At least one variable must remain.
a and b must not both be zero.
Class 9 · Maths · Chapter 4 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Linear Equations in Two Variables
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1. Read — Form ax + by + c = 0 · Exercise 4.1 · solutions · Exercise 4.2 · graph · Exercise 4.3 · parallels to the axes · Exercise 4.4, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
3. Mastery ★ — all of that lesson correct. Redo the wrong ones
4. The memory figure shows the line x + y = 7 passing through the plotted points (0, 7) and (7, 0).
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रूप ax + by + c = 0 · Textbook 4.2 · Exercise 4.1
The equation ax + by + c = 0 is a linear equation in two variables. a, b and c are real numbers. The condition is that a and b are not both zero. The power of x is 1 and the power of y is 1. x², y² or xy do not make the line of this chapter.
Write 2x + 3y = 12 as 2x + 3y − 12 = 0. Then a = 2, b = 3 and c = −12. The sign travels with the equation. Writing +12 would change the equation. In two variables, x = −5 is written x + 0·y + 5 = 0.
Question: Write y = 2 as ax + by + c = 0 and give a, b and c.
Answer: Write y − 2 = 0 as 0·x + 1·y − 2 = 0. So a = 0, b = 1 and c = −2. b is not zero, so the ban on both being zero is not broken.
When you take the sign of c, bring the equation to zero. In 2x + 3y = 12, c is −12, not +12.
a = 0 is allowed if b is not zero. y = 2 is linear. 0·x + 0·y + 5 = 0 is not linear.
At least one variable must remain.
a and b must not both be zero.
False — the power of x is 2. In a linear equation every variable has power 1.
12 comes to the left as a negative.
c = −12. The equation is 2x + 3y − 12 = 0.
The coefficient of y is 0.
x + 0·y + 5 = 0.
0·x + 3y − 4 = 0. a = 0, b = 3, c = −4. b is not zero, so a and b are not both zero. It is a linear equation in two variables.
हल और अनंत युग्म · Textbook 4.3 · Exercise 4.2
An ordered pair (x, y) that makes the equation true is a solution. In x + 2y = 6, put x = 2. Then 2 + 2y = 6, so y = 2. The pair (2, 2) is one solution. Put x = 0 and y = 3, the pair (0, 3). Put x = 4 and y = 1.
There are infinitely many such solutions. One variable is free and the other follows from it. Two different equations can pin one pair in a later class. In this chapter there is one equation.
Question: Find one solution of x + 2y = 6 when x = 6. Take 1 unit = 1 cm on the axes.
Equation: x + 2y = 6
Substitution: 6 + 2y = 6, so 2y = 0 and y = 0. The solution is (6, 0).
Unit: the point is on the x-axis, 6 cm to the right of the origin.
When four solutions are asked, give four different pairs. Writing one pair four times is not an example of infinitely many solutions.
Do not turn (2, 2) into the single number (2 + 2). A solution is a pair.
2 + 2·2 = 6.
(2, 2). 2 + 4 = 6.
False — (0, 3), (2, 2), (4, 1) and (6, 0) are all solutions. There are more.
2y = 6.
y = 3. The pair is (0, 3).
Put x = 0, then y = 7, so (0, 7) is a solution. Put x = 7, then y = 0, so (7, 0) is a solution. Both make the equation true, and they are two of infinitely many.
ग्राफ एक सरल रेखा है · Textbook 4.4 · Exercise 4.3
The solutions (0, 3), (2, 2), (4, 1) and (6, 0) of x + 2y = 6 sit on one straight line on the graph. That is the graph of every linear equation in two variables.
Two solutions are enough to draw the line. Keep a third solution for a check. If it lands on the line, the working is right. Write the scale on both axes.
Question: Through which two points would you draw the graph of x + y = 7? 1 unit = 1 cm.
Equation: x + y = 7
Substitution: when x = 0, y = 7, the point (0, 7). When y = 0, x = 7, the point (7, 0).
Unit: these intercepts are 7 cm on the y-axis and 7 cm on the x-axis. Join them.
Leaving the points unjoined is half an answer. Joining them with a straight line is the graph.
A curve, a parabola or a broken line is not the graph of this chapter. Power 1 means a line.
Infinitely many solutions lie on one line.
A straight line.
False — two solutions are enough. A third is good for a check, not compulsory.
y = 7.
(0, 7). Another easy point is (7, 0).
0 + 2·3 = 6.
(0, 3). At (3, 0), 3 is not 6.
(0, 3) and (6, 0) are two solutions. Plot them and join them with a straight line. That is the graph. For a check, (2, 2) should also lie on the line.
True — both pairs make the equation true.
