Linear Equations in Two Variables turns everyday links between two quantities, such as prices, fares or ages, into equations of the form ax + by + c = 0. One such equation has infinitely many solutions lying on a straight line. You read its slope and y-intercept from y = mx + c, and graph it, including cases where only whole-number values make sense. Then you solve a pair of equations by graphs, substitution and elimination, and decide whether a pair has one, none or infinitely many solutions. The Class 10 chapter on pairs builds directly on it.
Key Concepts
1. Linear Equations in Two Variables Through Practical Examples
Many everyday situations link two unknown quantities in a fixed way. A batter scores 40 runs using only fours and sixes. You do not know how many fours or how many sixes, but you do know the total. If the number of fours is x and the number of sixes is y, the runs from fours are 4x and the runs from sixes are 6y, so
4x + 6y = 40
This is one equation with two unknowns. Each unknown appears only to the power 1, and the two unknowns are never multiplied together. Such an equation is called a linear equation in two variables.
More examples from daily life:
- A pen costs Rs x and a notebook costs Rs y. Two pens and three notebooks cost Rs 110: 2x + 3y = 110.
- An auto charges Rs 25 for the first kilometre and Rs 15 for every kilometre after that. For a trip of x km the fare is y = 25 + 15(x – 1) = 15x + 10.
The standard form
Every linear equation in two variables can be written as
ax + by + c = 0
where a, b and c are real numbers and a and b are not both zero. To find a, b and c, move every term to the left side, changing the sign of any term that crosses the equals sign, and read the number in front of x, the number in front of y, and the constant.
| Given equation | Standard form | a | b | c |
|---|---|---|---|---|
| 3x = 2y + 7 | 3x – 2y – 7 = 0 | 3 | -2 | -7 |
| y = 4 | 0·x + 1·y – 4 = 0 | 0 | 1 | -4 |
| x = -3 | 1·x + 0·y + 3 = 0 | 1 | 0 | 3 |
| x/2 + y/3 = 1 | 3x + 2y – 6 = 0 (multiply by 6) | 3 | 2 | -6 |
y = 4 is still a linear equation in two variables: the coefficient of x is 0. The rule only forbids a and b from being zero at the same time. The standard form is not unique: 6x + 4y – 12 = 0 describes the same relationship as 3x + 2y – 6 = 0.
How to spot an equation that is not linear
Look at the variables, never at the numbers in front of them.
- x2 + y = 3 has x squared, so it is not linear.
- xy = 4 multiplies the two variables, which makes a term of degree 2.
- 2/x + y = 1 has x in the denominator, which is x to the power -1.
- √x + y = 2 has x to the power 1/2.
- √2 x + y = √3 is linear, because √2 is only a coefficient. x/4 – y/5 = 1 is linear too, because x/4 is (1/4)x.
One variable or two
As an equation in one variable, 2x + 3 = 0 has one solution, x = -3/2, a single point on the number line. As an equation in two variables it is 2x + 0·y + 3 = 0, and every pair (-3/2, y) is a solution: a whole line in the plane. Read which one the question asks for.
2. Solutions of a Linear Equation in Two Variables
A solution of a linear equation in two variables is a pair of values, one for x and one for y, that makes the two sides of the equation equal. It is written as an ordered pair (x, y), with the x-value first.
Checking a pair
Substitute both values and compare the two sides.
- For x + 3y = 7, the pair (1, 2) gives 1 + 6 = 7, so (1, 2) is a solution.
- The pair (2, 1) gives 2 + 3 = 5, and 5 ≠ 7, so (2, 1) is not a solution. The order of the numbers matters.
Why there are infinitely many solutions
Take 2x + y = 7. Choose any value of x, and the equation tells you the value of y that goes with it: y = 7 – 2x.
| x | 0 | 1 | 2 | 3.5 | -1 |
|---|---|---|---|---|---|
| y = 7 – 2x | 7 | 5 | 3 | 0 | 9 |
Since x can be any real number, the list never ends. One linear equation in two variables has infinitely many solutions; pinning down a single pair needs a second equation.
Finding a missing value
If a point with one unknown coordinate lies on the line, substitute the known coordinate and solve. Suppose (k, 2) is a solution of 3x – 2y = 11. Then 3k – 4 = 11, so 3k = 15 and k = 5.
3. Graph of a Linear Equation in Two Variables
Each solution (x, y) is a point on the Cartesian plane. When you plot several solutions of the same linear equation, they all fall on one straight line. This is why the equation is called linear.
Take x + 2y = 6, so y = (6 – x)/2.
| x | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| y | 3 | 2 | 1 | 0 |
Choosing even values of x keeps every y a whole number. The four points (0, 3), (2, 2), (4, 1) and (6, 0) lie on one line.
Every point on the line is a solution, and every solution is a point on the line. So to test a point, substitute it; no drawing is needed. (4, 1) lies on x + 2y = 6 because 4 + 2 = 6. (1, 4) does not, because 1 + 8 = 9.
Where the line crosses the axes
Every point on the x-axis has y = 0, and every point on the y-axis has x = 0.
- x-intercept point: put y = 0. For x + 2y = 6, x = 6, so the line meets the x-axis at (6, 0).
- y-intercept point: put x = 0. For x + 2y = 6, y = 3, so the line meets the y-axis at (0, 3).
These two points are the fastest way to draw a line. With the origin they form a right triangle: for x + 2y = 6 its area is (1/2) × 6 × 3 = 9 square units.
Special lines: x = k is parallel to the y-axis and y = k is parallel to the x-axis; x = 0 is the y-axis itself and y = 0 is the x-axis. A line such as y = 3x passes through (0, 0), (1, 3) and (-1, -3). Any equation ax + by = 0, with no constant, is satisfied by x = 0, y = 0, so its line always passes through the origin.
4. Slope-Intercept Form y = mx + c
When b ≠ 0, any linear equation ax + by + c = 0 can be rearranged to make y the subject:
by = -ax – c, so y = (-a/b)x + (-c/b)
This is the slope-intercept form, usually written y = mx + c. The CBSE course structure writes the same form as y = ax + b; only the letters differ. In these notes, m is the slope and c (in y = mx + c) is the y-intercept. Do not confuse this c with the constant c of the standard form. From the rearrangement above, m = -a/b and the y-intercept is -c/b.
