Linear Equations in Two Variables - Standard form of linear equations in two variables, finding multiple solutions, and plotting their graphs on the Cartesian plane

In Class 8, you mastered linear equations in one variable, such as $2x + 5 = 11$. Such equations have a unique solution because there is only one unknown quantity to determine. However, real-world phenomena rarely depend on a single factor. Consider scenarios where you purchase two different items at a market, measure temperature in different scales, or calculate a travel fare based on both a fixed entry fee and variable distance. In all these cases, two interdependent quantities vary together.

To model these situations mathematically, we use Linear Equations in Two Variables. Understanding this topic is foundational in secondary mathematics. It bridges simple arithmetic with analytical geometry and algebra, laying the groundwork for systems of linear equations in Class 10, linear programming in Class 12, and coordinate-based problem solving across physics and economics.

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In-Depth Conceptual Breakdown

1. Definition and Standard Form

An equation is said to be a linear equation in two variables if it can be written in the form:

$ax + by + c = 0$

where:

• $x$ and $y$ are the two unknown variables.
• $a$, $b$, and $c$ are real numbers ($\mathbb{R}$).
• $a$ and $b$ are not both zero. Mathematically, this condition is written as $a^2 + b^2 \neq 0$.

Why is it called "Linear"?

The term linear comes from the highest exponent (degree) of the variables in the equation. In $ax + by + c = 0$, the degree of both variable terms $x$ and $y$ is $1$. Algebraically, a first-degree polynomial equation in two variables generates a geometric curve that is always a straight line when plotted on a Cartesian plane.

Key Rules for Standard Form Representation:

1. All terms ($x$-term, $y$-term, and constant) must be brought to one side of the equal sign, leaving $0$ on the opposite side.
2. The term containing $x$ is typically written first, followed by the $y$-term, and then the constant term $c$.
3. Coefficients $a$, $b$, and $c$ retain their signs (+ or -).

Examples of Expressing Equations in Standard Form $ax + by + c = 0$:

Note: Equations like $x = -5$ or $y = 2$ are linear equations in one variable, but they can be expressed as linear equations in two variables by assigning a coefficient of $0$ to the missing variable.

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2. Solutions of a Linear Equation in Two Variables

A solution of a linear equation in two variables $ax + by + c = 0$ is an ordered pair of numbers $(x, y)$ such that when $x$ is substituted by the first value and $y$ by the second value, the Left Hand Side (LHS) of the equation becomes equal to the Right Hand Side (RHS).

Fundamental Property: Infinitely Many Solutions

Unlike a linear equation in one variable (which has a unique solution), a linear equation in two variables has infinitely many solutions.

Why? For any arbitrary value you choose for $x$, there exists a unique corresponding value for $y$ that satisfies the equation. Since there are infinitely many real numbers that $x$ can take, there are infinitely many pairs $(x, y)$ that satisfy $ax + by + c = 0$.

Algorithm for Finding Multiple Solutions:

To find solutions for an equation like $ax + by + c = 0$:

1. Express one variable explicitly in terms of the other. For instance, solve for $y$: $y = \frac{-c - ax}{b}$
2. Substitute convenient values for $x$ (such as $x = 0, 1, -1, 2$, etc.).
3. Calculate the corresponding value of $y$ for each $x$.
4. Write each pair as an ordered pair $(x, y)$.

Special Points (Intercepts):

• $y$-intercept: Set $x = 0$ to find where the equation crosses the $y$-axis.
• $x$-intercept: Set $y = 0$ to find where the equation crosses the $x$-axis.

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3. Plotting Graphs on the Cartesian Plane

The graphical representation of a linear equation in two variables is a straight line on the Cartesian plane.

Crucial Geometric Theorems:

1. Every point $(a, b)$ whose coordinates satisfy the equation $ax + by + c = 0$ lies on the line representing the equation.
2. Conversely, every point $(x_1, y_1)$ lying on the line represents a solution to the equation.
3. Any point that does not lie on the line is not a solution to the equation.

