique solution$; The ratios compare equations written as $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ (the ratios are the same if both are written as '= c'). Graph: lines cross at one point. Pair is consistent.
Coincident lines: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}} \Rightarrow infinitely many solutions$; Graph: one line on top of the other. Consistent and dependent.
Parallel lines: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}} \Rightarrow no solution$; Graph: lines never meet. Inconsistent.
Area of triangle with an axis: $Area = \frac{1}{2} \times base on axis \times ∣ coordinate of intersection perpendicular to that axis ∣$; With the x-axis, height is |y| of the intersection point; with the y-axis, height is |x|.

How to solve Graphical Method of Solving Linear Equations questions

Use this routine for any graph question on a pair of linear equations.

1Rewrite each equation as $y$ in terms of $x$ (or keep $a x + b y = c$ and use intercepts) so points are easy to compute.
2Make a table of three points for each line, choosing values that give whole-number coordinates.
3Draw axes with a clear, equal scale; plot the points and join them into straight lines extended across the grid.
4Label each line with its equation.
5Read the intersection point and write the answer as $x = \dots$ , $y = \dots$ ; verify it by substituting in both equations.
6If the lines are parallel or coincident, state no solution or infinitely many, and name the pair inconsistent or consistent (dependent).
7For a triangle question, find where each line meets the required axis, list the three vertices, and use $\frac{1}{2} \times base \times height$ with units.

Graphical Method of Solving Linear Equations 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.

Graphical method of solution

Suppose two mobile data plans are offered in your town. Plan P costs Rs 100 a month plus Rs 20 per extra GB used. Plan Q costs Rs 40 a month plus Rs 30 per extra GB. You want to know when both bills are equal.

Let $x$ be the extra GB used and $y$ the monthly bill in rupees. Write both equations, find two points on each line, and find where the lines intersect. Which plan is cheaper if you use 10 extra GB?

Try it first, then show the solutionSolution

Plan P: $y = 100 + 20 x$ ; Plan Q: $y = 40 + 30 x$ .
Points on P: $(0, 100)$ , $(3, 160)$ , $(6, 220)$ .
Points on Q: $(0, 40)$ , $(3, 130)$ , $(6, 220)$ .
Both lines pass through $(6, 220)$ , so they intersect there.
Check: $100 + 120 = 220$ and $40 + 180 = 220$ .
At $x = 10$ : P costs $100 + 200 = 300$ , Q costs $40 + 300 = 340$ .

Answer: Bills are equal (Rs 220) at 6 extra GB; for 10 extra GB, Plan P is cheaper at Rs 300 against Rs 340.

Memory hook: Two plans, two lines: where they cross, the bills match.

Graphical method of solution

Imagine you are laying a rangoli on a square grid in the school courtyard. Two straight chalk lines are drawn along $x + 2 y = 8$ and $x - y = 2$ , measured in metres, with the courtyard edge as the $x$ -axis.

Find graphically where the chalk lines cross, and find the area of the triangle enclosed by the two lines and the courtyard edge (the $x$ -axis).

Try it first, then show the solutionSolution

Points on $x + 2 y = 8$ : $(8, 0)$ , $(0, 4)$ , $(4, 2)$ .
Points on $x - y = 2$ : $(2, 0)$ , $(4, 2)$ , $(6, 4)$ .
Common point is $(4, 2)$ ; check $4 + 4 = 8$ , $4 - 2 = 2$ .
Vertices with $x$ -axis: $(8, 0)$ , $(2, 0)$ , $(4, 2)$ .
Base $= 8 - 2 = 6$ m, height $= 2$ m.
Area $= \frac{1}{2} \times 6 \times 2 = 6$ .

Answer: Lines cross at $(4, 2)$ ; enclosed area is $6 m^{2}$ .

Memory hook: Rangoli triangle: base on the edge, height from the crossing point.

Tricks to remember Graphical Method of Solving Linear Equations

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

I-P-C: Intersect, Parallel, Coincide
One, None, Infinite. Intersect gives one, Parallel gives none, Coincide gives infinite.
Use it when: Instantly match the graph picture to the number of solutions.
Three points, not two
Two points draw a line; the third point catches your mistake.
Use it when: Always tabulate three points so an error shows up as a bent line.
Zero-zero for intercepts
Put $x = 0$ for the $y$ -axis point, $y = 0$ for the $x$ -axis point.
Use it when: Quick plotting and quick triangle vertices.
Height comes from the crossing
Base lies on the axis; height is the crossing point's distance from that axis.
Use it when: Area questions with the x-axis (use |y|) or y-axis (use |x|).