What m and c mean
- Slope m tells you how much y changes when x increases by 1. If m = 2, y goes up by 2 for every step of 1 in x.
- y-intercept c is the value of y when x = 0. The line crosses the y-axis at (0, c).
A positive slope means the line rises from left to right, a negative slope means it falls, and m = 0 gives a horizontal line. A vertical line x = k (b = 0) cannot be written as y = mx + c. A larger value of m (ignoring sign) means a steeper line: y = 5x climbs faster than y = x.
Converting to slope-intercept form
Take 3x + 2y = 8. Move 3x across: 2y = -3x + 8. Divide every term by 2: y = -(3/2)x + 4. The slope is -3/2 and the y-intercept is 4.
Slope from two points
If a line passes through (x1, y1) and (x2, y2), its slope is
m = (y2 – y1)/(x2 – x1), that is, change in y divided by change in x.
Through (0, -1) and (2, 5): m = (5 – (-1))/(2 – 0) = 6/2 = 3. The point (0, -1) is on the y-axis, so c = -1 and the line is y = 3x – 1. Check with (2, 5): 3(2) – 1 = 5.
Parallel lines
Two lines with the same slope and different y-intercepts are parallel: they rise at the same rate and start at different heights, so they never meet. y = 2x + 1 and y = 2x – 5 are parallel. Two lines with the same slope and the same intercept are the same line.
5. Drawing Graphs When x and y Take Only Certain Values
In many real situations the variables can only be whole numbers. You cannot buy 2.5 pens or hit 1.3 sixes. The equation still describes a straight line, but only some points on that line make sense.
Return to the batter: 4x + 6y = 40, with x fours and y sixes. Dividing by 2 gives 2x + 3y = 20, so x = (20 – 3y)/2. For x to be a whole number, 20 – 3y must be even, so y must be even.
| y (sixes) | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| x (fours) | 10 | 7 | 4 | 1 |
Check one: 4(7) + 6(2) = 28 + 12 = 40. At y = 8, x would be -2, which is impossible. So the batter could have scored the 40 runs in exactly four ways, and the graph is four separate points: (10, 0), (7, 2), (4, 4) and (1, 6). They lie on the line 2x + 3y = 20, but you do not join them, because points between them (like x = 5.5) have no meaning here.
When a question says “x and y are whole numbers” or the context counts objects or people, draw dots only. When x and y can be any real numbers, draw the full line.
6. Pair of Linear Equations: Graphical Method and Nature of Solutions
Two linear equations in the same two variables, considered together, form a pair of linear equations in two variables. In general form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
A solution of the pair is an ordered pair (x, y) that satisfies both equations at once. On the graph, it is a point that lies on both lines.
The graphical method
Draw both lines on the same axes and read the point where they meet. For x + y = 5 and x – y = 1:
| x + y = 5 | |||
|---|---|---|---|
| x | 0 | 5 | 3 |
| y | 5 | 0 | 2 |
| x – y = 1 | |||
|---|---|---|---|
| x | 0 | 1 | 3 |
| y | -1 | 0 | 2 |
Both lines pass through (3, 2), so the solution is x = 3, y = 2. Check: 3 + 2 = 5 and 3 – 2 = 1.
The graphical method has a limit: a meeting point such as (1.37, 2.81) cannot be read exactly from graph paper. The algebraic methods fix this.
Three possibilities for two lines
| Lines | Number of solutions | Pair is called | Slopes and intercepts |
|---|---|---|---|
| Intersecting | Exactly one | Consistent | Different slopes |
| Parallel | None | Inconsistent | Same slope, different y-intercepts |
| Coincident (one on top of the other) | Infinitely many | Consistent (dependent) | Same slope, same y-intercept |
A pair is consistent if it has at least one solution and inconsistent if it has none.
Testing without drawing: the ratio test
Compare the ratios of the coefficients of the two equations, each written in the form ax + by + c = 0.
- a1/a2 ≠ b1/b2: intersecting lines, exactly one solution.
- a1/a2 = b1/b2 ≠ c1/c2: parallel lines, no solution.
- a1/a2 = b1/b2 = c1/c2: coincident lines, infinitely many solutions.
7. Algebraic Methods: Substitution and Elimination
Substitution method
- From one equation, write one variable in terms of the other. Pick the equation where a variable has coefficient 1 or -1 if you can.
- Substitute that expression into the other equation. You now have an equation in one variable.
- Solve it, then put the value back into the expression from step 1. Check both values in both equations.
Solve 2x + 3y = 11 and x – 2y = -12. From the second, x = 2y – 12. Substitute into the first: 2(2y – 12) + 3y = 11, so 4y – 24 + 3y = 11, 7y = 35 and y = 5. Then x = 2(5) – 12 = -2. Check: 2(-2) + 3(5) = -4 + 15 = 11, and -2 – 2(5) = -12. Both hold.
Elimination method
- Multiply one or both equations by suitable non-zero numbers so that one variable has the same coefficient (or coefficients that differ only in sign) in both.
- Subtract the equations if the coefficients are equal, or add them if they are opposite. That variable disappears.
- Solve, substitute back into either original equation, and check.
Solve 3x + 4y = 10 and 2x – 2y = 2. Multiply the second equation by 2: 4x – 4y = 4. Now the y-coefficients are +4 and -4, so add: 7x = 14 and x = 2. Then 3(2) + 4y = 10 gives 4y = 4 and y = 1. Check in the second equation: 2(2) – 2(1) = 2.
What the algebra tells you about the nature of the pair
Sometimes both variables disappear together.
- x + 2y = 4 and 2x + 4y = 12. Double the first: 2x + 4y = 8. Subtracting from the second gives 0 = 4, a false statement. The pair has no solution: the lines are parallel.