Step-by-Step Guide to Plotting the Graph:

Step 1: Simplify the Given Equation
  └── Rewrite equation in the form y = f(x) or x = f(y)

Step 2: Construct a Solution Table
  └── Choose at least 3 distinct values for x
  └── Calculate corresponding y values
  └── Tabulate ordered pairs (x1, y1), (x2, y2), (x3, y3)

Step 3: Draw Cartesian Axes
  └── Draw perpendicular axes X'OX and Y'OY on graph paper
  └── Choose an appropriate scale (e.g., 1 cm = 1 unit)

Step 4: Plot Points and Draw Line
  └── Mark the ordered pairs precisely on the grid
  └── Join all points with a ruler (verify they are collinear)
  └── Extend the line in both directions with arrows
  └── Label the line with its algebraic equation

Tip: Always find at least three solutions. While two points are geometrically sufficient to define a straight line, the third point serves as an essential check against arithmetic errors. If all three points do not lie on a single straight line (i.e., if they are not collinear), you have made a calculation or plotting error.

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4. Equations of Lines Parallel to Axes

1. Equation of the $x$-axis: On the $x$-axis, the $y$-coordinate of every point is $0$. Therefore, the equation of the $x$-axis is: $y = 0 \quad \text{or} \quad 0x + 1y + 0 = 0$

2. Equation of the $y$-axis: On the $y$-axis, the $x$-coordinate of every point is $0$. Therefore, the equation of the $y$-axis is: $x = 0 \quad \text{or} \quad 1x + 0y + 0 = 0$

3. Line Parallel to the $y$-axis: An equation of the form $x = a$ (where $a$ is a constant) represents a line parallel to the $y$-axis at a distance of $|a|$ units from it.

   • If $a > 0$, the line lies to the right of the $y$-axis.
   • If $a < 0$, the line lies to the left of the $y$-axis.

4. Line Parallel to the $x$-axis: An equation of the form $y = b$ (where $b$ is a constant) represents a line parallel to the $x$-axis at a distance of $|b|$ units from it.

   • If $b > 0$, the line lies above the $x$-axis.
   • If $b < 0$, the line lies below the $x$-axis.

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Summary Table of Equations and Graphical Behaviors

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Real-World Applications

Application 1: Household Budgeting and Purchasing

Suppose you visit a bookstore with $\text{₹}200$ to buy notebooks costing $\text{₹}40$ each ($x$) and pens costing $\text{₹}20$ each ($y$). The total expenditure model is:

$40x + 20y = 200$

Standard form: $40x + 20y - 200 = 0$ or $2x + y - 10 = 0$.

• If you buy $0$ notebooks ($x=0$), you can buy $10$ pens ($y=10$).
• If you buy $3$ notebooks ($x=3$), you can buy $4$ pens ($y=4$).
• If you buy $5$ notebooks ($x=5$), you can buy $0$ pens ($y=0$).

The solutions $(0,10)$, $(3,4)$, and $(5,0)$ show the trade-offs between purchasing notebooks and pens under a fixed budget.

Pens (y)
 ^
10 +  (0, 10)
 8 |    \
 6 |     \
 4 |      * (3, 4)
 2 |        \
 0 +---------*-----> Notebooks (x)
   0  1  2  3  4  5

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Application 2: Distance and Taxi Fare Structure

In many cities, taxi fares consist of a fixed minimum charge plus a variable charge per kilometer traveled. Suppose a city charges a fixed base fare of $\text{₹}15$ for the first kilometer and $\text{₹}8$ for every subsequent kilometer.

If total distance traveled is $x$ km and total fare is $\text{₹}y$:

• Distance after the 1st km = $(x - 1)$ km.
• Fare for remaining distance = $8(x - 1)$.
• Total fare equation: $y = 15 + 8(x - 1)$ $y = 15 + 8x - 8$ $8x - y + 7 = 0$

This linear model helps commuters calculate costs instantly for any distance $x$.

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Application 3: Temperature Scale Conversion

The relation between the Celsius scale ($C$) and the Fahrenheit scale ($F$) is expressed as a linear equation:

$F = \frac{9}{5}C + 32$

Taking $C$ on the $x$-axis and $F$ on the $y$-axis: $9x - 5y + 160 = 0$

• When $x = 0^\circ\text{C}$ (freezing point of water), $y = 32^\circ\text{F}$.
• When $x = 100^\circ\text{C}$ (boiling point of water), $y = 212^\circ\text{F}$.
• When $x = -40^\circ\text{C}$, $y = -40^\circ\text{F}$ (the point where both scales coincide).