Graphical Method of Solving Linear Equations: solved examples

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

Example 1: Reading the nature of lines · 1 mark · MCQBoard level

The pair of equations $3 x - 2 y = 6$ and $6 x - 4 y = 15$ represents two lines that are
(a) intersecting at exactly one point
(b) parallel
(c) coincident
(d) intersecting at two points

Try it first, then show the solutionSolution

$\frac{a _{1}}{a _{2}} = \frac{3}{6} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{- 2}{- 4} = \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{6}{15} = \frac{2}{5}$ .
First two ratios equal, third different, so lines are parallel.

Answer: (b) parallel

Example 2: Graphical solution and checking · 3 marks · Short answerBoard level

Solve graphically: $x + y = 6$ and $x - 2 y = 0$ . Verify your answer.

Try it first, then show the solutionSolution

Table for $x + y = 6$ : $(0, 6)$ , $(6, 0)$ , $(3, 3)$ . (½ mark)
Table for $x - 2 y = 0$ , i.e. $y = \frac{x}{2}$ : $(0, 0)$ , $(2, 1)$ , $(4, 2)$ . (½ mark)
Plot both lines on the same axes and label them. (1 mark)
The lines meet at $(4, 2)$ . (½ mark)
Check: $4 + 2 = 6$ and $4 - 4 = 0$ . (½ mark)

Answer: $x = 4$ , $y = 2$

Example 3: Triangle with the x-axis · 5 marks · Long answerHarder board level

Draw the graphs of $x - y + 1 = 0$ and $3 x + 2 y - 12 = 0$ . Find the vertices of the triangle formed by these lines and the $x$ -axis, and find its area.

Try it first, then show the solutionSolution

$x - y + 1 = 0 \Rightarrow y = x + 1$ : points $(- 1, 0)$ , $(0, 1)$ , $(2, 3)$ . (1 mark)
$3 x + 2 y = 12 \Rightarrow y = \frac{12 - 3 x}{2}$ : points $(4, 0)$ , $(0, 6)$ , $(2, 3)$ . (1 mark)
Draw and label both lines; they intersect at $(2, 3)$ . Check: $2 - 3 + 1 = 0$ , $6 + 6 - 12 = 0$ . (1 mark)
Vertices with $x$ -axis: $(- 1, 0)$ , $(4, 0)$ , $(2, 3)$ . (1 mark)
Base $= 4 - (- 1) = 5$ units, height $= 3$ units, area $= \frac{1}{2} \times 5 \times 3 = 7.5$ . (1 mark)

Answer: Vertices $(- 1, 0)$ , $(4, 0)$ , $(2, 3)$ ; area $= 7.5$ square units.

Example 4: Triangle with the y-axis · 5 marks · Long answerHarder board level

Solve graphically $2 x + y = 8$ and $x - y = 1$ . Find the vertices of the triangle formed by these lines and the $y$ -axis, and its area.

Try it first, then show the solutionSolution

Tables: $2 x + y = 8$ : $(0, 8)$ , $(4, 0)$ , $(3, 2)$ ; $x - y = 1$ : $(0, - 1)$ , $(1, 0)$ , $(3, 2)$ . (1 mark)
Plot both lines on the same axes with a clear scale and label them. (1 mark)
The lines meet at $(3, 2)$ , so $x = 3$ , $y = 2$ . Check: $6 + 2 = 8$ , $3 - 2 = 1$ . (1 mark)
Vertices with the $y$ -axis: $(0, 8)$ , $(0, - 1)$ , $(3, 2)$ . (1 mark)
Base $= 8 - (- 1) = 9$ units, height $= ∣ x ∣ = 3$ units, area $= \frac{1}{2} \times 9 \times 3 = 13.5$ . (1 mark)

Answer: $x = 3$ , $y = 2$ ; vertices $(0, 8)$ , $(0, - 1)$ , $(3, 2)$ ; area $= 13.5$ square units.

