CBSE Class 10 Maths: Linear Equations in Two Variables - Exercise 3.1

Understanding Linear Equations in Two Variables

A linear equation in two variables is an equation that can be written in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero. The 'two variables' refer to $x$ and $y$. The term 'linear' comes from the fact that when you plot all the possible solutions $(x, y)$ of such an equation on a Cartesian plane, they always form a straight line. Each point $(x, y)$ on this line is a solution to the equation.

When we deal with "a pair of linear equations in two variables" (also known as a system of linear equations), we are looking for values of $x$ and $y$ that satisfy both equations simultaneously. Exercise 3.1 primarily introduces you to two key aspects:

1. Forming Algebraic Equations: Translating a given word problem into a set of two linear equations.
2. Graphical Representation: Plotting these equations on a graph paper. Each equation will yield a straight line. The nature of these lines (intersecting, parallel, or coincident) tells us about the number of solutions the system has. For instance, if the lines intersect at a single point, that point $(x, y)$ is the unique solution to the system. If the lines are parallel, they never intersect, meaning there is no common solution. If the lines are coincident (one lies exactly on top of the other), they have infinitely many solutions.

Step-by-Step: Forming and Graphing Linear Equations

1. Identify Variables and Quantities — Carefully read the word problem. Identify the unknown quantities that you need to find. Assign distinct variables, typically $x$ and $y$, to these unknown quantities. Clearly define what each variable represents.
2. Formulate Algebraic Equations — Translate the conditions given in the word problem into two separate linear equations using the assigned variables. Each condition or statement in the problem will usually correspond to one equation. Ensure the equations are in the standard form $ax + by + c = 0$ or $ax + by = c$ for clarity.
3. Find Coordinates for Each Equation — For each linear equation, find at least two (preferably three) pairs of $(x, y)$ coordinates that satisfy the equation. To do this, choose a value for $x$ (or $y$), substitute it into the equation, and solve for the other variable. It's often helpful to choose values that lead to integer coordinates for easier plotting.
4. Plot Points and Draw Lines on Graph — Draw a Cartesian coordinate system with appropriate scales on the x-axis and y-axis. Plot the coordinates you found for the first equation. Use a ruler to draw a straight line passing through these points. Extend the line beyond the plotted points and label it with its equation. Repeat this process for the second equation on the same graph paper.
5. Interpret the Graphical Representation — Observe the relationship between the two lines you have drawn. Do they intersect at a single point? Are they parallel to each other? Or do they coincide (lie on top of each other)? This observation will determine the nature of the solution to the system of equations. In Ex 3.1, the primary goal is this representation, but understanding its implication is key.

Worked Example: Solving a Word Problem Graphically

• Problem: The cost of 4 pencils and 2 erasers is ₹60. Later, the cost of 2 pencils and 4 erasers is ₹40. Represent this situation algebraically and graphically. Step 1: Identify Variables Let $x$ be the cost of one pencil (in ₹). Let $y$ be the cost of one eraser (in ₹). Step 2: Formulate Algebraic Equations From the first condition: "The cost of 4 pencils and 2 erasers is ₹60." Equation 1: $4x + 2y = 60$ From the second condition: "Later, the cost of 2 pencils and 4 erasers is ₹40." Equation 2: $2x + 4y = 40$ Step 3: Find Coordinates for Each Equation For Equation 1: $4x + 2y = 60$ (can be simplified to $2x + y = 30$) Let's find three points: If $x = 0$: $2(0) + y = 30 \Rightarrow y = 30$. Point: $(0, 30)$ If $x = 10$: $2(10) + y = 30 \Rightarrow 20 + y = 30 \Rightarrow y = 10$. Point: $(10, 10)$ If $y = 0$: $2x + 0 = 30 \Rightarrow 2x = 30 \Rightarrow x = 15$. Point: $(15, 0)$ For Equation 2: $2x + 4y = 40$ (can be simplified to $x + 2y = 20$) Let's find three points: If $x = 0$: $0 + 2y = 20 \Rightarrow y = 10$. Point: $(0, 10)$ If $y = 0$: $x + 2(0) = 20 \Rightarrow x = 20$. Point: $(20, 0)$ If $x = 10$: $10 + 2y = 20 \Rightarrow 2y = 10 \Rightarrow y = 5$. Point: $(10, 5)$ Step 4: Plot Points and Draw Lines on Graph (Imagine plotting these points on a graph paper) Plot $(0, 30)$, $(10, 10)$, $(15, 0)$ and draw a line for $2x + y = 30$. Plot $(0, 10)$, $(20, 0)$, $(10, 5)$ and draw a line for $x + 2y = 20$. Step 5: Interpret the Graphical Representation Upon plotting, you will observe that the two lines intersect at a single point, which is $(13.33, 3.33)$ if you extend the lines accurately (or observe it closely if you pick values for precise intersection). This intersection point represents the unique solution where approximately $x = 13.33$ and $y = 3.33$ would satisfy both conditions. However, since cost should ideally be in whole numbers or two decimal places, this suggests that the numbers I chose for the problem might not lead to a 'nice' integer solution for a graphical method, which is common in real-world problems. The primary goal of Ex 3.1 is the setup and plotting, not necessarily finding exact integer solutions from the graph.