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Step-by-Step Solved Textbook Examples

Example 1: Expressing Equations in Standard Form

Problem: Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of $a$, $b$, and $c$ in each case:

1. $3x - \frac{y}{5} - 10 = 0$
2. $y = 2x + 7$
3. $2x = -5y$

Solution:

Part 1: $3x - \frac{y}{5} - 10 = 0$

• Rearranging terms to match $ax + by + c = 0$: $3x + \left(-\frac{1}{5}\right)y + (-10) = 0$
• Comparing with $ax + by + c = 0$: $\mathbf{a = 3, \quad b = -\frac{1}{5}, \quad c = -10}$

Alternatively, taking LCM of denominators ($5$): $15x - y - 50 = 0 \implies \mathbf{a = 15, \quad b = -1, \quad c = -50}$

Part 2: $y = 2x + 7$

• Transposing $y$ to the RHS: $2x - y + 7 = 0$
• Comparing with $ax + by + c = 0$: $\mathbf{a = 2, \quad b = -1, \quad c = 7}$

Part 3: $2x = -5y$

• Transposing $-5y$ to the LHS: $2x + 5y = 0$
• Expressing explicitly with constant $c = 0$: $2x + 5y + 0 = 0$
• Comparing with $ax + by + c = 0$: $\mathbf{a = 2, \quad b = 5, \quad c = 0}$

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Example 2: Finding Multiple Solutions

Problem: Find four distinct solutions for the linear equation $2x + 3y = 12$.

Solution:

Given equation: $2x + 3y = 12$.

Express $y$ in terms of $x$: $3y = 12 - 2x \implies y = \frac{12 - 2x}{3}$

Now, substitute convenient values for $x$:

1. Let $x = 0$: $y = \frac{12 - 2(0)}{3} = \frac{12}{3} = 4$ First solution: $(0, 4)$

2. Let $x = 3$: $y = \frac{12 - 2(3)}{3} = \frac{12 - 6}{3} = \frac{6}{3} = 2$ Second solution: $(3, 2)$

3. Let $x = 6$: $y = \frac{12 - 2(6)}{3} = \frac{12 - 12}{3} = 0$ Third solution: $(6, 0)$

4. Let $x = -3$: $y = \frac{12 - 2(-3)}{3} = \frac{12 + 6}{3} = \frac{18}{3} = 6$ Fourth solution: $(-3, 6)$

Final Answer:

Four solutions of the equation $2x + 3y = 12$ are $(0, 4)$, $(3, 2)$, $(6, 0)$, and $(-3, 6)$.

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Example 3: Plotting Graph and Point Verification

Problem: Draw the graph of the linear equation $x + 2y = 6$. From the graph, check whether the point $(2, 2)$ lies on the line.

Solution:

Step 1: Express $x$ in terms of $y$ (or $y$ in terms of $x$). $x = 6 - 2y$

Step 2: Generate a table of values.

• If $y = 0 \implies x = 6 - 2(0) = 6$
• If $y = 1 \implies x = 6 - 2(1) = 4$
• If $y = 2 \implies x = 6 - 2(2) = 2$
• If $y = 3 \implies x = 6 - 2(3) = 0$

Points to plot: $A(6, 0)$, $B(4, 1)$, $C(2, 2)$, $D(0, 3)$.

Step 3: Plotting and Drawing.

1. Draw axes $X'OX$ and $Y'OY$ on graph paper with scale: $1\text{ cm} = 1\text{ unit}$ on both axes.
2. Plot points $A(6,0)$, $B(4,1)$, $C(2,2)$, and $D(0,3)$.
3. Join the points with a straight line ruler and extend it on both sides.

Step 4: Verification of Point $(2,2)$.

• Graphically: The point $(2, 2)$ coincides exactly with point $C$ on the drawn line.
• Algebraically: $\text{LHS} = x + 2y = 2 + 2(2) = 2 + 4 = 6 = \text{RHS}$

Since LHS = RHS, the point $(2, 2)$ lies on the line $x + 2y = 6$ and is a valid solution.

