In a linear equation, the rate of change between variables remains constant. Therefore, equal changes in one variable produce equal changes in another variable.
This constant rate creates a straight-line graph.
Chapter 4 · CBSE · Class IX
🗒️ Defifnition
A linear equation is an algebraic equation in which the highest power of the variable is always 1. Such equations represent a straight line when plotted on a graph.
A linear equation may contain one variable, two variables, or more variables. However, in Class IX, we mainly study:
- Linear equations in one variable
- Linear equations in two variables
Important: The degree of a linear equation is always 1.
🔢 Formula
General Form of a Linear Equation
🏷️ Properties
Characteristics of Linear Equations
Characteristics of Linear Equations
Degree
Definition: The highest power of the variable in the equation.
Linear Equation: Degree = 1 (highest exponent is 1)
Contrast: \(x^2 + y = 5\) is quadratic (degree 2)
Linear Equation: Degree = 1 (highest exponent is 1)
Contrast: \(x^2 + y = 5\) is quadratic (degree 2)
Graph
Definition: Visual representation on Cartesian plane.
Linear Equation: Forms a straight line.
Example: \(y = 2x + 1\) → straight line with slope 2
Linear Equation: Forms a straight line.
Example: \(y = 2x + 1\) → straight line with slope 2
Variables
Definition: Unknown quantities represented by letters.
Types:
• One variable: \(2x = 8\) → \(x = 4\)
• Two variables: \(x + y = 10\)
Types:
• One variable: \(2x = 8\) → \(x = 4\)
• Two variables: \(x + y = 10\)
Solutions
Definition: Ordered pairs \((x,y)\) satisfying equation.
Verification: Substitute → LHS = RHS
Examples:
\(x + y = 5\):
• \((1,4)\): \(1+4=5\) ✓
• \((2,3)\): \(2+3=5\) ✓
• \((0,5)\): \(0+5=5\) ✓
Infinite solutions form the straight line.
Verification: Substitute → LHS = RHS
Examples:
\(x + y = 5\):
• \((1,4)\): \(1+4=5\) ✓
• \((2,3)\): \(2+3=5\) ✓
• \((0,5)\): \(0+5=5\) ✓
Infinite solutions form the straight line.
Representation
Definition: Equation showing relationship between quantities.
Real Examples:
• Distance = Speed × Time → \(d = st\)
• Total cost = Rate × Quantity → \(C = rq\)
• Age relation: \(x + y = 35\) (two people)
Applications: Physics formulas, economics, geometry problems.
Real Examples:
• Distance = Speed × Time → \(d = st\)
• Total cost = Rate × Quantity → \(C = rq\)
• Age relation: \(x + y = 35\) (two people)
Applications: Physics formulas, economics, geometry problems.
💡 Concept
Concept Behind Linear Equations
🤔 Did You Know?
Why Does a Linear Equation Form a Straight Line?
🗂️ Types / Category
Types of Linear Equations
Linear Equation in One Variable
General Equation: \( \small ax + b = 0 \)
Example: \(\small 2x + 5 = 11 \)
- Contains only one variable (x)
- Has exactly one solution: \( x = 3 \)
- Graph: Single point on number line
Linear Equation in Two Variables
General Equation: \( \small ax + by + c = 0 \)
Example: \(\small x + y = 7 \)
- Contains two variables (x, y)
- Has infinitely many solutions
- Examples: (1,6), (2,5), (3,4), (0,7)
- Graph: Straight line on Cartesian plane
📐 Derivation
Derivation of Standard Form
Consider a relationship between two variables:
\[\small
y=2x+3
\]
Move all terms to one side:
\[\small 2x-y+3=0\]
This becomes the standard form:
\[\small
ax+by+c=0
\]
Therefore:
- \(\small a=2\)
- \(\small b=-1\)
- \(\small c=3\)
✏️ Example
Determine whether the equation \(3x+5=11\) is linear or not.
If the highest power of variable is 1, then the equation is linear.
-
1Identify variable power
-
2Check highest exponent
-
3Conclude type of equation
Given equation:
\[\small
3x+5=11
\]
Highest power of \(x\) is 1.
Therefore, it is a linear equation.
✏️ Example
Find three solutions of:\[\small x+y=5\]
Substitute different values of one variable and calculate the other variable.
| \(x\) | \(y\) | Verification |
|---|---|---|
| 1 | 4 | \(1+4=5\) |
| 2 | 3 | \(2+3=5\) |
| 5 | 0 | \(5+0=5\) |
🌟 Importance
Importance for CBSE Board Aspirants
⚡ Exam Tip
❌ Common Mistakes
| Mistake | Correct Approach |
|---|---|
| Writing \(x^2+y=5\) as linear equation | It is not linear because power of \(x\) is 2. |
| Interchanging coordinates | Always write coordinates as \((x,y)\). |
| Incorrect graph scale | Maintain equal spacing on axes. |
| Forgetting axis labels | Always label \(x\)-axis and \(y\)-axis. |
📋 Case Study
A cab service charges a fixed amount of ₹50 plus ₹12 per kilometre travelled.
Let distance travelled be \(x\) km and total fare be \(y\).
Question
Form a linear equation representing the situation.
Solution
Fixed charge = ₹50
Charge per km = ₹12
Total fare:
\[\small
y=12x+50
\]
Therefore, the required linear equation is:
\[\small
12x-y+50=0
\]
📝 Summary
📘 Definition
🗂️ Types / Category
Standard Forms of Linear Equations
Linear Equation in One Variable
\[ \small ax + b = 0 \]
where \( a \neq 0 \), \( b \) are real constants, \( x \) is variable
\( 2x + 5 = 11 \) → \( x = 3 \) (unique solution)
- One solution on number line
- Graph: Single point
Linear Equation in Two Variables
\[ \small ax + by + c = 0 \]
where \( a, b \) not both zero
\( x + y = 7 \) → Solutions: (1,6), (2,5), (3,4)...
- Infinitely many solutions
- Graph: Straight line on Cartesian plane
Linear Equation in \( n \) Variables
\[ \small a_1x_1 + a_2x_2 + \cdots + a_nx_n + b = 0 \]
Linear combination of \( n \) variables = constant
\( 2x + 3y - z = 5 \) (3 variables)
- At least one coefficient \( a_i \neq 0 \)
- Forms hyperplane in n-dimensional space
🗒️ Components Of A Linear Equation
| Component | Description |
|---|---|
| Variable | A symbol representing an unknown quantity such as \(x,\;y,\;z\). |
| Coefficient | A numerical value multiplied with the variable, such as 5 in \(5x\). |
| Constant | A fixed numerical term without any variable. |
| Degree | The highest exponent of the variable, which is always 1 in a linear equation. |
✏️ Example
Examples of Linear Equations
Examples in One Variable
- \(2x+5=11\)
- \(7y-9=0\)
- \(4a+3=15\)
Examples in Two Variables
- \(x+y=8\)
- \(2x-3y+7=0\)
- \(5p+2q=12\)
Equations Which Are Not Linear
| Equation | Reason |
|---|---|
| \[x^2+3x+1=0\] | Power of variable is 2. |
| \[xy+5=0\] | Product of variables is present. |
| \[\sqrt{x}+2=0\] | Variable is inside square root. |
| \[\frac{1}{x}+4=0\] | Variable appears in denominator. |
🎨 SVG Diagram
Graphical Meaning of Linear Equations
Every linear equation in two variables represents a straight line on the Cartesian plane. Each point lying on the line satisfies the equation.