- x + 2y = 4 and 3x + 6y = 12. Triple the first: 3x + 6y = 12. Subtracting gives 0 = 0, a statement that is always true. The pair has infinitely many solutions: the lines coincide.
So all three methods, graphs, the ratio test and the algebra, always give the same verdict about a pair.
8. Modelling Real Situations With a Pair of Equations
Use the same four steps for every contextual problem.
- Name the unknowns. Write exactly what x and y stand for, with units. “Let the cost of one pen be Rs x.”
- Translate each condition into one equation. Two independent conditions give two equations.
- Solve by substitution, elimination or graphs.
- Check and conclude. Test the answer in the original words, not only in your equations, and answer the exact question asked.
Common problem types
| Type | How to set it up |
|---|---|
| Tickets or coins | One equation counts items (a + c = total number), the other counts money (price × number) |
| Ages | Shift both ages to the same moment: “5 years ago” means (x – 5) and (y – 5) |
| Two-digit numbers | Tens digit x, units digit y: number = 10x + y, reversed number = 10y + x |
| Comparing two plans | Write y = mx + c for each plan and set them equal to find the break-even point |
If your answer gives a negative number of people, a fraction of a ticket, or a two-digit number with a digit above 9, the model or the arithmetic is wrong. Go back and check.
Formula and Theorem Sheet
| Topic | Formula or Rule | When to use it |
|---|---|---|
| Standard form | ax + by + c = 0, a and b not both zero | Identifying a, b, c; ratio test |
| x-axis crossing | Put y = 0: point (-c/a, 0), a ≠ 0 | Fast graphing, area with the axes |
| y-axis crossing | Put x = 0: point (0, -c/b), b ≠ 0 | Fast graphing, area with the axes |
| Slope-intercept form | y = mx + c (written y = ax + b in the syllabus) | Reading slope and intercept; this c is the y-intercept, a different c from the standard form |
| Slope from standard form | m = -a/b, y-intercept = -c/b | Comparing two lines quickly |
| Slope from two points | m = (y2 – y1)/(x2 – x1) | Finding the equation of a line |
| Parallel lines | Same m, different c | No common point |
| Unique solution | a1/a2 ≠ b1/b2 | Intersecting lines, consistent |
| No solution | a1/a2 = b1/b2 ≠ c1/c2 | Parallel lines, inconsistent |
| Infinitely many solutions | a1/a2 = b1/b2 = c1/c2 | Coincident lines, consistent |
| Area of triangle cut off by a line and the axes | (1/2) × |x-intercept| × |y-intercept| | Graph questions |
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Important Definitions
- Linear equation in two variables: an equation that can be written in the form ax + by + c = 0, where a, b, c are real numbers and a and b are not both zero.
- Solution of a linear equation in two variables: an ordered pair (x, y) of values that makes the two sides of the equation equal.
- x-intercept: the x-coordinate of the point where a line crosses the x-axis (found by putting y = 0).
- y-intercept: the y-coordinate of the point where a line crosses the y-axis (found by putting x = 0). In y = mx + c it is c.
- Slope: the change in y for an increase of 1 in x; for two points on the line, (y2 – y1)/(x2 – x1). In y = mx + c it is m.
- Pair of linear equations in two variables: two linear equations in the same two variables x and y, considered together.
- Solution of a pair: an ordered pair (x, y) that satisfies both equations of the pair.
- Consistent pair: a pair that has at least one solution (intersecting or coincident lines).
- Inconsistent pair: a pair that has no solution (parallel lines).
Solved Examples
Example 1: Finding an unknown coefficient or coordinate
(a) If x = 2, y = 1 is a solution of 2x + 3y = k, find k. (b) If (k, 2) lies on the line 3x – 2y = 11, find k.
Solution: (a) Substitute: 2(2) + 3(1) = k, so k = 4 + 3 = 7.
(b) Substitute x = k and y = 2: 3k – 2(2) = 11, so 3k – 4 = 11, 3k = 15 and k = 5.
Example 2: Graph, intercepts and area
Draw the graph of 2x + y = 6. Find where it meets the axes, check whether (1, 4) lies on it, and find the area of the triangle it forms with the axes.
Solution: Write y = 6 – 2x.
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| y | 6 | 4 | 2 | 0 |
Plot (0, 6), (1, 4), (2, 2) and (3, 0) and join them with a straight line.
y-axis: x = 0 gives y = 6, so the point is (0, 6). x-axis: y = 0 gives 2x = 6, x = 3, so the point is (3, 0).
(1, 4): 2(1) + 4 = 6, so yes, it lies on the line (it is in the table).
The triangle has vertices (0, 0), (3, 0) and (0, 6). Area = (1/2) × 3 × 6 = 9 square units.
Example 3: Converting to slope-intercept form
Write 5x – 2y = 10 in the form y = mx + c. State its slope, its y-intercept and the point where it meets the x-axis.
Solution: 5x – 2y = 10 gives -2y = -5x + 10. Divide every term by -2: y = (5/2)x – 5.
Slope m = 5/2; y-intercept c = -5, so the line meets the y-axis at (0, -5).
x-axis: put y = 0: 0 = (5/2)x – 5, so x = 2. The line meets the x-axis at (2, 0).
Example 4: Graph when x and y take only whole-number values
Pencils cost Rs 3 each and erasers Rs 5 each. Riya spends exactly Rs 30. Write an equation, list every possible purchase, and describe the graph.
Solution: Let x be the number of pencils and y the number of erasers. Then 3x + 5y = 30, with x and y whole numbers.
Since 3x = 30 – 5y, the value 30 – 5y must be a multiple of 3. Try y = 0, 1, 2, and so on:
| y | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| 30 – 5y | 30 | 25 | 20 | 15 | 10 | 5 | 0 |
| x | 10 | no | no | 5 | no | no | 0 |
The possible purchases are (10, 0), (5, 3) and (0, 6). For y above 6 the value of x would be negative. The graph is three separate points on the line 3x + 5y = 30, and they are not joined.
Example 5: Solving a pair graphically
Solve graphically: x + 2y = 8 and 2x – y = 1.