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Example 4: Real-World Word Problem to Graph

Problem: The force ($F$) applied on a body is directly proportional to the acceleration ($a$) produced in it. Write an equation in two variables taking constant mass $m = 5\text{ kg}$, draw its graph, and find the force required when acceleration is $3\text{ m/s}^2$.

Solution:

Step 1: Formulate the linear equation. From Newton's Second Law of Motion: $\text{Force } (F) = \text{Mass } (m) \times \text{Acceleration } (a)$

Let Force $F$ be $y$ (in Newtons) and Acceleration $a$ be $x$ (in $\text{m/s}^2$). Given $m = 5\text{ kg}$:

$y = 5x \quad \implies \quad 5x - y = 0$

Step 2: Table of values.

• For $x = 0 \implies y = 5(0) = 0$
• For $x = 1 \implies y = 5(1) = 5$
• For $x = 2 \implies y = 5(2) = 10$
• For $x = 3 \implies y = 5(3) = 15$

Step 3: Finding Force for $a = 3\text{ m/s}^2$. Substitute $x = 3$ into $y = 5x$: $y = 5 \times 3 = 15\text{ N}$

Final Answer:

The required linear equation is $y = 5x$ (or $5x - y = 0$). The force required when acceleration is $3\text{ m/s}^2$ is $15\text{ N}$.

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Common Student Mistakes to Avoid

Misconception 1: Sign Errors in Standard Form $ax + by + c = 0$

• The Error: Students often forget to transpose signs when bringing terms from RHS to LHS. For instance, converting $3x = 2y - 5$ to $3x + 2y - 5 = 0$.
• Correct Approach: Remember that moving a term across the equal sign flips its sign: $3x - 2y + 5 = 0$ Here, $a = 3$, $b = -2$, and $c = 5$.

Misconception 2: Reversing Coordinates $(x, y) \rightarrow (y, x)$

• The Error: Plotting the ordered pair $(2, 5)$ as $2$ units along the $y$-axis and $5$ units along the $x$-axis.
• Correct Approach: Always strictly follow the ordered pair convention $(x, y)$:
  • First coordinate ($x$-abscissa): Move horizontally along the $x$-axis.
  • Second coordinate ($y$-ordinate): Move vertically parallel to the $y$-axis.

Misconception 3: Plotting Only Two Points

• The Error: Finding and plotting only two points when drawing a graph.
• Why it is risky: Any two points can be joined to form a straight line, even if one of your calculations contains a mistake!
• Correct Approach: Always calculate and plot at least 3 points. If all 3 points lie on a single straight line, your calculations are correct.

Misconception 4: Confusing $x = a$ and $y = b$ Orientation

• The Error: Assuming $x = 3$ is a horizontal line because the $x$-axis is horizontal.
• Correct Approach:
  • $x = a$ means $x$ remains fixed at $a$ for every value of $y$. Hence, $x = a$ is a vertical line parallel to the $y$-axis.
  • $y = b$ means $y$ remains fixed at $b$ for every value of $x$. Hence, $y = b$ is a horizontal line parallel to the $x$-axis.

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Practice Questions for Self-Assessment

Question 1

Express the linear equation $\frac{x}{3} + \frac{y}{4} = 1$ in standard form $ax + by + c = 0$. Also, find the points where this line intersects the $x$-axis and the $y$-axis.

Solution:

1. Take the LCM of denominators $3$ and $4$, which is $12$: $\frac{4x + 3y}{12} = 1 \implies 4x + 3y = 12$

2. Transpose $12$ to LHS: $4x + 3y - 12 = 0$ Standard form: $4x + 3y - 12 = 0$ ($a = 4, b = 3, c = -12$).

3. Intersection with $x$-axis: Set $y = 0$: $4x + 3(0) - 12 = 0 \implies 4x = 12 \implies x = 3$ Point of intersection with $x$-axis: $(3, 0)$.

4. Intersection with $y$-axis: Set $x = 0$: $4(0) + 3y - 12 = 0 \implies 3y = 12 \implies y = 4$ Point of intersection with $y$-axis: $(0, 4)$.

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Question 2

If the point $(3, 4)$ lies on the graph of the equation $3y = ax + 7$, find the value of $a$. Also, write two additional solutions for the resulting equation.