Example 5: Case study: tickets at a school fair · 4 marks · Case-basedBoard level

Imagine your school fair sells adult tickets at Rs $x$ and child tickets at Rs $y$ . One family pays Rs 100 for 1 adult and 2 children. Another pays Rs 140 for 2 adults and 1 child.
(i) Write the pair of equations. (1 mark)
(ii) Give two points with whole-number coordinates on each line. (1 mark)
(iii) Find the ticket prices from the graph and state whether the pair is consistent. (2 marks)

Try it first, then show the solutionSolution

(i) $x + 2 y = 100$ and $2 x + y = 140$ . (1 mark)
(ii) First line: $(100, 0)$ , $(40, 30)$ ; second line: $(70, 0)$ , $(40, 60)$ . (1 mark)
(iii) Try $x = 60$ : first line gives $y = 20$ ; second gives $y = 140 - 120 = 20$ . So both lines pass through $(60, 20)$ . (1 mark)
Check: $60 + 40 = 100$ , $120 + 20 = 140$ . Ratios $\frac{1}{2} \neq = \frac{2}{1}$ , so lines intersect: consistent, unique solution. (1 mark)

Answer: Adult ticket Rs 60, child ticket Rs 20; the pair is consistent with a unique solution.

Example 6: Coincident lines: three common points · 2 marks · Short answerBoard level

Do the lines $2 x + 3 y = 9$ and $4 x + 6 y = 18$ coincide? Give the graphical meaning and write three points common to both lines.

Try it first, then show the solutionSolution

$\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{3}{6} = \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{9}{18} = \frac{1}{2}$ . (½ mark)
All three ratios are equal, so the lines coincide: infinitely many solutions, pair consistent and dependent. (½ mark)
Points on $2 x + 3 y = 9$ : $(0, 3)$ , $(3, 1)$ , $(6, - 1)$ , since $0 + 9 = 9$ , $6 + 3 = 9$ , $12 - 3 = 9$ . (½ mark)
Same points in $4 x + 6 y = 18$ : $0 + 18 = 18$ , $12 + 6 = 18$ , $24 - 6 = 18$ . So all three are common. (½ mark)

Answer: Yes, the lines coincide; common points include $(0, 3)$ , $(3, 1)$ , $(6, - 1)$ ; infinitely many solutions.

Example 7: Assertion-Reason: coincident does not mean inconsistent · 1 mark · Assertion-reasonHOTS / competency

Assertion (A): The lines $3 x + 2 y = 6$ and $6 x + 4 y = 12$ are coincident, so the pair is inconsistent.
Reason (R): Coincident lines have infinitely many common points.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

Ratios: $\frac{3}{6} = \frac{2}{4} = \frac{6}{12} = \frac{1}{2}$ , so the lines are indeed coincident.
Coincident lines share infinitely many points, so the pair has solutions: it is consistent (dependent), not inconsistent. So A is false.
R is a true statement about coincident lines.

Answer: (d) A is false but R is true.

Example 8: Parallel lines: table and conclusion · 3 marks · Short answerBoard level

Draw the graphs of $x + 2 y = 4$ and $2 x + 4 y = 12$ . Show that the lines are parallel and write your conclusion about the solution.

Try it first, then show the solutionSolution

Table for $x + 2 y = 4$ : $(4, 0)$ , $(0, 2)$ , $(2, 1)$ . (½ mark)
Table for $2 x + 4 y = 12$ , i.e. $x + 2 y = 6$ : $(6, 0)$ , $(0, 3)$ , $(2, 2)$ . (½ mark)
Plot both lines on the same axes and label them; they run side by side and do not meet. (1 mark)
Ratios: $\frac{1}{2} = \frac{2}{4} \neq = \frac{4}{12} = \frac{1}{3}$ , confirming parallel lines. (½ mark)
Conclusion: no common point, so no solution; the pair is inconsistent. (½ mark)

Answer: The lines are parallel, so the pair has no solution and is inconsistent.

Example 9: Finding k from the type of graph · 3 marks · Short answerHarder board level

For what value of $k$ does the pair $k x + 2 y = 5$ and $3 x + y = 1$ have (i) no solution, (ii) infinitely many solutions? Describe what the graph looks like.