⚡ Exam Tip
📌 Note
Important Characteristics
💡 Concept
Concept Behind Straight Line Graph
🖼️ Figure
Graphical Representation
Graphical Representation of \(\small y=5x+3\)
👁️ Observation
Quick Understanding Table
✏️ Example
Determine whether the following equation is linear or not:\[x^2+3y=7\]
A linear equation must have variables raised only to power 1.
-
1Check powers of variables
-
2Identify whether any variable has exponent greater than 1
-
3Conclude the nature of equation
In the equation:
\[\small
x^2+3y=7
\]
The variable \(x\) has exponent 2.
Therefore, the equation is not linear.
🌟 Significance
Real-Life Significance of Linear Equations
Economics
Linear equations are used to calculate profit, loss, cost, and revenue relationships.
Physics
Equations of motion and uniform speed relationships often use linear equations.
Business
Businesses use linear models for budgeting and predicting expenses.
Daily Life
Taxi fare, mobile recharge, and shopping bills commonly follow linear relationships.
⚡ Exam Tip
❌ Common Mistakes
- Considering equations containing \(x^2\) or \(xy\) as linear equations.
- Ignoring the degree of variables while identifying linear equations.
- Confusing straight line graphs with curves.
- Writing incorrect ordered pairs while graph plotting.
- Forgetting that variables in denominator make equations non-linear.
📖 Introduction
🗂️ Types / Category
Forms of Representation
Standard Form
-
Linear Equation in One Variable:
\[\small ax+b=0\] -
Linear Equation in Two Variables:
\[\small ax+by+c=0\] -
Linear Equation in Three Variables:
\[\small ax+by+cz+d=0\]
where \(\small a,\;b,\;c,\;d\) are constants and variables have degree 1.
Example
\(\small 2x+3y-6=0\)
Slope-Intercept Form
The slope-intercept form is:
\[\small y=mx+c\]
where:
- \(m\) represents the slope of the line
- \(c\) represents the y-intercept
Meaning of Slope: Slope tells how rapidly the line rises or falls.
Point-Slope Form
The point-slope form is useful when:
- Slope of the line is known
- One point on the line is known
The formula is:
\[\small y-y_1=m(x-x_1)\]
where:
- \((x_1,y_1)\) is a known point
- \(m\) is the slope
📊 Comparison Table
| Form | Equation | Main Use |
|---|---|---|
| Standard Form | \(\small ax+by+c=0\) | General representation |
| Slope-Intercept Form | \(\small y=mx+c\) | Easy graph plotting |
| Point-Slope Form | \(\small y-y_1=m(x-x_1)\) | When slope and one point are known |
✏️ Example
Convert the equation \(\small 2x+y-5=0\) into slope-intercept form.
-
1Move variable terms appropriately
-
2Make \(y\) the subject
-
3Compare with \(y=mx+c\)
\[\small
2x+y-5=0
\]
Transposing terms:
\[\small
y=-2x+5
\]
Hence, slope:
\[\small
m=-2
\]
and y-intercept:
\[\small
c=5
\]
⚡ Exam Tip
❌ Common Mistakes
- Writing incorrect sign while transposing terms.
- Confusing slope with intercept.
- Forgetting brackets in point-slope form.
- Not simplifying the final equation properly.
❓ Question
Step-by-Step Solving for \(5x-9=-3x+19\)
📖 Introduction
🗺️ Roadmap
Roadmap of Solution
- Move all variable terms to one side.
- Move all constants to the opposite side
- Simplify the equation
- Divide by the coefficient of the variable
- Verify the solution
🧩 Solution
-
Add \(3x\) to both sides:
\[ \begin{aligned} 5x+3x-9&=19\\ 8x-9&=19 \end{aligned} \]
-
Move Constants to the Other Side
Add 9 to both sides:
-
\[
\begin{aligned}
8x&=19+9\\
8x&=28
\end{aligned}
\]
Divide by the Coefficient
-
Divide both sides by 8:
\[ \begin{aligned} x&=\frac{28}{8}\\ x&=\frac{7}{2}\\ x&=3.5 \end{aligned} \]
Verification of Answer
Substitute \(x=3.5\) into the original equation:
\[\small
5x-9=-3x+19
\]
Left-hand side:
\[\small
\begin{aligned}
5(3.5)-9\\
17.5-9\\
8.5
\end{aligned}
\]
Right-hand side:
\[\small
\begin{aligned}
-3(3.5)+19\\
-10.5+19\\
8.5
\end{aligned}
\]
Since both sides are equal, the solution is correct.
❌ Common Mistakes
- Forgetting to change sign while transposing terms.
- Incorrect addition of constants.
- Dividing only one side of the equation instead of both sides.
- Skipping verification of the final answer.
- Writing decimal answers incorrectly.
⚡ Exam Tip
📖 Introduction
📌 Note
General Form of a Linear Equation
🧩 Solution
Finding the Solution
To solve the equation:
\[\small
ax+b=0
\]
subtract \(b\) from both sides:
\[\small
ax=-b
\]
divide both sides by \(a\):
\[\small
x=-\frac{b}{a}
\]
🔬 Proof
Proof that Substitution Always Works
-
Step 1: Definition of Solution
A value is called a solution if substituting it into the equation makes the equation true.Substitute:
\[\small x=-\frac{b}{a} \]into:
\[\small ax+b=0 \] -
Step 2: Perform the Substitution
\[\small a\left(-\frac{b}{a}\right)+b=0 \]
-
Step 3: Simplify the Expression
\[\small -b+b=0 \]\[\small 0=0 \]
Since the equality is true, the substituted value satisfies the equation.
-
Step 4: Conclusion
Therefore:\[\small x=-\frac{b}{a} \]is always the correct solution of:\[\small ax+b=0 \]
🤔 Did You Know?
Why Substitution Always Works
Linear equations are built using:
- Addition
- Subtraction
- Multiplication
- Division
These operations follow the properties of equality. Therefore, every algebraic step preserves equality, making substitution mathematically valid.
Substitution checks whether the obtained value truly balances both sides of the equation.
ℹ️ Information
Mathematical Meaning of Substitution
Substitution confirms that the proposed value makes the equation balanced and true.
Since linear equations involve only first-degree terms, inverse operations always exist, ensuring that the substitution process remains valid and consistent.
📖 Introduction
📌 Note
General Form
🗂️ Types / Category
Main Methods of Solution
Substitution Method
In this method, one variable is expressed in terms of the other variable and substituted into
the second equation.
When to Use
When to Use
- When one equation can be simplified easily
- When coefficient of a variable is 1
Elimination Method
In this method, equations are added or subtracted to eliminate one variable.