रेखा पर बिंदु हल है · Textbook 4.4 · solution and point together
If a point lies on the graph, its coordinates are a solution of the equation. If a pair is a solution, the point lies on the line. A point off the line is not a solution. So a check means putting the coordinates into the equation.
Infinitely many lines can pass through one point. The point (1, 2) lies on x + y = 3, on y − x = 1, and also on y = 2x. One point does not fix one equation.
Question: Is (1, 2) a solution of x + y = 4? 1 unit = 1 cm.
Equation: x + y = 4
Substitution: 1 + 2 = 3, and 3 ≠ 4.
Answer: (1, 2) is not a solution. The point is not on the line. The point 1 cm right and 2 cm up from the origin is off this line.
Before writing “the point is on the line”, show whether the sum equals the equation.
The sentence “only one line passes through (1, 2)” is wrong. Give at least two equations in the example.
1 + 2 = 3.
x + y = 3.
False — the solutions and the points on the line are the same collection.
2 − 1 = 1.
1. y − x = 1.
(1, 2) will satisfy many equations.
Infinitely many. One point does not fix the equation alone.
3 + 2·1 = 5, so (3, 1) lies on the line and is a solution. Another equation is x + y = 4, because 3 + 1 = 4. One point can lie on infinitely many lines.
अक्ष और उनके समांतर रेखाएँ · Textbook 4.5 · Exercise 4.4
Every point on the x-axis has ordinate 0, so the equation of the x-axis is y = 0. It can also be written 0·x + 1·y = 0. The equation of the y-axis is x = 0.
x = a is a vertical line, parallel to the y-axis and a units away from it. y = a is a horizontal line, parallel to the x-axis. In x = a the variable y is free, and in y = a the variable x is free. So these too are linear equations with infinitely many solutions.
Question: Solve 2x + 1 = x − 3 and show it on the plane. 1 unit = 1 cm.
Solution: 2x − x = −3 − 1, so x = −4.
Plane: x + 0·y + 4 = 0. This is a line parallel to the y-axis, 4 cm to the left of the origin. Two solutions are (−4, 0) and (−4, 2). On the number line the same solution was the single point x = −4.
Do not write x = 3 as parallel to the x-axis. When x is fixed the line is vertical, so it is parallel to the y-axis.
On the number line x = −4 is a point. On the Cartesian plane it is a line. Name the plane in the answer.
10-second revision
On this axis the abscissa is zero.
x = 0. y = 0 is the x-axis.
True — the ordinate is fixed, so the line is horizontal.
The ordinate is zero.
y = 0.
x is fixed.
A vertical line parallel to the y-axis, 4 units to the left.
False — x = 4 is fixed and y may be any real number.
The graph is a horizontal line parallel to the x-axis, 2 cm below the x-axis. Two solutions are (0, −2) and (3, −2). x may be anything.
मूल से गुजरती रेखा y = mx · Textbook 4.6 · summary
Write y = mx as mx − y = 0. Here c = 0. Put x = 0 and y = 0. So the line passes through the origin. m tells the slope: when x grows by 1 unit, y changes by m units.
The force and acceleration example has the same form, y = kx. Here k is constant. The graph is a straight line through the origin. Change k and the slope changes, but the line still meets the origin.
Question: On y = 2x, write the point when x = 3. 1 unit = 1 cm. Show that the origin is also on the line.
Equation: y = 2x
Substitution: y = 2·3 = 6. The point is (3, 6), 3 cm right and 6 cm up from the origin. When x = 0, y = 0, so (0, 0) is also a solution.
y = 2x + 3 does not pass through the origin. (0, 0) is a solution only when the constant term is zero.
When the slope is asked, write “if x grows by one, how much y grows”. Only m = 2 is not enough if the meaning is missing.
10-second revision
c = 0.
(0, 0). 0 = 3·0.
False — when x = 0, y = 1, not the origin.
Twice.
6. The point is (3, 6).
The graph is a straight line through the origin. y = 4·2 = 8, so the point is (2, 8), 2 cm right and 8 cm up from the origin. A change in k changes the slope.
Pick a type. The 30 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
One variable must remain.
a and b cannot both be zero.
Bring it to zero.
2x + y − 5 = 0, so c = −5.
0 + 2·3 = 6.
(0, 3).
One variable stays free.
Infinitely many ordered pairs.
Join two solutions.
A straight line.
The third is a check.
Two.
2 = 2·1.
y = 2x.
The ordinate is zero.
y = 0.
A vertical line.
The y-axis.
y is fixed.
Horizontal, parallel to the x-axis.
The constant term is zero.
The origin (0, 0).
A point on the number line, a line on the plane.
The line x = −4, parallel to the y-axis.
Look at the power and the product.
xy = 6 is not linear.
True — they are not both zero. The y term remains.
