This MCQ module is based on: Algebraic Methods
TOPIC 7 OF 35
Algebraic Methods
🎓 Class 10
Mathematics
CBSE
Theory
Ch 3 — Pair of Linear Equations in Two Variables
⏱ ~35 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: Algebraic Methods
Targeting Class 10 level in Algebra, with Intermediate difficulty.
Upload images, PDFs, or Word documents to include their content in assessment generation.
3.3 Algebraic Methods — The Substitution Method
The graphical method gives us an accurate geometric picture, but it is not always precise — especially when the solution values are not integers, such as \(x = \tfrac{49}{22}, y = \tfrac{19}{22}\). For such cases we use algebraic methods. The first is the substitution method?.
Steps of the Substitution Method
Step 1: From one of the two equations, express one variable (say \(y\)) in terms of the other (\(x\)).Step 2: Substitute this expression for \(y\) into the other equation. You now have a linear equation in one variable \(x\).
Step 3: Solve for \(x\). Then back-substitute into the expression from Step 1 to find \(y\).
Worked Example 4 — Substitution
Solve: \(7x - 15y = 2\) ... (1) and \(x + 2y = 3\) ... (2).
Step 1 — from (2): \(x = 3 - 2y\).
Step 2 — substitute into (1): \(7(3 - 2y) - 15y = 2\) \(\Rightarrow 21 - 14y - 15y = 2 \Rightarrow -29y = -19\).
Step 3 — \(y = \dfrac{19}{29}\). Then \(x = 3 - 2 \times \dfrac{19}{29} = \dfrac{87 - 38}{29} = \dfrac{49}{29}\).
Verification: Substituting \(x = \tfrac{49}{29},\; y = \tfrac{19}{29}\) into (1): \(7 \cdot \tfrac{49}{29} - 15 \cdot \tfrac{19}{29} = \tfrac{343 - 285}{29} = \tfrac{58}{29} = 2\). ✓
What if the Pair has No Solution or Infinitely Many?
Remark
While using the substitution method, if all the variables cancel out and you obtain a true statement (e.g. \(0 = 0\)), the pair has infinitely many solutions — equations are dependent.If you obtain a false statement (e.g. \(0 = 9\)), the pair has no solution — the system is inconsistent.
Worked Example 5 — Age Problem
Aftab tells his daughter: "Seven years ago, I was seven times as old as you were. Also, three years from now, I shall be three times as old as you will be." Find their present ages.
Let Aftab's present age be \(s\) years and his daughter's be \(t\) years.
Seven years ago: \(s - 7 = 7(t - 7) \Rightarrow s - 7t = -42\) ... (1)
Three years from now: \(s + 3 = 3(t + 3) \Rightarrow s - 3t = 6\) ... (2)
Subtracting (2) from (1): \(-4t = -48 \Rightarrow t = 12\). Then \(s = 6 + 3(12) = 42\).
Answer: Aftab is 42 years old, and his daughter is 12 years old today.
Exercise 3.2 (Substitution Method)
Q1. Solve by substitution:
(i) \(x + y = 14\) and \(x - y = 4\) (ii) \(s - t = 3\) and \(\tfrac{s}{3} + \tfrac{t}{2} = 6\)
(iii) \(3x - y = 3\) and \(9x - 3y = 9\) (iv) \(0.2x + 0.3y = 1.3\) and \(0.4x + 0.5y = 2.3\)
(i) \(x + y = 14\) and \(x - y = 4\) (ii) \(s - t = 3\) and \(\tfrac{s}{3} + \tfrac{t}{2} = 6\)
(iii) \(3x - y = 3\) and \(9x - 3y = 9\) (iv) \(0.2x + 0.3y = 1.3\) and \(0.4x + 0.5y = 2.3\)
(i) \(x = y + 4\) → \((y + 4) + y = 14 \Rightarrow y = 5, x = 9\).
(ii) \(s = t + 3\) → \(\tfrac{t+3}{3} + \tfrac{t}{2} = 6 \Rightarrow 2(t+3) + 3t = 36 \Rightarrow 5t = 30 \Rightarrow t = 6, s = 9\).
(iii) Dividing (2) by 3 gives \(3x - y = 3\) which is equation (1). Dependent pair — infinitely many solutions.
(iv) Multiply by 10: \(2x + 3y = 13,\; 4x + 5y = 23\). From first: \(x = \tfrac{13-3y}{2}\). Sub: \(2(13 - 3y) + 5y = 23 \Rightarrow 26 - 6y + 5y = 23 \Rightarrow y = 3, x = 2\).
(ii) \(s = t + 3\) → \(\tfrac{t+3}{3} + \tfrac{t}{2} = 6 \Rightarrow 2(t+3) + 3t = 36 \Rightarrow 5t = 30 \Rightarrow t = 6, s = 9\).