When to Use
When to Use
- When coefficients are equal or can easily be made equal
- Useful for fast calculations
Graphical Method
In this method, graphs of both equations are drawn on the Cartesian plane. The intersection
point gives the solution.
Important Concept
Important Concept
- Intersecting lines give one unique solution
- Parallel lines give no solution
- Coincident lines give infinitely many solutions
Cross-Multiplication Method
This method uses determinant-like formulas to directly calculate values of variables.
Formula \[\small \frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1} \] When to Use
Formula \[\small \frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1} \] When to Use
- Useful for direct calculation
- Helpful in competitive examinations
📖 Introduction
🔄 Process
Step-by-Step Procedure
-
1Rearrange one equation to make one variable the subject.
-
2Substitute this expression into the other equation.
-
3Solve the resulting equation to obtain one variable.
-
4Substitute the obtained value back into either equation to find the second variable.
-
5Verify the solution in both equations.
✏️ Example
Solve the simultaneous equations:
\[\small 3x-4y=0\tag{1}\]
\[\small 9x-8y=12\tag{2}\]
-
1Express one variable in terms of the other
-
2Substitute into the second equation
-
3Solve for one variable
-
4Substitute back to find the remaining variable
-
Rearrange Equation (1)
From \[\small 3x-4y=0\]
-
Add \(\small 4y\) to both sides:
\[\small 3x=4y\]
-
Divide both sides by 3:
\[\small x=\frac{4y}{3}\]
-
Substitute value of \(\small x\) into Equation (2)
\[ \require{cancel} \begin{aligned} 9x-8y&=12\\ 9\left(\frac{4y}{3}\right)-8y&=12\\ \frac{\cancelto{3}{9}\times4y}{\cancelto{1}{3}}-8y&=12\\ 12y-8y&=12\\ 4y&=12\\ y&=\frac{\cancelto{3}{12}}{\cancelto{1}{4}}\\ y&=3 \end{aligned} \]
-
Find the Value of \(x\)
Substitute \(\small y=3\) into \(\small x=\frac{4y}{3}\)
-
\[
\begin{aligned}
x&=\frac{4\times3}{3}\\
x&=4
\end{aligned}
\]
Therefore, the solution is:
\[\small
x=4,\quad y=3
\]
Verification of Solution
Substitute:
\[\small
x=4,\quad y=3
\]
into Equation (1):
\[\small
\begin{aligned}
3x-4y&=0\\
3(4)-4(3)&=0\\
12-12&=0
\end{aligned}
\]
into Equation (2):
\[\small
\begin{aligned}
9x-8y&=12\\
9(4)-8(3)&=12\\
36-24&=12\\
12&=12
\end{aligned}
\]
Both equations are satisfied. Hence, the solution is verified.
✅ Advantages
Advantages of Substitution Method
- Simple and systematic method
- Useful when one variable has coefficient 1
- Reduces two equations into one equation
- Easy to understand conceptually
❌ Common Mistakes
- Forgetting brackets while substitution.
- Making sign errors during simplification.
- Incorrect cancellation in fractions.
- Substituting into the wrong equation.
- Forgetting to verify the final answer.
⚡ Exam Tip
📖 Introduction
🔄 Process
Step-by-Step Procedure
-
1Multiply one or both equations so that coefficients of one variable become equal.
-
2Add or subtract the equations to eliminate that variable.
-
3Solve the resulting equation for the remaining variable.
-
4Substitute the obtained value into either original equation to find the second variable.
-
5Verify the final solution in both equations.
✏️ Example
Solve the simultaneous equations:
\[\small 4a+5b=12 \tag{1}\]
\[\small 3a-5b=9 \tag{2}\]
-
1Observe coefficients of variables
-
2Eliminate one variable
-
3Solve for remaining variable
-
4Substitute back to find second variable
-
Eliminate Variable
\[\small b\]
-
Since coefficients of \(b\) are already equal and opposite
\[\small +5b \quad \text{and} \quad -5b\]
-
Add both equations:
\[\small\begin{aligned} 4a+5b+3a-5b&=12+9\\ 4a+3a+5b-5b&=21\\ 7a&=21\\ a&=3 \end{aligned} \]
-
Substitute
\[\small a=3\] into Equation (1):
-
\[\small \begin{aligned} 4a+5b&=12\\ 4\times3+5b&=12\\ 12+5b&=12\\ 5b&=12-12\\ 5b&=0\\ b&=0 \end{aligned} \]
Therefore, the solution is:
\[\small
a=3,\quad b=0
\]
Verification
Substitute:
\[\small
a=3,\quad b=0
\]
into Equation (1):
\[\small
\begin{aligned}
4a+5b&=12\\
4(3)+5(0)&=12\\
12+0&=12
\end{aligned}
\]
into Equation (2):
\[\small
\begin{aligned}
3a-5b&=9\\
3(3)-5(0)&=9\\
9-0&=9
\end{aligned}
\]
Both equations are satisfied. Hence, the obtained solution is correct.
🤔 Did You Know?
Why the Elimination Method Works
When equal and opposite coefficients are added together, the variable cancels out:
\[\small
+5b-5b=0
\]
This leaves only one variable in the equation, making it easier to solve.
Elimination reduces two-variable equations into a simpler one-variable equation.
✅ Advantages
Advantages of Elimination Method
- Faster for equations with suitable coefficients
- Reduces calculation complexity
- Avoids complicated substitutions
- Widely used in algebra and higher mathematics
❌ Common Mistakes
- Forgetting signs while adding or subtracting equations.
- Eliminating the wrong variable accidentally.
- Making arithmetic mistakes during simplification.
- Not substituting the obtained value correctly.
- Forgetting verification of the final answer.
⚡ Exam Tip
📖 Introduction
🔄 Process
Step-by-Step Procedure
-
1Convert both equations into slope-intercept form: \[\small y=mx+c\]
-
2Find at least two points for each equation.
-
3Plot the points carefully on the Cartesian plane.
-
4Draw straight lines passing through the plotted points.
-
5Identify the intersection point of the two lines.
✏️ Example
Solve graphically:
\[\small x+y=5 \tag{1}\]
\[\small 2x-y=4 \tag{2}\]
-
Convert into \(\small y=mx+c\) Form
-
Rearranging Equation (1):
\[\small \begin{align} x+y&=5\\ y&=5-x \end{align} \tag{3} \]
-
Rearranging Equation (2):
\[ \begin{align} 2x-y&=4\\ y&=2x-4 \end{align} \tag{4} \]
-
Plot the Graphs
-
Plot the graphs of Equations (3) and (4) on the Cartesian plane.
Graphical Method for Solving Simultaneous Linear Equations
Interpretaion
Identify the Intersection Point.
The two lines intersect at: \[\small (3,2)\]
The two lines intersect at: \[\small (3,2)\]
Therefore, the solution is:
\[\small
x=3,\quad y=2
\]
💡 Concept
Concept Behind the Graphical Method
✅ Advantages
Advantages of Graphical Method
- Provides visual understanding of equations.
- Helps identify number of solutions easily.
- Useful in coordinate geometry concepts.