Try it first, then show the solutionSolution

$\frac{a _{1}}{a _{2}} = \frac{k}{3}$ , $\frac{b _{1}}{b _{2}} = \frac{2}{1} = 2$ , $\frac{c _{1}}{c _{2}} = \frac{5}{1} = 5$ . (½ mark)
(i) No solution needs $\frac{k}{3} = 2 \neq = 5$ : $k = 6$ , and $2 \neq = 5$ holds. (1 mark)
(ii) Infinitely many solutions need $\frac{k}{3} = 2$ and $2 = 5$ together. The second is false, so no value of $k$ works. (1 mark)
Graph: for $k = 6$ the lines are parallel; for any $k \neq = 6$ they intersect at one point; they are never coincident. (½ mark)

Answer: (i) $k = 6$ (parallel lines); (ii) no value of $k$ , the lines can never coincide.

Example 10: Word problem: form the equations, then graph · 5 marks · Long answerBoard level

The sum of two numbers is 10 and the larger exceeds the smaller by 2. Taking the larger number as x and the smaller as y, form a pair of linear equations, solve it graphically, and find the area of the triangle formed by the two lines and the $y$ -axis.

Try it first, then show the solutionSolution

Let the larger number be $x$ and the smaller $y$ : $x + y = 10$ and $x - y = 2$ . (1 mark)
Tables: $x + y = 10$ : $(0, 10)$ , $(10, 0)$ , $(6, 4)$ ; $x - y = 2$ : $(0, - 2)$ , $(2, 0)$ , $(6, 4)$ . (1 mark)
Draw both lines; they meet at $(6, 4)$ . Check: $6 + 4 = 10$ , $6 - 4 = 2$ . (1 mark)
Vertices with the $y$ -axis: $(0, 10)$ , $(0, - 2)$ , $(6, 4)$ . (1 mark)
Base $= 10 - (- 2) = 12$ units, height $= 6$ units, area $= \frac{1}{2} \times 12 \times 6 = 36$ . (1 mark)

Answer: The numbers are 6 and 4; area of the triangle $= 36$ square units.

Example 11: Case study: two auto services with a hidden constant · 4 marks · Case-basedHOTS / competency

Suppose two auto services run in your town. For a ride of $x$ km, Service A charges Rs $y = 15 x + 5$ . Service B charges Rs $y = k x + 30$ , and a 4 km ride with B costs Rs 70.
(i) Find $k$ . (1 mark)
(ii) Write both equations in the form $a x + b y = c$ and give two points on each line. (1 mark)
(iii) From the graph, find the distance at which both fares are equal, and say which service is cheaper for an 8 km ride. (2 marks)

Try it first, then show the solutionSolution

(i) $70 = 4 k + 30 \Rightarrow 4 k = 40 \Rightarrow k = 10$ . (1 mark)
(ii) A: $15 x - y = - 5$ , points $(1, 20)$ , $(3, 50)$ ; B: $10 x - y = - 30$ , points $(0, 30)$ , $(4, 70)$ . (1 mark)
(iii) Both lines pass through $(5, 80)$ : $75 + 5 = 80$ and $50 + 30 = 80$ . So the fares are equal at 5 km, Rs 80. (1 mark)
At 8 km: A $= 120 + 5 = 125$ , B $= 80 + 30 = 110$ . B is cheaper by Rs 15. (1 mark)

Answer: $k = 10$ ; fares are equal (Rs 80) at 5 km; for 8 km, Service B is cheaper (Rs 110 against Rs 125).

Where students lose marks in Graphical Method of Solving Linear Equations

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

Mistake: Plotting only two points
One wrong point silently gives a wrong line and wrong solution.
Wrong working
For $2 x + y = 8$ : $(0, 8)$ , $(3, 1)$ joined; reads crossing with $x - y = 1$ at a wrong point.
Right working
$(0, 8)$ , $(4, 0)$ , $(3, 2)$ are collinear; $(3, 1)$ was a slip since $6 + 1 \neq = 8$ .
Do this: Always use three points and check each in the equation.
Usually costs: Up to 1-2 marks for wrong graph and solution.
Mistake: Wrong base for a triangle with the $y$ -axis
Students take $x$ -intercepts when the question says $y$ -axis.
Wrong working
For $2 x + y = 8$ and $x - y = 1$ with the $y$ -axis, base $= 4 - 1 = 3$ .
Right working
Base on $y$ -axis $= 8 - (- 1) = 9$ ; height $= 3$ ; area $= 13.5$ .
Do this: Underline which axis the question names before finding vertices.
Usually costs: 1-2 marks in the area part.
Mistake: Subtracting a negative intercept wrongly
Base is the distance between intercepts, not their difference with sign slips.
Wrong working
Intercepts $- 1$ and $4$ : base $= 4 - 1 = 3$ .
Right working
Base $= 4 - (- 1) = 5$ .
Do this: Base $=$ larger coordinate minus smaller, keeping signs.
Usually costs: 1 mark.
Mistake: Not stating the conclusion for parallel lines
Drawing is done but the required word is missing.
Wrong working
Graph shows two parallel lines; nothing written.
Right working
Lines are parallel, so no solution; the pair is inconsistent.
Do this: Always write shape, number of solutions and consistent/inconsistent.
Usually costs: ½ to 1 mark.

