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Introduction and Graphical Method

🎓 Class 10 Mathematics CBSE Theory Ch 3 — Pair of Linear Equations in Two Variables ⏱ ~35 min

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3.1 Introduction

Consider a familiar village fair scene — a Giant Wheel and a ring-toss game (Hoopla). A child watches people enjoy rides on the wheel priced at one unit per ride, and plays Hoopla where each ring costs a fixed amount. Suppose the total number of rides taken by the child on the Giant Wheel is x, and the number of Hoopla throws is y. If the entrance and ride budget give us one equation in x and y, and an independent counting condition gives another, we get a pair of linear equations in two variables^?.

General Form

A pair of linear equations in two variables x and y may be written as \[ a_1 x + b_1 y + c_1 = 0 \] \[ a_2 x + b_2 y + c_2 = 0 \] where \(a_1, b_1, c_1, a_2, b_2, c_2\) are real numbers such that \(a_1^2 + b_1^2 \neq 0\) and \(a_2^2 + b_2^2 \neq 0\).

Example situation: Akhila went to a fair with Rs. 20. She wanted to ride the Giant Wheel and also play Hoopla. The number of rides she took on the Giant Wheel was half the number of times she played Hoopla. If each ride costs Rs. 3 and a game of Hoopla costs Rs. 4, how would you find the number of rides and games?

Let the number of rides be x and the number of Hoopla games be y. Then:

\( y = \tfrac{1}{2}x \) ...(1)

\( 3x + 4y = 20 \) ...(2)

Can we find the solutions of this pair of equations? There are several ways. Let us begin with graphical representation.

3.2 Graphical Method of Solution of a Pair of Linear Equations

A pair of linear equations which has no solution is called an inconsistent^? pair. A pair which has a solution is called consistent. When the equations have infinitely many common solutions, they are called dependent^? (which is a special case of consistent).

Geometrically, when two straight lines are drawn on the Cartesian plane for a pair of linear equations, exactly one of the following happens:

The two lines intersect at one point. The pair has a unique solution (consistent).
The two lines are parallel. The pair has no solution (inconsistent).
The two lines coincide (one lies on the other). The pair has infinitely many solutions (dependent & consistent).

Fig 3.1 — Three cases for a pair of lines in the plane.

Ratio Test for the Three Cases

Let two equations be \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\). Compare the ratios \(\dfrac{a_1}{a_2},\; \dfrac{b_1}{b_2},\; \dfrac{c_1}{c_2}\):

Ratio Comparison	Graphical	Algebraic Meaning
\(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\)	Intersecting lines	Unique solution (consistent)
\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\)	Coincident lines	Infinitely many solutions (dependent, consistent)
\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\)	Parallel lines	No solution (inconsistent)

Why the ratio test works

If the coefficients of x and y are proportional, the slopes of the two lines are identical. If in addition the constant term is in the same ratio, the two equations are the same line (coincident). If the constant term breaks the ratio, the lines are parallel but separate. If the x-y coefficients are not proportional, the slopes differ and the lines must meet at exactly one point.

Worked Example 1 — Unique Solution

Check graphically whether the pair \( x + 3y = 6 \) and \( 2x - 3y = 12 \) is consistent.

Find two solutions of each equation to plot the lines.

\(x\)	0	6
\(y = \dfrac{6-x}{3}\)	2	0

\(x\)	0	3
\(y = \dfrac{2x-12}{3}\)	-4	-2

Lines intersect at (6, 0). The pair has unique solution \(x=6, y=0\); consistent.

Worked Example 2 — No Solution (Parallel)

Check graphically whether the pair \( 5x - 8y + 1 = 0 \) and \( 3x - \tfrac{24}{5}y + \tfrac{3}{5} = 0 \) is consistent.

Multiplying equation (2) throughout by 5 gives \( 15x - 24y + 3 = 0 \), i.e. \(5x - 8y + 1 = 0\) — this is exactly equation (1). So the two equations represent the same line (coincident). The pair has infinitely many solutions and is consistent.

Worked Example 3 — Real-World Modelling

Champa went to a "Sale" to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased." Help her friends find how many pants and skirts Champa bought.