Solution:
| x + 2y = 8, so y = (8 – x)/2 | ||||
|---|---|---|---|---|
| x | 0 | 2 | 4 | 8 |
| y | 4 | 3 | 2 | 0 |
| 2x – y = 1, so y = 2x – 1 | ||||
|---|---|---|---|---|
| x | 0 | 1 | 2 | 3 |
| y | -1 | 1 | 3 | 5 |
Both tables contain (2, 3), so the lines intersect at (2, 3). The solution is x = 2, y = 3. Check: 2 + 6 = 8 and 4 – 3 = 1.
Example 6: Nature of solutions and a missing coefficient
(a) Without solving, state the nature of each pair: (i) 3x + 2y = 7 and x – y = 4 (ii) 2x – 4y = 5 and x – 2y = 1. (b) Find k for which kx + 3y = 5 and 6x + 9y = 7 has no solution.
Solution: (a)(i) a1/a2 = 3/1 = 3 and b1/b2 = 2/(-1) = -2. They differ, so there is exactly one solution (intersecting lines).
(ii) In standard form: 2x – 4y – 5 = 0 and x – 2y – 1 = 0. a1/a2 = 2, b1/b2 = (-4)/(-2) = 2, c1/c2 = (-5)/(-1) = 5. The first two are equal and the third differs, so no solution (parallel lines).
(b) For no solution we need a1/a2 = b1/b2: k/6 = 3/9 = 1/3, so k = 2. Check the third ratio: c1/c2 = (-5)/(-7) = 5/7, which is not 1/3. So at k = 2 the lines are parallel and the pair has no solution.
Example 7: Substitution method
Solve 3x – y = 7 and 2x + 5y = 16 by substitution.
Solution: From the first equation, y = 3x – 7. Substitute into the second:
2x + 5(3x – 7) = 16, so 2x + 15x – 35 = 16, 17x = 51, and x = 3.
Then y = 3(3) – 7 = 2. So x = 3, y = 2.
Check: 3(3) – 2 = 7 and 2(3) + 5(2) = 6 + 10 = 16.
Example 8: Elimination method
Solve 5x + 3y = 35 and 2x – 4y = -12 by elimination.
Solution: To eliminate y, make its coefficients 12 and -12. Multiply the first equation by 4 and the second by 3:
20x + 12y = 140 and 6x – 12y = -36.
Add: 26x = 104, so x = 4. Substitute into the first equation: 5(4) + 3y = 35, so 3y = 15 and y = 5.
So x = 4, y = 5. Check in the second equation: 2(4) – 4(5) = 8 – 20 = -12.
Example 9: Cost of pens and notebooks
3 pens and 2 notebooks cost Rs 110. 2 pens and 5 notebooks cost Rs 220. Find the cost of one pen and of one notebook.
Solution: Let one pen cost Rs p and one notebook Rs n. Then
3p + 2n = 110 … (1) and 2p + 5n = 220 … (2)
Multiply (1) by 2 and (2) by 3: 6p + 4n = 220 and 6p + 15n = 660. Subtract the first from the second: 11n = 440, so n = 40.
From (1): 3p + 80 = 110, so 3p = 30 and p = 10.
A pen costs Rs 10 and a notebook costs Rs 40. Check with (2): 2(10) + 5(40) = 20 + 200 = 220.
Example 10: Ages of a father and son
The sum of the ages of a father and his son is 50 years. Five years ago, the father was seven times as old as the son. Find their present ages.
Solution: Let the father’s present age be f years and the son’s be s years.
f + s = 50 … (1)
Five years ago their ages were (f – 5) and (s – 5), so f – 5 = 7(s – 5), which gives f – 5 = 7s – 35, that is f – 7s = -30 … (2)
Subtract (2) from (1): 8s = 80, so s = 10. Then f = 50 – 10 = 40.
The father is 40 years old and the son is 10 years old. Check: five years ago they were 35 and 5, and 35 = 7 × 5.
Example 11: A two-digit number
The sum of the digits of a two-digit number is 9. The number formed by reversing the digits is 27 more than the original number. Find the number.
Solution: Let the tens digit be x and the units digit be y. The number is 10x + y and the reversed number is 10y + x.
x + y = 9 … (1)
(10y + x) – (10x + y) = 27, so 9y – 9x = 27, that is y – x = 3 … (2)
Add (1) and (2): 2y = 12, so y = 6, and x = 3. The number is 36. Check: 3 + 6 = 9, and 63 – 36 = 27.
Competency-Based Questions (with answers)
1. Case-based: Choosing a data plan
A mobile company offers two monthly plans. Plan A has a fixed charge of Rs 150 and costs Rs 10 per GB used. Plan B has a fixed charge of Rs 50 and costs Rs 20 per GB used. Let x be the data used in GB and y the monthly bill in rupees.
(a) Write the equation for each plan. Plan A: y = 10x + 150. Plan B: y = 20x + 50.
(b) For what usage do the two plans cost the same, and what is that cost? 10x + 150 = 20x + 50 gives 100 = 10x, so x = 10 GB. The cost is 10(10) + 150 = Rs 250.
(c) Which plan should a user who needs 15 GB choose? Plan A: 10(15) + 150 = Rs 300. Plan B: 20(15) + 50 = Rs 350. Plan A is cheaper by Rs 50.
2. Case-based: School annual day tickets
For the school annual day, adult tickets cost Rs 100 and child tickets cost Rs 60. On the first evening, 50 tickets were sold for a total of Rs 4200.
(a) Form a pair of linear equations. Let a be the number of adult tickets and c the number of child tickets. a + c = 50 and 100a + 60c = 4200.
(b) Solve the pair. From the first, c = 50 – a. Substitute: 100a + 60(50 – a) = 4200, so 100a + 3000 – 60a = 4200, 40a = 1200 and a = 30. Then c = 20.
(c) Interpret the answer. 30 adult tickets and 20 child tickets were sold. Check: 30 × 100 + 20 × 60 = 3000 + 1200 = 4200.