False — a solution is an ordered pair (x, y).
True — and every solution is on the line.
False — x = 0 is the y-axis. The x-axis is y = 0.
True — when x = 0, y = 0.
False — (1, 2) lies on infinitely many equations, such as x + y = 3 and y = 2x.
True — x is fixed and the line is vertical.
False — all the solutions of one equation lie on the same line.
ax.
a is the coefficient of x.
1.
4 + 2y = 6.
Straight.
Not a curve.
0.
The abscissa is zero.
The y-axis.
Vertical.
The x-axis.
y = 0.
7.
0 + 7.
−4.
2x − x = −3 − 1.
x = 0 is the y-axis, y = 0 the x-axis, x = 4 vertical, y = 4 horizontal.
A solution is a pair, the graph is a line, y = mx meets the origin, and both zero is forbidden.
Assertion (A): x + 2y = 6 has infinitely many solutions.
Reason (R): Once one variable is chosen the other follows, and the choices are infinite.
Both are true and R is the correct reason.
Assertion (A): The graph is a straight line.
Reason (R): x = 0 is the equation of the y-axis.
Both are true, but R does not explain why the graph is a line.
Assertion (A): y = 0 is the x-axis.
Reason (R): On the x-axis the abscissa is always zero.
A is true. R is false — on the x-axis the ordinate is zero.
Assertion (A): xy = 9 is a linear equation in two variables.
Reason (R): In the linear form every variable has power 1 and a product of the variables is not a term.
A is false. R is true — this is why xy = 9 is not linear.
Assertion (A): (3, 0) does not lie on x + 2y = 6.
Reason (R): 3 + 2·0 = 3, which is not equal to 6.
Both are true and R is the correct check.
Power 1 and separate terms are linear. x² and xy are not linear.
x = 0 is the y-axis. y = −2 and y = 0 are horizontal. y = 3x passes through the origin.
ax + by + c = 0, where a, b, c are real and a, b are not both zero.
(0, 5) and (5, 0). 0 + 5 = 5 and 5 + 0 = 5.
A straight line.
x = 0 is the y-axis. y = 0 is the x-axis.
Putting x = 0 gives y = 0, so (0, 0) is a solution.
2x + (−1)y + (−4) = 0, so a = 2, b = −1, c = −4. When x = 2, 4 − y = 4, so y = 0. The point (2, 0) is on the x-axis, 2 cm to the right of the origin.
Plot (0, 7) and (7, 0) and join them. Check: when x = 3, y = 4, and (3, 4) should lie on the same line because 3 + 4 = 7.
On the number line it is one point. On the plane x + 0·y + 4 = 0 is a line parallel to the y-axis, 4 units left of the origin. y takes infinitely many values.
The solutions are (0, 3), (2, 2), (4, 1) and (6, 0). They lie on one straight line. Two points are enough to draw the line. At (1, 1), 1 + 2 = 3 ≠ 6, so the point is not on the line.
x + y = 3, y − x = 1 and y = 2x are all true at (1, 2). One point can lie on infinitely many lines. Two solutions of one equation fix that one line.
y = 0 is the x-axis, solution (2, 0). x = 0 is the y-axis, solution (0, 5). x = 3 is parallel to the y-axis, solution (3, 1). y = −2 is parallel to the x-axis, solution (1, −2). y = 2x passes through the origin, solution (1, 2).
This model set is for practice. It is not a question from any year’s annual examination. Annual questions will be added only when a source page is available.
Move terms so the right side is zero.
x + y − 3 = 0. c = −3.
A horizontal line.
The x-axis.
2.
The third is a check.
True — 6 + 0 = 6.
(0, 0) and (1, 3). The graph is a straight line through the origin.
These are competency-based practice questions. They are not copies of a CBSE paper.
If x is fixed the line is upright.
x = 5 is vertical, so it is parallel to the y-axis.
One equation, two variables.
There are infinitely many solutions and they lie on a straight line.
Assertion (A): (2, 2) does not lie on x + 2y = 6.
Reason (R): A pair is a solution only when the coordinates make the equation true.
A is false — 2 + 4 = 6, so the point is on the line. R is true.
False — the third solution is a check of the same line.
y = 2·3 + 10 = 16 litres. When x = 0, y = 10, not the origin. Because the constant term is 10, the line does not pass through the origin.
3 + 5 = 8 and 5 + 3 = 8, so both are solutions. They are two of the infinitely many pairs. Both points lie on the same straight line.
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What you learned
| What | Keep this |
|---|---|
| Standard form | ax + by + c = 0 |
| Solutions | infinitely many pairs |
| Graph | a straight line |
| x = 0 | the y-axis |
| y = 0 | the x-axis |
| y = mx | through the origin |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.