Solution:

1. Since $(3, 4)$ lies on $3y = ax + 7$, $x = 3$ and $y = 4$ must satisfy the equation: $3(4) = a(3) + 7$ $12 = 3a + 7$ $3a = 12 - 7 = 5 \implies a = \frac{5}{3}$

2. Substitute $a = \frac{5}{3}$ back into the equation: $3y = \frac{5}{3}x + 7 \implies 9y = 5x + 21 \implies 5x - 9y + 21 = 0$

3. Find two additional solutions:

   • Let $x = 0$: $5(0) - 9y + 21 = 0 \implies 9y = 21 \implies y = \frac{21}{9} = \frac{7}{3}$ Solution: $\left(0, \frac{7}{3}\right)$
   • Let $y = 0$: $5x - 9(0) + 21 = 0 \implies 5x = -21 \implies x = -\frac{21}{5} = -4.2$ Solution: $(-4.2, 0)$

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Question 3

Draw the graph of $2x + y = 6$. Calculate the area of the triangle formed by this line, the $x$-axis, and the $y$-axis.

Solution:

1. Express $y$ in terms of $x$: $y = 6 - 2x$

2. Table of values:

   • For $x = 0 \implies y = 6$ $\rightarrow$ Point $A(0, 6)$ on $y$-axis
   • For $x = 3 \implies y = 0$ $\rightarrow$ Point $B(3, 0)$ on $x$-axis
   • For $x = 2 \implies y = 2$ $\rightarrow$ Point $C(2, 2)$

3. Triangle formation: The line intersects the $y$-axis at $A(0,6)$ and the $x$-axis at $B(3,0)$. Together with the origin $O(0,0)$, it forms a right-angled triangle $\triangle AOB$.

   • Base $OB = 3\text{ units}$
   • Height $OA = 6\text{ units}$

4. Area calculation: $\text{Area of } \triangle AOB = \frac{1}{2} \times \text{Base} \times \text{Height}$ $\text{Area} = \frac{1}{2} \times 3 \times 6 = 9\text{ sq. units}$

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Question 4

Write the linear equation that is parallel to the $x$-axis and passes through the point $(-2, -3)$. Express it in standard form in two variables.

Solution:

1. Any line parallel to the $x$-axis has the general equation $y = b$.
2. Since the line passes through $(-2, -3)$, its $y$-coordinate is fixed at $-3$.
3. Thus, the equation is $y = -3$.
4. Converting $y = -3$ to standard form $ax + by + c = 0$: $0x + 1y + 3 = 0$

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Exam Revision & Frequently Asked Questions (FAQs)

Q1: How many solutions does a linear equation in two variables have, and why?

Answer: A linear equation in two variables has infinitely many solutions. Geometrically, a linear equation represents a continuous straight line on the Cartesian plane. Since a line consists of an infinite number of collinear points, every single point on that line represents an ordered pair $(x, y)$ that satisfies the equation.

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Q2: What is the general equation of the x-axis and y-axis?

Answer:

• Equation of $x$-axis: $y = 0$ (or $0x + 1y + 0 = 0$), because the $y$-coordinate of every point lying on the $x$-axis is zero.
• Equation of $y$-axis: $x = 0$ (or $1x + 0y + 0 = 0$), because the $x$-coordinate of every point lying on the $y$-axis is zero.

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Q3: How can we algebraically test if a point lies on the graph of a linear equation without drawing it?

Answer: Substitute the $x$-coordinate and $y$-coordinate of the point into the LHS of the given linear equation.

• If $\text{LHS} = \text{RHS}$, the point lies on the line (it is a valid solution).
• If $\text{LHS} \neq \text{RHS}$, the point does not lie on the line (it is not a solution).

Example: Does $(1, 2)$ lie on $3x + 2y = 7$? $\text{LHS} = 3(1) + 2(2) = 3 + 4 = 7 = \text{RHS}$ Yes, $(1, 2)$ lies on the line.

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Q4: What is the nature of the graph for equations of the form $y = mx$ or $ax + by = 0$?

Answer: Any linear equation where the constant term $c = 0$ (i.e., $ax + by = 0$ or $y = mx$) always passes through the origin $(0, 0)$.

Proof: Substituting $x = 0$ gives $a(0) + b y = 0 \implies y = 0$. Therefore, $(0,0)$ is always a solution for such equations.