- Makes interpretation of linear relationships easier.
⚠️ Limitations
Limitations of Graphical Method
- Exact values may not always be obtained from graphs.
- Time-consuming compared to algebraic methods.
- Drawing accurate graphs requires careful scaling.
- Difficult for equations involving large numbers.
❌ Common Mistakes
- Plotting incorrect coordinates.
- Using unequal scales on axes.
- Drawing curved lines instead of straight lines.
- Forgetting to label axes properly.
- Misreading the intersection point.
⚡ Exam Tip
📖 Introduction
📌 Note
General Form of Equations
🔢 Formula
Cross Multiplication Formula
🔄 Process
Step-by-Step Procedure
-
1Arrange both equations in standard form: \[\small ax+by+c=0 \]
-
2Identify coefficients carefully.
-
3Apply the cross multiplication formula.
-
4Simplify the ratios to find values of \(\small x\) and \(\small y\).
-
5Verify the solution in the original equations.
✅ Advantages
Advantages of Cross Multiplication Method
- Fast and direct method
- Reduces lengthy algebraic calculations
- Useful in competitive examinations
- Helpful when equations are already in standard form
⚠️ Limitations
Limitations of the Method
- Applicable only to two-variable linear equations.
- Mistakes in coefficients lead to incorrect answers.
- Large coefficients may increase calculation complexity.
- Requires careful handling of signs.
❌ Common Mistakes
- Forgetting to write equations in standard form.
- Interchanging coefficients incorrectly.
- Ignoring negative signs during cross multiplication.
- Using wrong denominator expression.
- Forgetting to verify the final solution.
⚡ Exam Tip
🗺️ Overview
Overview of Solution Methods
Simultaneous equations can be solved using different algebraic and graphical techniques. Each method has its own advantages depending on the structure of the equations.
Choosing the correct method helps simplify calculations and improves problem-solving speed.
Important: All methods ultimately produce the same solution if calculations are
performed correctly.
📊 Comparison Table
Substitution Method
Elimination Method
Graphical Method
Cross-Multiplication Method
| Method | Key Idea | When to Use | Example Result |
|---|---|---|---|
| Substitution Method | Replace one variable using another equation | Useful when one coefficient is 1 or easily isolated | \(\small x=4,\;y=3\) |
| Elimination Method | Add or subtract equations to eliminate one variable | Best when coefficients are equal or can be made equal | \(\small a=3,\;b=0\) |
| Graphical Method | Plot graphs and identify the intersection point | Useful for visual understanding | \(\small (3,2)\) |
| Cross-Multiplication Method | Apply direct determinant-like ratios | Fast for two-variable linear systems | Calculated algebraically |
🤔 Did You Know?
How to Choose the Correct Method
Choose Substitution Method
- When a variable can be isolated easily
- When coefficients are simple
Choose Elimination Method
- When coefficients are already equal
- When quick cancellation is possible
Choose Graphical Method
- For visual interpretation
- To understand geometric meaning
Choose Cross-Multiplication Method
- For direct formula-based solving
- Useful in competitive exams
🌟 Importance
❌ Common Mistakes
- Choosing an unnecessarily lengthy method.
- Making sign mistakes during elimination.
- Incorrect plotting in graphical method.
- Using wrong coefficients in cross multiplication.
- Forgetting verification of solutions.
⚡ Exam Tip
❓ Question
Solve the following system of equations using the method of elimination by substitution:
\[\small x+y=7\]
\[\small 3x-2y=11\]
💡 Concept
🗺️ Roadmap
- Express one variable from the first equation
- Substitute into the second equation
- Solve for one variable
- Substitute back to find the second variable
- Verify the final answer
🧩 Solution
-
Write the Given Equations
\[\small \begin{aligned} x+y&=7\\ 3x-2y&=11 \end{aligned} \]
-
Express \(\small x\) in Terms of \(\small y\)
-
From the first equation:
\[\small x+y=7\]
-
Subtract \(\small y\) from both sides:
\[\small x=7-y\]
-
Substitute
\[x=7-y\]
-
into
\[\small 3x-2y=11\]
-
\[\small \begin{aligned} 3(7-y)-2y&=11\\ 21-3y-2y&=11 \end{aligned} \]
-
Combine Like Terms
\[\small \begin{aligned} -5y&=11-21\\ -5y&=-10\\ y&=2 \end{aligned} \]
-
Find the Value of \(\small x\)
-
Substitute
\[\small y=2\]
-
into
\[\small x=7-y\]
-
\[\small \begin{aligned} x&=7-2\\ x&=5 \end{aligned} \]
Therefore, the solution is:
\[
x=5,\quad y=2
\]
Verification
Substitute:
\[\small
x=5,\quad y=2
\]
into the first equation:
\[\small
\begin{aligned}
x+y&=7\\
5+2&=7\\
7&=7
\end{aligned}
\]
into the second equation:
\[\small
\begin{aligned}
3x-2y&=11\\
3(5)-2(2)&=11\\
15-4&=11\\
11&=11
\end{aligned}
\]
Both equations are satisfied. Hence, the solution is correct.
❌ Common Mistakes
- Forgetting brackets during substitution.
- Making sign mistakes while simplifying expressions.
- Incorrectly combining like terms.
- Forgetting to substitute the obtained value back.
- Skipping verification of the final answer.