Common Mistakes and Exam Tips for Ex 3.1

To score well in problems involving linear equations and their graphical representation, be mindful of these common pitfalls:

1. Incorrect Equation Formation: This is the most frequent mistake. Always read the word problem carefully, identifying what each variable represents and how the conditions relate them. For example, if a problem states 'twice the age,' ensure you write '2x', not 'x + 2'. Double-check your equations against the problem statement.
2. Arithmetic Errors: When finding coordinates for plotting, simple calculation mistakes can lead to incorrect points and ultimately, incorrect lines. Always re-verify your $(x, y)$ pairs by substituting them back into the original equation.
3. Inaccurate Plotting: Use proper graph paper and a sharp pencil. Ensure your scale on both axes is consistent and clearly marked. Each unit on the graph should represent the same value throughout. Plot points precisely, and use a ruler to draw straight lines extending beyond the points. Label your lines with their respective equations.
4. Choosing Difficult Coordinates: While any two points define a line, choosing points where $x=0$ or $y=0$ (intercepts) often simplifies calculations. Also, try to choose values that lead to whole numbers or easily plottable fractions to avoid inaccuracies.
5. Not Labelling Axes: Always label your x-axis and y-axis (e.g., 'Cost of Pencils (₹)', 'Number of Bats') and indicate the scale used (e.g., '1 unit = ₹50'). This makes your graph clear and understandable.

Practice Questions with Solutions

• Q: Akhil buys 5 notebooks and 3 pens for ₹120. His friend, Sumit, buys 2 notebooks and 6 pens of the same kind for ₹90. Represent this situation algebraically and graphically. A: Step 1: Let $x$ be the cost of one notebook and $y$ be the cost of one pen. Step 2: From Akhil's purchase: $5x + 3y = 120$. From Sumit's purchase: $2x + 6y = 90$. Step 3: For $5x + 3y = 120$: $(0, 40)$, $(24, 0)$, $(15, 15)$. For $2x + 6y = 90$ (or $x + 3y = 45$): $(0, 15)$, $(45, 0)$, $(15, 10)$. Step 4: Plot these points and draw the lines. Observe their intersection. Final answer: Algebraic representation: $5x + 3y = 120$ and $2x + 6y = 90$. Graphical representation involves plotting these lines.
• Q: The difference between two numbers is 5. If twice the first number is 10 more than the second number, represent this situation algebraically and graphically. A: Step 1: Let the first number be $x$ and the second number be $y$. Step 2: From the first condition: $x - y = 5$. From the second condition: $2x = y + 10 \Rightarrow 2x - y = 10$. Step 3: For $x - y = 5$: $(0, -5)$, $(5, 0)$, $(10, 5)$. For $2x - y = 10$: $(0, -10)$, $(5, 0)$, $(10, 10)$. Step 4: Plot these points and draw the lines. Observe their intersection. Final answer: Algebraic representation: $x - y = 5$ and $2x - y = 10$. Graphical representation involves plotting these lines.
• Q: A taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105, and for a journey of 15 km, the charge paid is ₹155. Represent this situation algebraically and graphically. A: Step 1: Let the fixed charge be $₹x$ and the charge per km be $₹y$. Step 2: For 10 km: $x + 10y = 105$. For 15 km: $x + 15y = 155$. Step 3: For $x + 10y = 105$: $(5, 10)$, $(105, 0)$, $(15, 9)$. For $x + 15y = 155$: $(5, 10)$, $(155, 0)$, $(20, 9)$. Step 4: Plot these points and draw the lines. Observe their intersection. Final answer: Algebraic representation: $x + 10y = 105$ and $x + 15y = 155$. Graphical representation involves plotting these lines.
• Q: The perimeter of a rectangular garden is 36 meters. Its length is 4 meters more than its width. Represent this situation algebraically and graphically. A: Step 1: Let the length of the garden be $l$ meters and the width be $w$ meters. Step 2: Perimeter is $2(l+w) = 36 \Rightarrow l+w = 18$. Length is 4 more than width: $l = w + 4 \Rightarrow l - w = 4$. Step 3: For $l + w = 18$: $(0, 18)$, $(18, 0)$, $(9, 9)$. For $l - w = 4$: $(0, -4)$, $(4, 0)$, $(10, 6)$. Step 4: Plot these points (using $l$ on x-axis and $w$ on y-axis) and draw the lines. Final answer: Algebraic representation: $l + w = 18$ and $l - w = 4$. Graphical representation involves plotting these lines.

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