Let the number of pants be x and skirts be y. Then:

\( y = 2x - 2 \) ...(1) \( y = 4x - 4 \) ...(2)

\(x\)	2	0
\(y=2x-2\)	2	-2

\(x\)	0	1
\(y=4x-4\)	-4	0

Fig 3.2 — The two lines intersect at (1, 0). So Champa bought 1 pant... wait, check: at \(x=1\), line 1 gives \(y=0\), line 2 gives \(y=0\). But this is the common point. Actually we solve: \(2x-2=4x-4 \Rightarrow x=1, y=0\). Re-verify the problem — see solution below.

Solution (verified algebraically): From (1) and (2): \(2x-2 = 4x-4 \Rightarrow 2x = 2 \Rightarrow x = 1\). But the textbook setup expects the pair to yield a practical positive answer; in fact solving carefully: the figure from NCERT shows intersection at \(x=1, y=0\). Substituting back confirms both equations. So Champa bought 1 pant and 0 skirts if we take the equations literally — however the classic NCERT variant asks for \(x, y\) where the lines meet; graphically verifying gives a unique solution, confirming the pair is consistent.

Activity: Plot-and-Predict the Line Pair

Materials: graph paper, ruler, pencil.

Take the pair \(x + y = 5\) and \(2x + 2y = 10\). Compute two solutions of each.
Plot them on the same graph paper.
Before plotting the second line, predict whether the lines will be intersecting, parallel, or coincident using the ratio test.
Verify your prediction from the plot.
Repeat with \(x - y = 3\) and \(3x - 3y = 15\). What do you observe?

Pair 1: \(\tfrac{1}{2}=\tfrac{1}{2}=\tfrac{5}{10}\) — all three ratios equal, so coincident; infinite solutions.
Pair 2: \(\tfrac{1}{3}=\tfrac{-1}{-3}\neq\tfrac{3}{15}\) — first two equal, third different: parallel; no solution.

Interactive: Classify a Pair by Coefficients

a₁

b₁

c₁

a₂

b₂

c₂

Enter coefficients and click Classify.

Exercise 3.1

Q1. Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Let boys = \(x\), girls = \(y\).
\(x + y = 10\) and \(y = x + 4\), i.e. \(-x + y = 4\).
Plotting: lines intersect at \((3, 7)\). So 3 boys and 7 girls.

(ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.

Let pencil = \(x\), pen = \(y\). Then \(5x+7y=50\) and \(7x+5y=46\). Adding: \(12x+12y=96 \Rightarrow x+y=8\). Subtracting: \(-2x+2y=4 \Rightarrow y-x=2\). Solve: \(x=3, y=5\). Pencil Rs. 3, Pen Rs. 5.

Competency-Based Questions

Scenario: A ticket kiosk at a fairground sells combo tickets. A combo of 2 Giant Wheel rides and 3 Hoopla games costs Rs. 17. Another combo of 3 Giant Wheel rides and 2 Hoopla games costs Rs. 18. The fair-manager wants to know the price of one ride and one game.

Q1. Form the pair of linear equations and determine whether the system is consistent by the ratio test.

L3 Apply

Let ride = \(x\), game = \(y\). \(2x+3y=17\) and \(3x+2y=18\). Here \(\tfrac{a_1}{a_2}=\tfrac{2}{3}\) and \(\tfrac{b_1}{b_2}=\tfrac{3}{2}\). Since \(\tfrac{2}{3}\neq\tfrac{3}{2}\), lines intersect — system is consistent with unique solution.

Q2. Interpret: what does a coincident pair of linear equations represent in a real-world pricing context like this?

L4 Analyse

Coincident lines mean one equation is a scalar multiple of the other — the "two" offers carry exactly the same pricing information (e.g., "2 rides + 3 games = Rs. 17" and "4 rides + 6 games = Rs. 34" are the same deal). Any (x, y) satisfying the first works for the second. The manager would learn nothing new from the second offer.

Q3. Evaluate: a rival stall offers \(2x+3y=17\) and \(4x+6y=30\). Is it consistent? Justify using the ratio test and state whether the customer can deduce individual prices.

L5 Evaluate

\(\tfrac{2}{4}=\tfrac{3}{6}=\tfrac{1}{2}\) but \(\tfrac{-17}{-30}=\tfrac{17}{30}\neq\tfrac{1}{2}\). Ratios \(\tfrac{a_1}{a_2}=\tfrac{b_1}{b_2}\neq\tfrac{c_1}{c_2}\), so lines are parallel, pair is inconsistent. The two offers contradict each other — the customer cannot pin down prices; the stall has a pricing error.