⚡ Exam Tip
❓ Question
Solve using the elimination by substitution method:
\[\small \frac{x+7}{5}-\frac{2x-y}{4}=3y-5\]
\[\small \frac{4x-3}{6}+\frac{5y-7}{2}=18-5x\]
💡 Concept
🗺️ Roadmap
- Remove fractions using LCM
- Simplify both equations
- Express one variable in terms of the other
- Substitute into the second equation
- Solve for both variables
🧩 Solution
-
Simplify the First Equation
\[\small \frac{x+7}{5}-\frac{2x-y}{4}=3y-5\]
-
Take LCM of 5 and 4:
\[\small 20\]
-
\[\small \begin{aligned} \frac{4(x+7)-5(2x-y)}{20}&=3y-5\\ \frac{4x+28-10x+5y}{20}&=3y-5 \end{aligned} \]
-
Multiply both sides by 20:
\[\small \begin{aligned} 4x-10x+5y+28&=60y-100\\ -6x+5y-60y&=-100-28\\ -6x-55y&=-128 \end{aligned} \]
-
Express \(\small x\) in terms of \(\small y\):
\[\small \begin{aligned} -6x&=55y-128\\ x&=\frac{128-55y}{6} \end{aligned} \]
-
Simplify the Second Equation
\[\small \frac{4x-3}{6}+\frac{5y-7}{2}=18-5x\]
-
Take LCM of 6 and 2:
\[\small 6\]
-
\[\small \begin{aligned} \frac{4x-3+15y-21}{6}&=18-5x\\ 4x+15y-24&=108-30x \end{aligned} \]
-
Collect like terms:
\[\small \begin{aligned} 4x+30x+15y&=108+24\\ 34x+15y&=132 \end{aligned} \]
-
Substitute the Value of \(\small x\)<
-
Substitute
\[\small x=\frac{128-55y}{6}\]
-
into
\[\small 34x+15y=132\]
-
\[\small \begin{aligned} 34\left(\frac{128-55y}{6}\right)+15y&=132\\ \frac{17}{3}(128-55y)+15y&=132 \end{aligned} \]
-
Multiply both sides by 3
\[\small \begin{aligned} 17(128-55y)+45y&=396\\ 2176-935y+45y&=396\\ 2176-890y&=396 \end{aligned} \]
-
Solve for \(\small y\)
\[\small \begin{aligned} 890y&=2176-396\\ 890y&=1780\\ y&=\frac{1780}{890}\\ y&=2 \end{aligned} \]
-
Find the Value of \(\small x\)
-
Substitute
\[\small y=2\]
-
into
\[\small x=\frac{128-55y}{6}\]
-
\[\small \begin{aligned} x&=\frac{128-55(2)}{6}\\ x&=\frac{128-110}{6}\\ x&=\frac{18}{6}\\ x&=3 \end{aligned} \]
Therefore, the solution is:
\[
x=3,\quad y=2
\]
Verification
Substitute:
\[\small
x=3,\quad y=2
\]
into the first equation:
\[\small
\begin{aligned}
\frac{3+7}{5}-\frac{2(3)-2}{4}&=3(2)-5\\
\frac{10}{5}-\frac{4}{4}&=6-5\\
2-1&=1
\end{aligned}
\]
into the second equation:
\[\small
\begin{aligned}
\frac{4(3)-3}{6}+\frac{5(2)-7}{2}&=18-5(3)\\
\frac{9}{6}+\frac{3}{2}&=3\\
\frac{3}{2}+\frac{3}{2}&=3\\
3&=3
\end{aligned}
\]
Both equations are satisfied. Hence, the obtained solution is correct.
🧠 Remember
Important Learning Points
❌ Common Mistakes
- Incorrect LCM while removing fractions.
- Forgetting brackets during multiplication.
- Sign mistakes while simplifying equations.
- Calculation errors in fractions.
- Skipping verification of the solution.
⚡ Exam Tip
❓ Question
Solve using the method of elimination by equating coefficients:
\[\small 3x-4y=10\]
\[\small 5x-3y=24\]
💡 Concept
🗺️ Roadmap
- Write both equations clearly
- Equalize coefficients of one variable
- Eliminate the variable by subtraction
- Solve for the remaining variable
- Substitute back to obtain the second variable
🧩 Solution
-
Write the Given Equations
\[\small \begin{align} 3x-4y&=10 \tag{1}\\ 5x-3y&=24 \tag{2} \end{align} \]
-
Equalize Coefficients of \(\small x\)
-
Multiply Equation (1) by 5 and Equation (2) by 3
\[\small \begin{aligned} 3x-4y&=10 \quad\times5\\ 5x-3y&=24 \quad\times3 \end{aligned} \]
-
\[\small \begin{aligned} 15x-20y&=50\\ 15x-9y&=72 \end{aligned} \]
-
Eliminate Variable \(\small x\)
-
Subtract the second equation from the first:
\[\small \begin{aligned} 15x-20y&=50\\ -(15x-9y&=72)\\ \hline -11y&=-22 \end{aligned} \]
-
Divide both sides by \(-11\):
\[\small \begin{aligned} y&=\frac{-22}{-11}\\ y&=2 \end{aligned} \]
-
Find the Value of \(\small x\)
-
Substitute
\[\small y=2\]
-
into Equation (1)
\[ \begin{aligned} 3x-4(2)&=10\\ 3x-8&=10\\ 3x&=18\\ x&=\frac{18}{3}\\ x&=6 \end{aligned} \]
Therefore, the solution is:
\[\small
\boxed{x=6,\quad y=2}
\]
Verification
Substitute:
\[\small
x=6,\quad y=2
\]
into Equation (1):
\[\small
\begin{aligned}
3x-4y&=10\\
3(6)-4(2)&=10\\
18-8&=10\\
10&=10
\end{aligned}
\]
into Equation (2):
\[\small
\begin{aligned}
5x-3y&=24\\
5(6)-3(2)&=24\\
30-6&=24\\
24&=24
\end{aligned}
\]
Both equations are satisfied. Hence, the obtained solution is correct.
🧠 Remember
Important Learning Points
❌ Common Mistakes
- Forgetting to multiply every term of the equation.
- Making sign mistakes during subtraction.
- Incorrect simplification after elimination.
- Substituting incorrect values into equations.
- Forgetting verification of the final answer.
⚡ Exam Tip
❓ Question
Solve by the method of cross multiplication:
\[\small 3x+y=13\]
\[\small x-3y+9=0\]
💡 Concept
🗺️ Roadmap
- Convert equations into standard form
- Identify coefficients carefully
- Apply the cross multiplication formula
- Solve for \(\small x\) and \(\small y\)
- Verify the obtained solution
🔢 Formula
Cross Multiplication Formula
🧩 Solution
- Write Equations in Standard Form
- \[\small \begin{aligned} 3x+y-13&=0\\ x-3y+9&=0 \end{aligned} \]
- Identify Coefficients
-
Coefficient Equation 1 Equation 2 \(a\) \(a_1=3\) \(a_2=1\) \(b\) \(b_1=1\) \(b_2=-3\) \(c\) \(c_1=-13\) \(c_2=9\) - Apply the Formula
- \[\small \frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1} \]
- Substitute the values
- \[\small \begin{aligned} \frac{x}{(1)(9)-(-3)(-13)} &= \frac{y}{(-13)(1)-(9)(3)} = \frac{1}{(3)(-3)-(1)(1)} \end{aligned} \]
- \[\small \begin{aligned} \frac{x}{9-39} &= \frac{y}{-13-27} = \frac{1}{-9-1} \end{aligned} \]
- \[\small \frac{x}{-30} = \frac{y}{-40} = \frac{1}{-10} \]
- Find the Value of \(\small x\)
- \[ \begin{aligned} \frac{x}{-30} &= \frac{1}{-10}\\ x &= \frac{30}{10}\\ x &=3 \end{aligned} \]
- Find the Value of \(\small y\)
-
\[\small
\begin{aligned}
\frac{y}{-40}
&=
\frac{1}{-10}\\
y
&=
\frac{40}{10}\\
y
&=4
\end{aligned}
\]
Therefore, the solution is: \[ \boxed{x=3,\quad y=4} \]
🗒️ Verification
Verification of Answer
Substitute:
\[\small
x=3,\quad y=4
\]
into the first equation:
\[\small
\begin{aligned}
3x+y&=13\\
3(3)+4&=13\\
9+4&=13\\
13&=13
\end{aligned}
\]
into the second equation:
\[\small
\begin{aligned}
x-3y+9&=0\\
3-3(4)+9&=0\\
3-12+9&=0\\
0&=0
\end{aligned}
\]
Both equations are satisfied. Hence, the solution is verified.
🧠 Remember
Important Learning Points
❌ Common Mistakes
- Forgetting to convert equations into standard form.