3. Case-based: Filling and draining tanks
Tank P holds 600 litres and is being drained at 25 litres per minute. At the same moment, tank Q, which holds 100 litres, is being filled at 25 litres per minute. Let x be the time in minutes and y the volume of water in litres.
(a) Write an equation for each tank. Tank P: y = 600 – 25x (linear decay). Tank Q: y = 100 + 25x (linear growth).
(b) When will the tanks hold equal amounts, and how much? 600 – 25x = 100 + 25x gives 500 = 50x, so x = 10 minutes. Each tank then holds 100 + 250 = 350 litres.
(c) When will tank P be empty? 600 – 25x = 0 gives x = 24 minutes. The graph of tank P crosses the x-axis at (24, 0).
4. Assertion-Reason
Assertion (A): The graph of y = 3 is a line parallel to the x-axis. Reason (R): y = 3 can be written as 0·x + 1·y – 3 = 0, so it is a linear equation in two variables.
Options: (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.
Answer: (b). Both statements are true. R only shows that y = 3 is linear. The line is parallel to the x-axis because every point on it has y-coordinate 3.
5. Assertion-Reason
Assertion (A): The pair 2x + 3y = 5 and 4x + 6y = 10 has infinitely many solutions. Reason (R): For this pair, a1/a2 = b1/b2 = c1/c2.
Answer: (a). The ratios are 2/4, 3/6 and (-5)/(-10), all equal to 1/2. The second equation is twice the first, so the lines coincide and every point on one is on the other. R is the correct explanation of A.
6. Assertion-Reason
Assertion (A): The linear equation 2x + 5y = 10 has a unique solution. Reason (R): The graph of a linear equation in two variables is a straight line.
Answer: (d). A is false: (5, 0), (0, 2) and (2.5, 1) are all solutions, and there are infinitely many. R is true: every point on the straight line is a solution, which is exactly why there are infinitely many.
7. Data reasoning: A savings table
Aman records the money in his savings box at the end of each day:
| Day (x) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Money in Rs (y) | 5 | 8 | 11 | 14 | 17 |
(a) Is the relationship linear? Give a reason. Yes. Each time x increases by 1, y increases by the same amount, 3.
(b) Write the equation in the form y = mx + c. The slope is 3 and the value at x = 0 is 5, so y = 3x + 5.
(c) How much will he have at the end of day 10, and on which day will he first have Rs 50? Day 10: 3(10) + 5 = Rs 35. For Rs 50: 3x + 5 = 50 gives x = 15, so at the end of day 15.
8. Error analysis
A student solved x + y = 7 and x – y = 3 as follows: “Subtracting, 2y = 10, so y = 5 and x = 2.” Find the error and give the correct solution.
Answer: Subtracting the second equation from the first gives (x + y) – (x – y) = 7 – 3, that is 2y = 4. The student added the right-hand sides while subtracting the left-hand sides. Correct working: y = 2 and x = 5. Check: 5 + 2 = 7 and 5 – 2 = 3.
Important Questions for Exams
1-Mark Questions
Q1. Write 2x = y in the form ax + by + c = 0 and state a, b, c.
2x – y + 0 = 0, so a = 2, b = -1, c = 0.
Q2. Is (2, -1) a solution of 3x + 2y = 4?
3(2) + 2(-1) = 6 – 2 = 4. Yes, it is a solution.
Q3. Write the slope and y-intercept of y = -4x + 7.
Slope -4, y-intercept 7.
Q4. How many solutions does the equation 3x – 2y = 6 have?
Infinitely many.
Q5. Write the equation of the x-axis.
y = 0.
2-Mark Questions
Q6. Find k if (3, k) lies on the line 4x – y = 5.
4(3) – k = 5, so 12 – k = 5 and k = 7.
Q7. Find the points where 3x + 4y = 12 meets the two axes.
Put y = 0: 3x = 12, x = 4, giving (4, 0). Put x = 0: 4y = 12, y = 3, giving (0, 3).
Q8. Without solving, state whether 3x – y = 2 and 9x – 3y = 6 has one, none or infinitely many solutions.
In standard form, 3x – y – 2 = 0 and 9x – 3y – 6 = 0. a1/a2 = 3/9 = 1/3, b1/b2 = (-1)/(-3) = 1/3, c1/c2 = (-2)/(-6) = 1/3. All equal, so infinitely many solutions (coincident lines).
Q9. Write 6x + 3y = 9 in the form y = mx + c and state the slope.
3y = -6x + 9, so y = -2x + 3. Slope -2, y-intercept 3.
3-Mark Questions
Q10. Solve by substitution: x + y = 10 and 2x – y = 11.
From the first, y = 10 – x. Substitute: 2x – (10 – x) = 11, so 3x – 10 = 11, 3x = 21, x = 7. Then y = 10 – 7 = 3. Check: 7 + 3 = 10, and 14 – 3 = 11. Answer: x = 7, y = 3.
Q11. Solve by elimination: 4x + 3y = 18 and 3x – 2y = 5.
Multiply the first by 2 and the second by 3: 8x + 6y = 36 and 9x – 6y = 15. Add: 17x = 51, x = 3. Then 4(3) + 3y = 18, 3y = 6, y = 2. Check: 3(3) – 2(2) = 9 – 4 = 5. Answer: x = 3, y = 2.
Q12. For what value of k does the pair 2x + ky = 7 and 4x + 6y = 3 have no solution?
No solution needs a1/a2 = b1/b2 ≠ c1/c2. So 2/4 = k/6, giving k = 3. Check: c1/c2 = (-7)/(-3) = 7/3, which is not equal to 1/2. So k = 3.
5-Mark Questions
Q13. Solve graphically: x – y = 1 and 2x + y = 8.
Table for x – y = 1 (y = x – 1): (0, -1), (1, 0), (3, 2). Table for 2x + y = 8 (y = 8 – 2x): (0, 8), (4, 0), (3, 2). Plot both lines on the same axes. They intersect at (3, 2), so x = 3, y = 2. Check: 3 – 2 = 1 and 6 + 2 = 8.