Exam tips

Label every line with its equation and mark the intersection point with its coordinates.
Write the point tables in your answer sheet; examiners award marks for them even if the graph is slightly off.
Verify the read-off solution by substitution in both equations; it takes one line.
In triangle questions, list the three vertices clearly before computing area, and give the answer in square units.
Use the ratio test to confirm your graph shows the right type of lines.

Practice questions on Graphical Method of Solving Linear Equations

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 · MCQHarder board level
For which value of $k$ does the pair $k x + 3 y = 6$ and $2 x + y = 2$ have no solution?
(a) $6$
(b) $3$
(c) $- 6$
(d) There is no such value of $k$
Hint
No solution needs $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ but $\frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ . Check the constants too.
Answer
(d). Parallel lines need $\frac{k}{2} = \frac{3}{1}$ , so $k = 6$ . But then $\frac{c _{1}}{c _{2}} = \frac{6}{2} = 3$ as well, so all three ratios are equal and the lines coincide (infinitely many solutions). For any other k the lines intersect. So no value of k gives no solution.
Q2 · 3 marks · Short answerBoard level
Solve graphically: $x + y = 4$ and $2 x - y = 2$ .
Hint
Use intercepts for both lines, then read the crossing point.
Answer
Points: $(0, 4)$ , $(4, 0)$ , $(2, 2)$ and $(0, - 2)$ , $(1, 0)$ , $(2, 2)$ . Lines meet at $(2, 2)$ : $x = 2$ , $y = 2$ .
Q3 · 5 marks · Long answerBoard level
Draw the graphs of $x + 3 y = 6$ and $2 x - 3 y = 12$ . Find the vertices of the triangle formed by these lines and the $y$ -axis, and its area.
Hint
Find the $y$ -intercepts of both lines, then the intersection; the height is the $x$ -coordinate of the intersection.
Answer
Points: $(6, 0)$ , $(0, 2)$ , $(3, 1)$ and $(6, 0)$ , $(0, - 4)$ , $(3, - 2)$ . Intersection $(6, 0)$ . Vertices $(0, 2)$ , $(0, - 4)$ , $(6, 0)$ . Base $6$ , height $6$ , area $18$ square units.
Q4 · 3 marks · Short answerBoard level
Show graphically that the pair $x - 2 y = 2$ and $2 x - 4 y = 8$ has no solution. Write two points on each line.
Hint
Plot the intercepts; check whether the lines ever meet.
Answer
Line 1: $(2, 0)$ , $(0, - 1)$ , $(4, 1)$ . Line 2: $(4, 0)$ , $(0, - 2)$ , $(6, 1)$ . Lines are parallel (ratios $\frac{1}{2} = \frac{1}{2} \neq = \frac{1}{4}$ ), so no solution; inconsistent.
Q5 · 3 marks · Short answerBoard level
The lines $y = x$ and $x + y = 8$ are drawn with the $x$ -axis. Find graphically the vertices of the triangle formed and its area.
Hint
$y = x$ passes through the origin, which is one vertex.
Answer
Lines meet at $(4, 4)$ . Vertices $(0, 0)$ , $(8, 0)$ , $(4, 4)$ . Area $= \frac{1}{2} \times 8 \times 4 = 16$ square units.
Q6 · 2 marks · Short answerBoard level
Aman drew $x + y = 5$ and $2 x + 2 y = 10$ and wrote: 'The lines meet at $(2, 3)$ , so the solution is $x = 2$ , $y = 3$ .' Find the error and state the correct conclusion with its graphical meaning.
Hint
Compare all three ratios.
Answer
Ratios $\frac{1}{2} = \frac{1}{2} = \frac{5}{10}$ , so the lines coincide. $(2, 3)$ is only one of infinitely many solutions. The pair is consistent and dependent with infinitely many solutions.
Q7 · 3 marks · Short answerHarder board level
The line $a x + y = 7$ passes through the point $(2, 3)$ . Find $a$ , and then solve graphically the pair $a x + y = 7$ and $x - y = 2$ .