Q4. Design: create two linear equations in \(x\) (apple price) and \(y\) (banana price) such that the system is dependent (coincident). Explain how a shopper would interpret the two deals.

L6 Create

Example: \(3x + 2y = 20\) and \(6x + 4y = 40\). Multiply first by 2 to get second — same pricing deal presented in doubled quantity. A shopper would realise both offers express the same per-unit combined price, so either can be chosen; there are infinitely many (x, y) pairs consistent with them and individual prices cannot be uniquely identified without a third, independent condition.

Assertion–Reason Questions

Assertion (A): The pair \(2x + 3y = 7\) and \(4x + 6y = 14\) has infinitely many solutions.
Reason (R): For a dependent system, \(\tfrac{a_1}{a_2}=\tfrac{b_1}{b_2}=\tfrac{c_1}{c_2}\).

(a) Both true, R explains A

(b) Both true, R does not explain A

(d) A false, R true

(a) — ratios all equal \(\tfrac{1}{2}\); lines coincide; R correctly explains A.

Assertion (A): The pair \(x + 2y = 5\) and \(2x + 4y = 12\) has no solution.
Reason (R): The coefficient ratios for \(x\) and \(y\) are equal but the constant ratio differs, which means the two lines are parallel.

(a) Both true, R explains A

(b) Both true, R does not explain A

(d) A false, R true

(a) — \(\tfrac{1}{2}=\tfrac{2}{4}\neq\tfrac{5}{12}\), parallel lines; R explains A.

Assertion (A): Every pair of linear equations has a unique solution.
Reason (R): Two lines on a plane always meet at exactly one point.

(a) Both true, R explains A

(b) Both true, R does not explain A

(d) A false, R true

(d) is closest — but actually both A and R are false. A is false because parallel and coincident pairs exist. R is false because parallel lines never meet and coincident lines meet at every point. Correct answer: neither holds; mark as (both false).

Frequently Asked Questions

What is a pair of linear equations in two variables?

A pair of linear equations consists of two equations of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, where x and y are variables. NCERT Class 10 Chapter 3 studies methods to solve such pairs.

How do you solve a pair of linear equations graphically?

Draw each equation's line on the same coordinate plane. The solution is the intersection point. If lines meet at one point, there's a unique solution; if they coincide, infinitely many; if parallel, none. NCERT Class 10 Chapter 3 teaches this.

What is a consistent pair of equations?

A pair is consistent if it has at least one solution. Intersecting lines give a consistent pair with a unique solution; coincident lines give a consistent pair with infinitely many solutions. NCERT Class 10 Maths Chapter 3 defines this.

What is an inconsistent pair?

An inconsistent pair has no solution. Graphically, the two lines are parallel and never meet. Algebraically, a1/a2 = b1/b2 != c1/c2. NCERT Class 10 Chapter 3 lists this criterion.

When do two lines coincide?

Two lines coincide when their coefficients are proportional: a1/a2 = b1/b2 = c1/c2. The equations represent the same line, so every point on it is a solution. NCERT Class 10 Maths Chapter 3 details this case.

Why is graphical method important in Class 10?

The graphical method gives visual intuition about solutions and the consistency of equation pairs. It motivates algebraic methods and connects algebra with geometry. NCERT Class 10 Chapter 3 begins with this approach.

Frequently Asked Questions — Pair of Linear Equations in Two Variables

What is Introduction and Graphical Method in NCERT Class 10 Mathematics?

Introduction and Graphical Method is a key concept covered in NCERT Class 10 Mathematics, Chapter 3: Pair of Linear Equations in Two Variables. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Introduction and Graphical Method step by step?

To solve problems on Introduction and Graphical Method, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 3: Pair of Linear Equations in Two Variables?

The essential formulas of Chapter 3 (Pair of Linear Equations in Two Variables) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Introduction and Graphical Method important for the Class 10 board exam?

Introduction and Graphical Method is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Introduction and Graphical Method?

Common mistakes in Introduction and Graphical Method include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Introduction and Graphical Method?

End-of-chapter NCERT exercises for Introduction and Graphical Method cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 3, and solve at least one previous-year board paper to consolidate your understanding.

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