- Writing incorrect signs for coefficients.
- Mixing numerator and denominator terms.
- Calculation mistakes during simplification.
- Not verifying the final answer.
⚡ Exam Tip
❓ Question
Solve
\[\small \frac{7}{x}+\frac{8}{y}=2\]
\[\small \frac{2}{x}+\frac{13}{y}=22\]
💡 Concept
🗺️ Roadmap
Solution Roadmap
- Remove denominators by cross multiplication
- Form algebraic equations
- Relate the variables
- Substitute into one equation
- Find the values of both variables
🗒️ Soution
- Write the Given Equations
- \[\small \begin{align} \frac{7}{x}+\frac{8}{y}&=2 \tag{A}\\ \frac{2}{x}+\frac{13}{y}&=22 \tag{B} \end{align} \]
- Remove Denominators
- Multiply Equation (A) by \(\small xy\)
- \[\small \begin{align} \frac{7}{x}+\frac{8}{y}&=2\\ 7y+8x&=2xy \tag{1} \end{align} \]
- Multiply Equation (B) by \(\small xy\)
- \[\small \begin{align} \frac{2}{x}+\frac{13}{y}&=22\\ 2y+13x&=22xy \tag{2} \end{align} \]
- Divide Equation (1) by Equation (2)
- \[\small \begin{aligned} \frac{7y+8x}{2y+13x} &= \frac{2xy}{22xy}\\ \frac{7y+8x}{2y+13x} &= \frac{1}{11} \end{aligned} \]
- Cross multiply
- \[\small \begin{aligned} 11(7y+8x)&=2y+13x\\ 77y+88x-2y-13x&=0\\ 75y+75x&=0 \end{aligned} \]
- Divide both sides by 75:
- \[\small \begin{aligned} y+x&=0\\ y&=-x \end{aligned} \]
- Substitute into Equation (A)
- Substitute \(\small y=-x\) into
- \[\small \frac{7}{x}+\frac{8}{y}=2\]
- \[\small \begin{aligned} \frac{7}{x}+\frac{8}{-x}&=2\\ \frac{7-8}{x}&=2\\ \frac{-1}{x}&=2 \end{aligned} \]
- Cross multiply
- \[\small \begin{aligned} 2x&=-1\\ x&=-\frac{1}{2} \end{aligned} \]
- Find the Value of \(\small y\)
- Since
- \[\small y=-x\]
- therefore,
- \[ \begin{aligned} y&=-\left(-\frac{1}{2}\right)\\ y&=\frac{1}{2} \end{aligned} \]
-
Therefore, the solution is: \[ \boxed{x=-\frac{1}{2},\quad y=\frac{1}{2}} \]
🧠 Remember
❌ Common Mistakes
- Forgetting to multiply all terms by \(xy\).
- Making sign errors while simplifying fractions.
- Incorrect division of equations.
- Ignoring restrictions on denominators.
- Skipping verification of final answers.
⚡ Exam Tip
NCERT Class IX · Chapter 4
Linear Equations in
Two Variables
A complete AI-powered learning engine — concepts, formulas, solver, practice, and interactive tools all in one place.
📐 8 Core Concepts
🧮 12 Formulas
📝 30+ Practice Qs
🎯 5 Interactive Modules
📚 Core Concepts
Eight foundational ideas from NCERT Class IX Chapter 4 — explained clearly with examples.
Concept 1
What is a Linear Equation in Two Variables?
An equation of the form ax + by + c = 0, where a, b, c are real numbers and a, b are not both zero, is called a linear equation in two variables x and y.
Standard Form
ax + by + c = 0 where a² + b² ≠ 0
Why "linear"? Because the highest power of each variable is 1. The graph of such an equation is always a straight line.
Examples: 2x + 3y = 6 ✓ | x – y = 0 ✓ | x² + y = 5 ✗ (not linear)
Concept 2
Solutions of a Linear Equation
A pair of values (x₀, y₀) that satisfies ax + by + c = 0 is called a solution. A linear equation in two variables has infinitely many solutions.
Example
For 2x + y = 5:
• x=1 → y=3 → (1,3) ✓
• x=0 → y=5 → (0,5) ✓
• x=2 → y=1 → (2,1) ✓
• x=½ → y=4 → (½,4) ✓
All are valid solutions — and there are infinitely many more!
Key Idea
Every point (x, y) on the graph of the line IS a solution, and every solution corresponds to a point on that line.
The set of all solutions forms the line itself — an infinite collection of ordered pairs.
Concept 3
Graph of a Linear Equation
The graph of ax + by + c = 0 is a straight line. To draw it, we need at least two points (we usually find three for accuracy).
-
Find x-interceptPut y = 0 in the equation and solve for x. This gives the point where the line crosses the x-axis: (x₀, 0).
-
Find y-interceptPut x = 0 in the equation and solve for y. This gives the point where the line crosses the y-axis: (0, y₀).
-
Find a third point (optional but good practice)Choose any convenient x value, find corresponding y, and verify all three points are collinear.
-
Plot and joinPlot the points on the coordinate plane and draw a straight line through them. Extend with arrows on both ends.
Concept 4
Lines Parallel to Axes
Line Parallel to X-Axis
y = k (k is a constant)
The line y = k is horizontal, runs parallel to the x-axis at height k. Every point has y-coordinate = k regardless of x.
Example: y = 3 passes through (0,3), (1,3), (–2,3), etc.
Line Parallel to Y-Axis
x = h (h is a constant)
The line x = h is vertical, runs parallel to the y-axis at h. Every point has x-coordinate = h regardless of y.
Example: x = –2 passes through (–2,0), (–2,1), (–2,5), etc.
Special Cases: y = 0 is the x-axis itself. x = 0 is the y-axis itself. These are the two coordinate axes.
Concept 5
Equation of a Line Through the Origin
If c = 0, the equation becomes ax + by = 0, which always passes through the origin (0, 0).
Line Through Origin
y = mx where m = –a/b (the slope)
Such lines pass through origin and have slope m. Examples: y = 2x, y = –x, y = (3/2)x.
The x-intercept and y-intercept of such a line are both zero — the line passes through (0, 0).
Concept 6
Representing Real-Life Situations
Linear equations model real-world relationships. The key is identifying the two variables and expressing their relationship.
Example 1 (Cost problem):
A shopkeeper sells apples at ₹a each and oranges at ₹b each. If the total cost of x apples and y oranges is ₹100:
ax + by = 100
Example 2 (Age problem):
Sum of ages of father (x) and son (y) is 45 years:
x + y = 45
Example 3 (Distance-Speed):
A train travels at speed s km/h. Distance covered in t hours = d km:
s·t = d → s·t – d = 0
Concept 7
Table of Values Method
To find solutions systematically, create a table by assigning convenient values to one variable and computing the other.