Hint
Substitute the point first to get the equation, then tabulate both lines.
Answer
$2 a + 3 = 7$ , so $a = 2$ . Points on $2 x + y = 7$ : $(0, 7)$ , $(2, 3)$ , $(3, 1)$ ; on $x - y = 2$ : $(2, 0)$ , $(0, - 2)$ , $(3, 1)$ . Lines meet at $(3, 1)$ : $x = 3$ , $y = 1$ .
Q8 · 3 marks · Short answerHarder board level
Find $p$ and $q$ so that $2 x + 3 y = 7$ and $p x + (p + q) y = 28$ represent coincident lines. Then write three points common to both lines.
Hint
Set $\frac{2}{p} = \frac{3}{p + q} = \frac{7}{28}$ .
Answer
$\frac{2}{p} = \frac{1}{4}$ gives $p = 8$ ; $\frac{3}{p + q} = \frac{1}{4}$ gives $p + q = 12$ , so $q = 4$ . Lines $2 x + 3 y = 7$ and $8 x + 12 y = 28$ ; common points $(2, 1)$ , $(5, - 1)$ , $(- 1, 3)$ .
Q9 · 5 marks · Long answerHarder board level
The length of a rectangular garden is 4 m more than its breadth, and half its perimeter is 20 m. Taking the length as x and the breadth as y, form a pair of linear equations, solve it graphically, and find the area of the triangle formed by the two lines and the $x$ -axis (1 unit on the graph = 1 m).
Hint
Let length be $x$ and breadth $y$ ; half the perimeter is $x + y$ .
Answer
$x - y = 4$ and $x + y = 20$ . Points: $(4, 0)$ , $(0, - 4)$ , $(12, 8)$ and $(20, 0)$ , $(0, 20)$ , $(12, 8)$ . Length 12 m, breadth 8 m. Triangle vertices $(4, 0)$ , $(20, 0)$ , $(12, 8)$ ; area $= \frac{1}{2} \times 16 \times 8 = 64 m^{2}$ .
Q10 · 3 marks · Short answerHOTS / competency
Riya says: 'The lines $3 x + 6 y = 9$ and $x + 2 y = k$ coincide for every value of $k$ , because $\frac{3}{1} = \frac{6}{2}$ .' Find her error, find the value of $k$ for which they coincide, and describe the graph when $k = 5$ .
Hint
Coincidence needs all three ratios equal, not just two.
Answer
She ignored the constants. Coincident needs $\frac{3}{1} = \frac{6}{2} = \frac{9}{k}$ , so $k = 3$ . For $k = 5$ , $\frac{9}{5} \neq = 3$ , so the lines are parallel: no solution, inconsistent.
Q11 · 4 marks · Long answerHarder board level
The lines $2 x + 3 y = 12$ and $k x - y = 18$ meet at a point on the x-axis.
(i) Find that point and the value of $k$ .
(ii) Draw both lines and find the vertices of the triangle they form with the y-axis.
(iii) Find the area of this triangle.
Hint
A point on the x-axis has y = 0; it must lie on both lines.
Answer
(i) On the x-axis $y = 0$ , so from $2 x + 3 y = 12$ , $x = 6$ : the point is $(6, 0)$ . It lies on $k x - y = 18$ , so $6 k = 18$ , $k = 3$ . (1)
(ii) $2 x + 3 y = 12$ : $(0, 4)$ , $(6, 0)$ , $(3, 2)$ . $3 x - y = 18$ : $(0, - 18)$ , $(6, 0)$ , $(5, - 3)$ . Vertices with the y-axis: $(0, 4)$ , $(0, - 18)$ and $(6, 0)$ . (2)
(iii) Base along the y-axis $= 4 - (- 18) = 22$ , height $= 6$ , area $= \frac{1}{2} \times 22 \times 6 = 66$ square units. (1)
Q12 · 3 marks · Short answerHarder board level
The lines $x - y = 0$ and $x + y = k$ , where $k > 0$ , form a triangle with the x-axis whose area is 16 square units. Find $k$ , the vertices of the triangle, and the point where the two lines meet.
Hint
Write the vertices in terms of k first, then use $\frac{1}{2} \times$ base $\times$ height.
Answer
The lines meet where $x = y$ and $2 x = k$ , at $(\frac{k}{2}, \frac{k}{2})$ . With the x-axis the vertices are $(0, 0)$ , $(k, 0)$ and $(\frac{k}{2}, \frac{k}{2})$ . Area $= \frac{1}{2} \times k \times \frac{k}{2} = \frac{k ^{2}}{4} = 16$ , so $k^{2} = 64$ and $k = 8$ (as $k > 0$ ). Vertices $(0, 0)$ , $(8, 0)$ , $(4, 4)$ ; the lines meet at $(4, 4)$ .