Q14. A fraction becomes 1/3 when 2 is subtracted from the numerator, and it becomes 1/2 when 6 is subtracted from the denominator. Find the fraction.
Let the fraction be x/y. First condition: (x – 2)/y = 1/3, so 3x – 6 = y, that is 3x – y = 6 … (1). Second condition: x/(y – 6) = 1/2, so 2x = y – 6, that is 2x – y = -6 … (2). Subtract (2) from (1): x = 12. From (1): 36 – y = 6, so y = 30. The fraction is 12/30. Check: (12 – 2)/30 = 10/30 = 1/3, and 12/(30 – 6) = 12/24 = 1/2. Do not reduce it to 2/5: that fraction fails both conditions.
Q15. A taxi charges a fixed amount plus a fixed rate per kilometre. A 10 km ride costs Rs 105 and a 15 km ride costs Rs 155. Find the fixed charge and the rate per km, write the fare equation in the form y = mx + c, and find the fare for 25 km.
Let the fixed charge be Rs f and the rate Rs k per km. f + 10k = 105 … (1) and f + 15k = 155 … (2). Subtract (1) from (2): 5k = 50, so k = 10. From (1): f = 105 – 100 = 5. Fixed charge Rs 5, rate Rs 10 per km. For x km the fare is y = 10x + 5, with slope 10 (rupees per km) and y-intercept 5 (the fixed charge). For 25 km: y = 10(25) + 5 = Rs 255.
Common Mistakes and Examiner Tips
- Saying one linear equation in two variables has a unique solution. It has infinitely many. A unique solution needs a pair of equations whose lines intersect.
- Calling y = 4 or x = -3 “not an equation in two variables”. They are, with a = 0 or b = 0. Only a = b = 0 together is ruled out.
- Forgetting to change the sign when a term crosses the equals sign. 3x = 2y – 4 becomes 3x – 2y + 4 = 0, so c = +4, not -4. Write the full standard form before reading a, b, c.
- Dividing only some terms when making y the subject. From 3x + 2y = 8, 2y = -3x + 8 gives y = -(3/2)x + 4. Dividing only the 8 by 2 gives the wrong line. Divide every term.
- Getting the slope sign wrong. For ax + by + c = 0 the slope is -a/b, with a minus sign. For 2x + y = 6 the slope is -2, so the line falls.
- Joining points when x and y must be whole numbers. If the context counts pens, people or runs, draw separate dots on the line and do not join them.
- Using the ratio test with equations in different forms. Put both equations in the form ax + by + c = 0 first, with c on the left. Mixing ax + by = c with ax + by + c = 0 flips the sign of c and can give a wrong verdict.
- Checking only a1/a2 = b1/b2 for “no solution”. Also show c1/c2 is different; if it is equal, there are infinitely many solutions.
- Adding one side and subtracting the other in elimination. Whatever you do to the left sides, do the same to the right sides. Write the operation (“Subtract (2) from (1)”) so the examiner can follow your step.
- Stopping at x and y in word problems. Write the final answer in words with units and check it against the original sentences.
Quick Revision Points
- A linear equation in two variables has the form ax + by + c = 0, with a, b, c real and a, b not both zero.
- y = k and x = k are linear equations in two variables, with one coefficient equal to zero.
- A solution is an ordered pair (x, y) that makes both sides equal. Order matters.
- One linear equation in two variables has infinitely many solutions.
- x = k is parallel to the y-axis; y = k is parallel to the x-axis. x = 0 is the y-axis and y = 0 is the x-axis.
- Slope-intercept form: y = mx + c (the syllabus writes y = ax + b). m is the slope, c is the y-intercept.
- From ax + by + c = 0 with b ≠ 0: slope = -a/b, y-intercept = -c/b.
- Slope through two points: (y2 – y1)/(x2 – x1). Positive slope rises, negative slope falls, zero slope is horizontal.
- When x and y can only be whole numbers, the graph is a set of separate points on the line.
- A pair of linear equations is two linear equations in the same two variables; its solution satisfies both.
- Intersecting lines: one solution, consistent. Parallel lines: no solution, inconsistent. Coincident lines: infinitely many solutions, consistent (dependent).
- Ratio test: a1/a2 ≠ b1/b2 gives one solution; a1/a2 = b1/b2 ≠ c1/c2 gives none; all three equal gives infinitely many.
- Substitution: make one variable the subject, substitute into the other equation, solve, back-substitute.
- Elimination: make one variable’s coefficients equal or opposite, subtract or add, solve, back-substitute.
- In elimination, 0 = a non-zero number means no solution; 0 = 0 means infinitely many.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams and internal assessment. In the CBSE Class IX Mathematics course structure for 2026-27, Linear Equations in Two Variables sits in Unit II, Algebra, which carries 20 marks out of the 80-mark theory paper. That 20 is shared by four chapters: Introduction to Polynomials, Sequences and Progressions, Exploring Algebraic Identities, and Linear Equations in Two Variables. The course structure does not give a separate mark for this chapter. The unit is allotted 66 teaching periods. The remaining 20 marks of the year are internal assessment (pen-paper test and multiple assessment 10, portfolio 5, lab practical 5).
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
The CBSE question paper design weights questions by typology: about 54% remembering and understanding (for example, writing an equation in standard form, checking a solution, stating the nature of a pair), about 24% applying (drawing graphs, solving by substitution or elimination) and about 22% analysing, evaluating and creating (modelling a real situation, comparing plans, finding and correcting errors).
The course structure lists these key concepts: linear equations in two variables through practical examples, graphical representation of solutions, the slope-intercept form, graphs when x and y take only certain values, pairs of linear equations and their graphical solution, consistency and inconsistency, and the substitution and elimination methods. The same ideas return in the Class 10 chapter Pair of Linear Equations in Two Variables.
Class 9 Maths – swipe through all 10 cards to understand the whole chapter.
The Standard Form
A linear equation in two variables can always be written in one tidy shape.
Move every term to the left, changing signs as terms cross the equals sign.
- 3x = 2y + 7 becomes 3x – 2y – 7 = 0: a = 3, b = -2, c = -7.