(iii) Dividing (2) by 3 gives \(3x - y = 3\) which is equation (1). Dependent pair — infinitely many solutions.
(iv) Multiply by 10: \(2x + 3y = 13,\; 4x + 5y = 23\). From first: \(x = \tfrac{13-3y}{2}\). Sub: \(2(13 - 3y) + 5y = 23 \Rightarrow 26 - 6y + 5y = 23 \Rightarrow y = 3, x = 2\).
Q2. Solve \(2x + 3y = 11\) and \(2x - 4y = -24\); hence find \(m\) if \(y = mx + 3\).
Subtract: \(7y = 35 \Rightarrow y = 5\). Then \(2x = 11 - 15 = -4 \Rightarrow x = -2\). Now \(5 = m(-2) + 3 \Rightarrow m = -1\).
3.4 Elimination Method
In the elimination method?, we eliminate one of the two variables directly by adding or subtracting suitable multiples of the two equations. This is often quicker than substitution when fractions would otherwise arise.
Steps of the Elimination Method
Step 1: Multiply both equations by suitable non-zero constants so that the coefficients of one variable become numerically equal.Step 2: Add or subtract the equations to eliminate that variable. You obtain a single linear equation in the other variable.
Step 3: Solve for the remaining variable; back-substitute to find the eliminated one.
Step 4: If the coefficients of both variables cancel and you get \(0 = 0\), the pair is dependent (infinite solutions). If you get a non-zero constant = 0, the pair is inconsistent (no solution).
Worked Example 6 — Elimination
Solve: \(9x - 4y = 2000\) ... (1) and \(7x - 3y = 2000\) ... (2).
Multiply (1) by 3 and (2) by 4: \(27x - 12y = 6000\) and \(28x - 12y = 8000\).
Subtracting: \(-x = -2000 \Rightarrow x = 2000\). Then \(9(2000) - 4y = 2000 \Rightarrow 4y = 16000 \Rightarrow y = 4000\).
Answer: \(x = 2000,\; y = 4000\).
Worked Example 7 — Sum-of-Digits
The sum of a two-digit number and the number obtained by reversing its digits is 66. The digits of the number differ by 2. Find the number. How many such numbers are there?
Let the tens digit be \(x\) and units digit be \(y\). The number is \(10x + y\); the reversed number is \(10y + x\).
Sum condition: \((10x + y) + (10y + x) = 66 \Rightarrow 11(x + y) = 66 \Rightarrow x + y = 6\) ... (1)
Digit-difference: \(x - y = 2\) OR \(y - x = 2\) (either digit could be larger) ... (2)
Case A: \(x + y = 6\) and \(x - y = 2\) → \(x = 4, y = 2\); number is 42.
Case B: \(x + y = 6\) and \(y - x = 2\) → \(x = 2, y = 4\); number is 24.
Answer: There are two such numbers — 42 and 24.
Activity: Race Between Two Methods
Pair up. Partner A uses substitution; Partner B uses elimination. Same problem.
- Pick this pair: \(3x + 4y = 10,\; 2x - 2y = 2\).
- Both partners solve independently — one by substitution, the other by elimination.
- Time each person. Compare the number of algebraic steps used.
- Repeat with: \(0.5x + 0.7y = 0.74,\; 0.3x - 0.5y = -0.38\). Which method is faster now? Why?
Problem 1 answer: \(x=2, y=1\). Both methods work smoothly; elimination is slightly faster because \(2y\) is easily eliminated.
Problem 2 (multiply by 100): \(50x + 70y = 74,\; 30x - 50y = -38\). Elimination is cleaner — equalise coefficients of \(x\) or \(y\) and subtract. Answer: \(x = 0.5, y = 0.7\). Rule of thumb: elimination beats substitution when decimal or fraction coefficients would create messy expressions.
Problem 2 (multiply by 100): \(50x + 70y = 74,\; 30x - 50y = -38\). Elimination is cleaner — equalise coefficients of \(x\) or \(y\) and subtract. Answer: \(x = 0.5, y = 0.7\). Rule of thumb: elimination beats substitution when decimal or fraction coefficients would create messy expressions.
3.5 Equations Reducible to a Pair of Linear Equations
Some equations are not linear but can be transformed into a linear pair by a clever substitution — typically by letting \(\tfrac{1}{x} = u\) and \(\tfrac{1}{y} = v\), or similar.
Worked Example 8 — Reciprocals
Solve: \(\dfrac{2}{x} + \dfrac{3}{y} = 13\) and \(\dfrac{5}{x} - \dfrac{4}{y} = -2\).