Q13 · 4 marks · Case-basedHOTS / competency
Consider the lines $x + y = 3$ , $2 x - y = 3$ and $3 x - 2 y = k$ .
(i) Solve the first two equations graphically. (1 mark)
(ii) Find $k$ so that all three lines pass through one point. (1 mark)
(iii) Take $k = 6$ instead. Find where the third line meets each of the first two lines, and explain why the three lines now enclose a triangle. (2 marks)
Hint
A point common to all three lines must be the solution of the first two.
Answer
(i) $x + y = 3$ : $(0, 3)$ , $(3, 0)$ , $(2, 1)$ ; $2 x - y = 3$ : $(0, - 3)$ , $(1.5, 0)$ , $(2, 1)$ . They meet at $(2, 1)$ . (1)
(ii) The third line must pass through $(2, 1)$ : $6 - 2 = k$ , so $k = 4$ . (1)
(iii) With $3 x - 2 y = 6$ : with $x + y = 3$ , $3 x - 2 (3 - x) = 6$ gives $x = 2.4$ , $y = 0.6$ ; with $2 x - y = 3$ , $y = 2 x - 3$ gives $3 x - 4 x + 6 = 6$ , so $x = 0$ , $y = - 3$ . The three meeting points $(2, 1)$ , $(2.4, 0.6)$ and $(0, - 3)$ are different, and no two of the lines are parallel (their ratios $\frac{a}{b}$ differ), so the lines enclose a triangle with these vertices. (2)
Q14 · 4 marks · Case-basedHOTS / competency
Case study: Imagine two water tanks on a school roof. Tank A holds 200 L and is filled at 30 L per minute, so after $x$ minutes it holds $y = 200 + 30 x$ litres. Tank B holds 500 L and is drained so that $y = 500 - k x$ ; after 2 minutes it holds 440 L.
(i) Find $k$ . (1 mark)
(ii) Write both equations in the form $a x + b y = c$ and give two points on each line. (1 mark)
(iii) Find graphically when the tanks hold equal amounts and how much each holds then. Which tank holds more after 10 minutes? (2 marks)
Hint
Use the 2-minute reading to find $k$ before drawing anything.
Answer
(i) $440 = 500 - 2 k$ , so $k = 30$ . (ii) $30 x - y = - 200$ : $(0, 200)$ , $(2, 260)$ ; $30 x + y = 500$ : $(0, 500)$ , $(2, 440)$ . (iii) Lines meet at $(5, 350)$ : equal after 5 minutes, 350 L each. After 10 minutes A holds 500 L and B holds 200 L, so Tank A holds more.

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.What does the intersection point of two lines represent?Show answer
The unique solution of the pair.
2.Parallel lines mean how many solutions?Show answer
None; the pair is inconsistent.
3.Coincident lines mean how many solutions?Show answer
Infinitely many; consistent and dependent.
4.Where does $3 x + 4 y = 12$ meet the $y$ -axis?Show answer
At $(0, 3)$ .
5.Minimum points needed to draw a line, and how many should you plot?Show answer
Two are enough; plot three to check.

Graphical Method of Solving Linear Equations: frequently asked questions

How many points should I plot for each line in the graphical method?

Two points fix a line, but plot three. The third point checks your arithmetic, because if it is not in line you know there is an error.

What if the lines cross at a fractional point?

A graph cannot read fractions reliably, so board questions usually give integer solutions. If you suspect a fraction, solve by substitution or elimination to confirm.

Is graph paper needed in the board exam?

Graph questions are answered on graph paper provided in the exam, with a clear scale. Draw neatly with a pencil and ruler and label the lines.

Is the cross-multiplication method part of this chapter?

No. Cross-multiplication and equations reducible to linear form are not in the 2026-27 syllabus.