- y = 4 is 0·x + 1·y – 4 = 0, still linear in two variables.
- xy = 4, x2 + y = 3 and 2/x + y = 1 are not linear.
Infinitely Many Solutions
A solution is an ordered pair (x, y) that makes both sides equal.
Order matters: (1, 2) and (2, 1) are different points.
- Pick any x and the equation fixes y, so the list never ends.
- Check a pair by substituting both values.
- (k, 2) on 3x – 2y = 11 gives 3k – 4 = 11, so k = 5.
The Graph Is a Straight Line
Every solution is a point on one line, and every point on the line is a solution.
Plot three points: two fix the line, the third checks your arithmetic.
- x + 2y = 6 meets the axes at (6, 0) and (0, 3).
- Triangle with the axes: (1/2) × 6 × 3 = 9 square units.
- x = k is parallel to the y-axis, y = k to the x-axis.
Slope-Intercept Form
Making y the subject shows how steep the line is and where it starts.
The syllabus writes it as y = ax + b. This c is the y-intercept, not the constant in ax + by + c = 0.
- 3x + 2y = 8 gives y = -(3/2)x + 4: slope -3/2, intercept 4.
- Positive slope rises, negative falls, zero is horizontal.
- Same slope, different intercepts: parallel lines.
Slope in Real Life
A constant rate of change gives a straight-line model.
The intercept is the starting value; the slope is the rate.
- Data plan y = 10x + 150: Rs 10 per GB plus Rs 150 fixed.
- Tank y = 500 – 20x drains 20 L a minute, empty at x = 25.
- Through (2, 5) and (4, 9): m = 2, line y = 2x + 1.
When Only Some Values Make Sense
If x and y count objects, the graph is separate dots on the line.
Do not join the dots when fractions have no meaning.
- Whole-number solutions: (10, 0), (7, 2), (4, 4), (1, 6).
- 3x + 5y = 30 with pencils and erasers: (10, 0), (5, 3), (0, 6).
- Say ‘whole numbers only’ in your answer to explain the dots.
Nature of a Pair
Two lines can cross once, never meet, or lie on top of each other.
a1/a2 = b1/b2 ≠ c1/c2 → none; all three equal → infinitely many.
- Intersecting: one solution, consistent.
- Parallel: no solution, inconsistent.
- Coincident: infinitely many solutions, consistent (dependent).
Substitution Method
Make one variable the subject and put it into the other equation.
Best when a variable already has coefficient 1 or -1.
- 2x + 5(3x – 7) = 16 gives 17x = 51, so x = 3.
- Back-substitute: y = 3(3) – 7 = 2.
- Check in both original equations: 6 + 10 = 16.
Elimination Method
Match the coefficients of one variable, then add or subtract.
0 = 4 (false) means no solution; 0 = 0 means infinitely many.
- Opposite coefficients: add. Equal coefficients: subtract.
- Do the same operation to both sides of each equation.
- x = 2, then 3(2) + 4y = 10 gives y = 1.
Word Problems With a Pair
Name the unknowns, turn each condition into one equation, solve, check.
Answer in words with units and test it in the original story.
- Tickets: one equation counts tickets, the other counts money.
- Ages: shift both ages to the same moment before equating.
- Plans cost the same where the two lines cross.
📝 Practice Linear Equations in Two Variables - 10 questions
Concept-check questions · with answers & explanations
Start →Close ✕
Why D: y = 5 is the same as 0·x + 1·y – 5 = 0, so a = 0, b = 1, c = -5. The form only needs a and b not BOTH zero. Every pair with y = 5, such as (0, 5) and (3, 5), is a solution.
Why not A: a = 1, b = 0, c = -5 gives x – 5 = 0, that is x = 5, a different line.
Why not B: 0·x + y + 5 = 0 means y = -5.
Why not C: A missing variable simply has coefficient 0.
Remember: Missing variable means coefficient 0; y = 5 still counts as an equation in two variables.
Why B: The condition is 5p + 20n = 60, so p = 12 – 4n. Both p and n must be whole numbers with p not negative, so n can be 0, 1, 2 or 3, giving p = 12, 8, 4 and 0. The pairs are (12, 0), (8, 1), (4, 2) and (0, 3): 4 pairs. On a graph these are 4 separate points on the line, not the whole line.
Why not A: 3 misses one of the cases where she buys none of one item, (12, 0) or (0, 3).
Why not C: The line has infinitely many real points, but fractions of a pen or notebook cannot be bought.
Why not D: 12 is only the largest number of pens possible, not the number of pairs.
Remember: In a real context, list only whole-number, non-negative values; the graph becomes separate points.
Why A: Step 1 (y-intercept): (0, 4) lies on the y-axis, so c = 4. Step 2 (slope = change in y / change in x): m = (0 – 4)/(2 – 0) = -2. Step 3 (write and rearrange): y = -2x + 4, that is 2x + y = 4. Check (2, 0): 2(2) + 0 = 4.
Why not B: x + 2y = 4 passes through (0, 2) and (4, 0), the intercepts reversed.
Why not C: 2x – y = 4 is y = 2x – 4, with slope +2 and intercept -4.
Why not D: x + y = 4 passes through (0, 4) but meets the x-axis at (4, 0), not (2, 0).
Remember: Check your final equation with BOTH given points.
Why B: Step 1 (slope-intercept form): the first gives y = -(k/2)x + 5/2 and the second gives y = -3x + 1. Step 2 (parallel lines have equal slopes): -k/2 = -3, so k = 6. Step 3 (check intercepts): 5/2 and 1 differ, so with k = 6 the lines are parallel and distinct, and the pair has no solution.
Why not A: 3/2 comes from comparing the coefficients the wrong way round.
Why not C: k = -6 gives slope +3, while the second line has slope -3.
Why not D: Choosing k to match the slopes makes the lines parallel.
Remember: For no solution, make the slopes equal and confirm the intercepts are different.