For 3x + 2y = 12:
| x | y = (12 – 3x) / 2 | Point | Verify |
|---|---|---|---|
| 0 | 6 | (0, 6) | 3(0)+2(6)=12 ✓ |
| 2 | 3 | (2, 3) | 3(2)+2(3)=12 ✓ |
| 4 | 0 | (4, 0) | 3(4)+2(0)=12 ✓ |
| -2 | 9 | (-2, 9) | 3(-2)+2(9)=12 ✓ |
Concept 8
Slope and Intercepts
Slope-Intercept Form
y = mx + c
m = slope, c = y-intercept
m = slope, c = y-intercept
Convert ax + by + c = 0 to this form by isolating y:
y = (–a/b)x + (–c/b)
Slope m = –a/b | y-intercept = –c/b
Intercept Form
x/p + y/q = 1
p = x-intercept, q = y-intercept
p = x-intercept, q = y-intercept
If you know x-intercept p and y-intercept q:
Line passes through (p, 0) and (0, q).
Slope = –q/p
🔣 All Formulas at a Glance
Every formula and relationship you need for this chapter — colour-coded and clearly labelled.
Core Forms
1 — General / Standard Form
ax + by + c = 0 (a² + b² ≠ 0; a, b, c ∈ ℝ)
2 — Slope-Intercept Form
y = mx + c (m = slope, c = y-intercept)
3 — Intercept Form
x/a + y/b = 1 (a = x-intercept, b = y-intercept; neither = 0)
4 — Point-Slope Form
y – y₁ = m(x – x₁) (passes through (x₁, y₁) with slope m)
5 — Two-Point Form
(y – y₁)/(y₂ – y₁) = (x – x₁)/(x₂ – x₁) (passes through (x₁,y₁) and (x₂,y₂))
Intercepts & Conversions
6 — x-Intercept
Set y = 0: x = –c/a
7 — y-Intercept
Set x = 0: y = –c/b
8 — Slope from Standard Form
m = –a/b (b ≠ 0)
9 — Slope from Two Points
m = (y₂ – y₁) / (x₂ – x₁) (x₁ ≠ x₂)
Special Lines
10 — Horizontal Line
y = k (slope = 0)
11 — Vertical Line
x = h (slope = undefined)
12 — Through Origin
y = mx (c = 0)
Quick Reference Table
| Form | Equation | Slope | y-intercept | x-intercept |
|---|---|---|---|---|
| Standard | ax + by + c = 0 | –a/b | –c/b | –c/a |
| Slope-Intercept | y = mx + c | m | c | –c/m |
| Intercept | x/p + y/q = 1 | –q/p | q | p |
| Through Origin | y = mx | m | 0 | 0 |
| Horizontal | y = k | 0 | k | none (if k≠0) |
| Vertical | x = h | ∞ | none | h |
🤖 AI Step-by-Step Solver
Enter any linear equation in two variables and get a complete, step-by-step solution with graph.
Equation Input
Enter the equation in the form ax + by + c = 0. Provide the coefficients a, b, c.
✨ Solution Steps
Find a Specific Solution
Given the equation above, check if a specific (x, y) pair is a solution, or find y for a given x.
📝 Practice Questions with Solutions
Concept-wise practice questions with complete step-by-step solutions. All original — not from the textbook.
Set A — Identifying & Writing Equations
Q1. The sum of two numbers is 28 and one number is twice the other. Write this as a linear equation in two variables and find three solutions.
▾
Step-by-Step Solution
- Define VariablesLet the two numbers be x and y.
- Write the equationsSum: x + y = 28. One is twice the other: y = 2x. Combined: x + 2x = 28 → 3x = 28. But as a single linear equation in two variables: x + y = 28 (we use both conditions simultaneously to find solutions).
- Substitute y = 2x into x + y = 28x + 2x = 28 → 3x = 28 → x = 28/3. Then y = 56/3. So the specific pair is (28/3, 56/3).
- Find three solutions of x + y = 28(10, 18), (14, 14), (20, 8) — all satisfy x + y = 28.
Answer: Equation: x + y = 28 | Solutions: (10,18), (14,14), (20,8)
Q2. A taxi charges ₹50 as a fixed base fare plus ₹15 per kilometre. Write the cost C as a linear equation in terms of distance d, and find C when d = 8 km.
▾
Step-by-Step Solution
- Identify VariablesLet C = total cost (₹), d = distance (km).
- Frame the equationC = 50 + 15d → rearranged: 15d – C + 50 = 0
- Substitute d = 8C = 50 + 15(8) = 50 + 120 = ₹170
Equation: C – 15d – 50 = 0 | When d = 8: C = ₹170
Q3. Write the equation 5y = 3x – 7 in standard form ax + by + c = 0 and identify a, b, c.
▾
- Rearrange to Standard Form5y = 3x – 7 → 3x – 5y – 7 = 0
- Read coefficientsComparing with ax + by + c = 0: a = 3, b = –5, c = –7
- Verify a² + b² ≠ 03² + (–5)² = 9 + 25 = 34 ≠ 0 ✓
3x – 5y – 7 = 0 | a = 3, b = –5, c = –7
Set B — Finding & Verifying Solutions
Q4. For the equation 4x – 3y = 12, verify whether (3, 0), (0, –4), and (6, 4) are solutions.
▾
- Test (3, 0)LHS = 4(3) – 3(0) = 12 – 0 = 12 = RHS ✓ → Solution
- Test (0, –4)LHS = 4(0) – 3(–4) = 0 + 12 = 12 = RHS ✓ → Solution
- Test (6, 4)LHS = 4(6) – 3(4) = 24 – 12 = 12 = RHS ✓ → Solution
All three points (3,0), (0,–4), and (6,4) are solutions and lie on the line.
Q5. Find the value of k if (2, k) is a solution of 3x + 5y = 21. Also find the value of y when x = –3.
▾
- Substitute (2, k)3(2) + 5k = 21 → 6 + 5k = 21 → 5k = 15 → k = 3
- Find y when x = –33(–3) + 5y = 21 → –9 + 5y = 21 → 5y = 30 → y = 6
k = 3 | When x = –3, y = 6
Q6. The point (p, p+1) lies on the line 2x – y = 3. Find p and write the actual solution point.
▾
- Substitute into equation2(p) – (p+1) = 3 → 2p – p – 1 = 3 → p – 1 = 3 → p = 4
- Find the pointx = 4, y = 4+1 = 5 → Point is (4, 5)
- Verify2(4) – 5 = 8 – 5 = 3 ✓
p = 4, Solution point = (4, 5)
Set C — Graphing & Intercepts
Q7. Draw the graph of 2x + y = 6. Find the area of the triangle formed by the line and both coordinate axes.
▾
- Find x-intercept (set y=0)2x + 0 = 6 → x = 3 → Point A(3, 0)
- Find y-intercept (set x=0)0 + y = 6 → y = 6 → Point B(0, 6)
- Plot and join A(3,0) and B(0,6)Draw the line extending on both sides.
- Find the triangle areaTriangle formed by the line and axes has vertices at O(0,0), A(3,0), B(0,6).
Area = ½ × base × height = ½ × 3 × 6 = 9 square units
x-intercept: (3,0) | y-intercept: (0,6) | Triangle area = 9 sq. units
Q8. Determine the equation of a line passing through (2, 5) and (4, 9). Write it in standard form.