Why A: From the first, x = 4 – 2y. Substituting: 2(4 – 2y) + 4y = 10, so 8 – 4y + 4y = 10, that is 8 = 10. Both variables vanish and a false statement remains, so no pair satisfies both equations. The lines y = -(1/2)x + 2 and y = -(1/2)x + 5/2 are parallel.
Why not B: The leftover statement 8 = 10 is false, not always true.
Why not C: (2, 1) fits x + 2y = 4, but 4 + 4 = 8, not 10.
Why not D: (0, 2) fits x + 2y = 4, but 0 + 8 = 8, not 10.
Remember: Variables vanish: a false statement means no solution, a true statement means infinitely many.
Why C: The costs are P = 200 + 5x and Q = 100 + 10x. They are equal when 200 + 5x = 100 + 10x, so 5x = 100 and x = 20 hours (both Rs 300). Q has the larger slope, so after 20 hours it grows faster. At 30 hours, P = 350 and Q = 400, so P is cheaper beyond 20 hours.
Why not A: At 10 hours P = 250 and Q = 200, so Q is cheaper below 20 hours.
Why not B: At 20 hours both cost Rs 300, so neither is cheaper.
Why not D: The lower rate lets P catch up and overtake after 20 hours.
Remember: Find where the two lines meet, then test one value on each side.
Why C: Step 1 (split the journey): the first km costs Rs 30 and the remaining (x – 1) km cost Rs 12 each. Step 2 (form and simplify the equation): y = 30 + 12(x – 1) = 30 + 12x – 12 = 12x + 18. Check: x = 1 gives 12 + 18 = 30, and x = 3 gives 30 + 12 + 12 = 54 = 12(3) + 18.
Why not A: This charges Rs 30 for every kilometre.
Why not B: This charges Rs 12 for every km including the first, plus Rs 30, so 1 km would cost Rs 42.
Why not D: 42 = 30 + 12 adds the two charges into one rate.
Remember: Test a fare model with x = 1; it must give the stated first-kilometre fare.
Why C: Step 1 (substitute the point): 3(2) – k = 4, so 6 – k = 4. Step 2 (solve the one-variable equation): k = 6 – 4 = 2. Check: 3(2) – 2 = 4.
Why not A: 10 comes from writing k = 6 + 4.
Why not B: -2 comes from 6 + k = 4, treating -y as +y.
Why not D: 6 is only the value of 3x; the subtraction was never done.
Remember: A point on a line fits its equation; substitute the known coordinate and solve for the unknown one.
Why D: Step 1 (slope-intercept form): with slope 3, the line is y = 3x + c. Step 2 (the point lies on the line): 5 = 3(1) + c, so c = 2. The line is y = 3x + 2. Check: at x = 1, y = 3 + 2 = 5.
Why not A: 5 is the y-value at x = 1, not at x = 0, so it is not the intercept.
Why not B: This has slope 5 and intercept 3.
Why not C: From 5 = 3 + c, c = 5 – 3 = 2, not -2.
Remember: Only a point with x = 0 gives the intercept directly; otherwise substitute and solve for c.
Why A: From 4x – 2y + 6 = 0, 2y = 4x + 6, so y = 2x + 3. This is the same as the first equation: same slope 2 and same y-intercept 3. The two equations describe one line, so every point on it is a common solution.
Why not B: The slopes are equal, but so are the intercepts, so the lines are not separate parallel lines.
Why not C: The equations look different but reduce to the same line.
Why not D: Two straight lines cannot share exactly two points.
Remember: Same slope and same intercept means the same line, so infinitely many solutions.
Chapter Navigation
Previous: Sequences and Progressions Class 9 Notes
Related Chapters in Class 9 Maths
- Coordinate Geometry Class 9 Notes
- Introduction to Polynomials Class 9 Notes
- Number System Class 9 Notes
- Exploring Algebraic Identities Class 9 Notes
- Circles Class 9 Notes
- Area and Perimeter Class 9 Notes
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Frequently Asked Questions
It is an equation that can be written as ax + by + c = 0, where a, b and c are real numbers and a and b are not both zero. Each variable appears only to the power 1 and the variables are never multiplied together. For example, 2x + 3y = 110 is linear, while xy = 4 and x² + y = 3 are not.
Infinitely many. For any value of x you choose, the equation gives a matching value of y. For 2x + y = 7, the pairs (0, 7), (1, 5), (2, 3) and (3.5, 0) are all solutions. Plotted together, all the solutions form one straight line, and every point on that line is a solution.
Yes. It can be written as 0·x + 1·y - 4 = 0, so a = 0, b = 1 and c = -4. The definition only rules out a and b both being zero. Its graph is a straight line parallel to the x-axis, 4 units above it, made of all points whose y-coordinate is 4.
Rearrange the equation into the form y = mx + c by making y the subject and dividing every term by the coefficient of y. Then m is the slope and c is the y-intercept. For 3x + 2y = 8, 2y = -3x + 8 gives y = -(3/2)x + 4, so the slope is -3/2 and the y-intercept is 4. In general the slope is -a/b.
Write both equations as ax + by + c = 0 and compare ratios. If a₁/a₂ ≠ b₁/b₂, the lines intersect and there is exactly one solution. If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel and there is no solution. If all three ratios are equal, the lines coincide and there are infinitely many solutions.
Use substitution when one variable already has coefficient 1 or -1, or is already the subject, such as y = 3x - 7. Use elimination when the coefficients are larger numbers, multiplying the equations so one variable has equal or opposite coefficients and then subtracting or adding. Both methods always give the same answer, so pick the one with less arithmetic.
When the variables count whole objects, such as pens, tickets or sixes hit, only whole-number solutions make sense. For 3x + 5y = 30 with pencils and erasers, the only purchases are (10, 0), (5, 3) and (0, 6). These points lie on the line, but points between them have no meaning, so the graph is drawn as separate dots.
Class 9 has no board exam, so the chapter is tested in school exams. It belongs to Unit II, Algebra, which carries 20 marks of the 80-mark paper, shared with three other chapters. Expect questions on standard form, checking solutions, graphs and intercepts, slope-intercept form, the nature of a pair, substitution and elimination, and word problems on costs, ages and numbers.