▾
- Find slope mm = (9–5)/(4–2) = 4/2 = 2
- Use point-slope formy – 5 = 2(x – 2) → y – 5 = 2x – 4 → y = 2x + 1
- Convert to standard form2x – y + 1 = 0 | a=2, b=–1, c=1
y = 2x + 1 or 2x – y + 1 = 0
Q9. The x-intercept of a line is 4 and its y-intercept is –3. Find the equation and the slope of the line.
▾
- Use intercept formx/4 + y/(–3) = 1 → x/4 – y/3 = 1
- Multiply through by 123x – 4y = 12 → standard form: 3x – 4y – 12 = 0
- Find slopem = –a/b = –3/(–4) = 3/4
Equation: 3x – 4y – 12 = 0 | Slope = 3/4
Set D — Word Problems & Applications
Q10. A water tank has 200 litres and is being drained at 8 litres/min. Write a linear equation and find after how many minutes will 40 litres remain.
▾
- Define variablesLet t = time in minutes, W = water remaining in litres.
- Write the equationW = 200 – 8t → Standard form: 8t + W – 200 = 0
- Solve for W = 4040 = 200 – 8t → 8t = 160 → t = 20 minutes
Equation: 8t + W = 200 | After 20 minutes, 40 litres will remain.
Q11. A worker earns ₹x per day for regular hours and ₹y per hour for overtime. In a week, he works 5 regular days and 3 overtime hours earning ₹2700. Write the equation. If x = 450, find y.
▾
- Frame equation5x + 3y = 2700
- Substitute x = 4505(450) + 3y = 2700 → 2250 + 3y = 2700 → 3y = 450 → y = 150
Equation: 5x + 3y = 2700 | y = ₹150 per overtime hour
💡 Tips & Tricks
Smart shortcuts, memory aids, and exam strategies to boost your score.
Exam Shortcuts
Quick Intercept Method: To find x-intercept, cover/replace y with 0 in the equation and solve. To find y-intercept, replace x with 0. You only need these two points to draw any non-vertical, non-horizontal line.
Slope Memory Aid: "Rise over Run" — slope = (change in y)/(change in x). A positive slope goes up-right, a negative slope goes down-right. Zero slope = horizontal, undefined slope = vertical.
Checking Solutions Fast: Always substitute both values back into the original equation. If LHS = RHS, you're right. This takes 10 seconds and saves full marks.
Through Origin Shortcut: If c = 0 in ax + by + c = 0, the line definitely passes through (0, 0). No need to calculate — the origin is always a solution.
Choosing Good Values for Table: Pick x = 0 (gives y-intercept), y = 0 (gives x-intercept), and one more easy value. Avoid fractions if possible for neat graphs — choose x so that the numerator is divisible by the coefficient of y.
Concept Tricks
Converting Between Forms: To go from ax + by + c = 0 to y = mx + c form: isolate y. Move ax and c to the other side, then divide by b. Remember b ≠ 0 for this to work.
Area of Triangle from Line: If a line cuts x-axis at (a,0) and y-axis at (0,b), the triangle with the origin has area = ½|a||b|. This is a very common exam question!
Infinite Solutions Reasoning: A linear equation in two variables has infinitely many solutions because for every real value of x, there is a unique corresponding real value of y. The line extends infinitely in both directions.
Identifying Line Type at a Glance: If only y appears (no x): horizontal line. If only x appears (no y): vertical line. If both appear and c = 0: line through origin. Otherwise: a general oblique line.
Word Problem Strategy
- Read and IdentifyRead the problem twice. Identify the two unknown quantities — these become your variables x and y.
- Write the RelationshipExpress the given conditions mathematically connecting x and y. One condition → one equation.
- StandardiseConvert to ax + by + c = 0. This makes it easier to read off coefficients and intercepts.
- Answer the Specific QuestionThe question may ask for the equation, specific values, or the graph. Answer exactly what is asked.
⚠️ Common Mistakes to Avoid
Mistakes that students commonly make — know them, recognise them, and never lose marks on them.
Mistake 1
Forgetting c can be zero
Student writes: "ax + by + c = 0 is valid only when c ≠ 0"
c can be zero! When c=0, the line passes through the origin. The constraint is a² + b² ≠ 0, meaning a and b cannot both be zero simultaneously.
Mistake 2
Confusing "No Solution" with "Infinitely Many"
Student thinks a single equation in two variables has "no unique solution" and writes "no solution".
A single linear equation in two variables has infinitely many solutions. "No solution" applies only to a pair of inconsistent linear equations (Chapter 3 of Class X).
Mistake 3
Wrong Sign in Standard Form
Given 2x – 3y = 6, student writes a=2, b=3, c=6 forgetting that standard form has "+ c" with c moved to left side.
Rearrange correctly: 2x – 3y – 6 = 0 → a=2, b=–3, c=–6. Always move everything to left side.
Mistake 4
Drawing Graph with Only 2 Points
Student finds only the two intercepts and joins them without verifying a third point or extending the line beyond the intercepts.
Always find a third point to confirm collinearity. Also extend the line beyond the plotted points with arrows on both ends to indicate the line extends to infinity.
Mistake 5
Slope of Horizontal vs Vertical Lines
Student says "x = 3 has slope 0" and "y = 2 has undefined slope."
It's the reverse: y = k (horizontal) has slope = 0. x = h (vertical) has slope = undefined (infinite). Remember: a horizontal line doesn't rise at all → slope zero.
Mistake 6
Incorrect Intercept Reading
For 3x + 2y = 6, student writes x-intercept = 6 and y-intercept = 6 (reading the constant directly).
Set y=0: 3x=6 → x-intercept=2. Set x=0: 2y=6 → y-intercept=3. You must substitute and solve, not just read off the constant.
🎮 Interactive Modules
Five hands-on tools to build deep intuition — explore, play, and understand.
Live Equation Grapher
Enter any equation ax + by + c = 0 and instantly see its graph with intercepts and slope labelled.
Slope Explorer
Adjust the slope slider and y-intercept to see how the line changes in real-time.
Solution Verifier
Enter an equation and a point — instantly find out if the point lies on the line.
Solution Table Builder
Generate a full table of 8 solutions for any linear equation automatically.
Word Problem Lab
Generate a fresh word problem from a random real-life scenario with full solution. Click to get a new problem each time!
🏆 Knowledge Quiz
10 multiple-choice questions covering the full chapter. Instant feedback after each answer.
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ACADEMIA AETERNUM
तमसो मा ज्योतिर्गमय · Est. 2025
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Linear Equations in Two Variables Class 9 Notes
Linear Equations in Two Variables Class 9 Notes — Complete Notes & Solutions · academia-aeternum.com
The study of Linear Equations in Two Variables builds upon your understanding of linear equations in one variable, extending it to equations involving two unknowns. A linear equation in two variables is an equation that can be expressed in the form \(ax+by+c=0\), where \(a,\;b,\text{ and }c\) are real constants, and \(a,\;b\) are not both zero. The solution of such an equation is a pair of values — one for \(x\) and one for \(y\) — that together make the equation true. Graphically, each…
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