Let \(u = \tfrac{1}{x},\; v = \tfrac{1}{y}\). Equations become \(2u + 3v = 13\) and \(5u - 4v = -2\).
Multiply first by 4, second by 3: \(8u + 12v = 52\), \(15u - 12v = -6\). Add: \(23u = 46 \Rightarrow u = 2\). Then \(v = 3\).
So \(x = \tfrac{1}{u} = \tfrac{1}{2},\; y = \tfrac{1}{v} = \tfrac{1}{3}\).
Worked Example 9 — Boat and Stream
Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Let Ritu's still-water speed be \(x\) km/h and the current's speed be \(y\) km/h.
Downstream speed = \(x + y = \tfrac{20}{2} = 10\) ... (1). Upstream speed = \(x - y = \tfrac{4}{2} = 2\) ... (2).
Add: \(2x = 12 \Rightarrow x = 6\). Then \(y = 4\).
Answer: Rowing speed = 6 km/h; current speed = 4 km/h.
Competency-Based Questions
Scenario: A school canteen offers two combos. Combo A is 3 samosas and 2 juice boxes for Rs. 80. Combo B is 2 samosas and 3 juice boxes for Rs. 75. The canteen manager wants to publish individual item prices.
Q1. Use the elimination method to find the price of one samosa and one juice box.
L3 ApplyLet samosa = \(x\), juice = \(y\). Then \(3x+2y=80\) and \(2x+3y=75\). Multiply first by 3, second by 2: \(9x+6y=240,\;4x+6y=150\). Subtract: \(5x=90 \Rightarrow x=18\). Then \(y=(80-54)/2=13\). Samosa Rs. 18, Juice Rs. 13.
Q2. Analyse: would the substitution method lead to messy fractions here? Compare step-counts.
L4 AnalyseFrom \(3x+2y=80 \Rightarrow y = \tfrac{80-3x}{2}\). Substituting in second: \(2x + 3 \cdot \tfrac{80-3x}{2} = 75 \Rightarrow 4x + 3(80 - 3x) = 150 \Rightarrow -5x = -90 \Rightarrow x = 18\). Both methods work, but substitution introduced a fraction step. Elimination avoids fractions and is cleaner here. Step-count is comparable (~3 steps each).
Q3. Evaluate: a new deal says 5 samosas and 5 juice boxes cost Rs. 150. Is this consistent with the earlier prices? Justify.
L5 EvaluateCheck: \(5(18) + 5(13) = 90 + 65 = 155\), not 150. So the new deal is inconsistent with the earlier prices — it offers a Rs. 5 discount or contains an error. Taking just the new deal with Combo A gives equations \(3x + 2y = 80,\; 5x + 5y = 150\) which reduce to unique solution \(x = 10, y = 25\) — different from above, confirming the three deals cannot all hold together.
Q4. Create a real-life reciprocal-type problem (similar to Example 8) and give its solution using substitution \(u=1/x, v=1/y\).
L6 CreateProblem: Two pipes can fill a tank. If pipe A alone takes \(x\) hours and pipe B alone takes \(y\) hours, together they fill the tank in 6 hours; if A runs 3 hours then B continues, B alone filling the remainder takes 10 hours. Work-rate gives \(\tfrac{1}{x} + \tfrac{1}{y} = \tfrac{1}{6}\) and \(\tfrac{3}{x} + \tfrac{10}{y} = 1\). Let \(u=\tfrac{1}{x}, v=\tfrac{1}{y}\): \(u + v = \tfrac{1}{6}, \; 3u + 10v = 1\). Solve: \(u = \tfrac{1}{10},\; v = \tfrac{1}{15}\). So A takes 10 hrs, B takes 15 hrs.
Assertion–Reason Questions
Assertion (A): Substituting for a variable in one equation reduces a pair of linear equations to a single equation in one variable.
Reason (R): Any two simultaneous linear equations can be solved by expressing one variable in terms of the other.
Reason (R): Any two simultaneous linear equations can be solved by expressing one variable in terms of the other.
(a) — the substitution step literally performs the reduction that R describes.
Assertion (A): Solving \(\tfrac{2}{x} + \tfrac{3}{y} = 13\) and \(\tfrac{5}{x} - \tfrac{4}{y} = -2\) by letting \(u=1/x, v=1/y\) yields \(x = 1/2,\; y = 1/3\).
Reason (R): Reducible equations become linear in new variables chosen to match the non-linear structure.
Reason (R): Reducible equations become linear in new variables chosen to match the non-linear structure.
(a) — verified in Worked Example 8; R provides the principle behind A.
Assertion (A): When the elimination step produces \(0 = 9\), the pair has infinitely many solutions.
Reason (R): A contradiction \(0 = 9\) indicates the two lines coincide.
Reason (R): A contradiction \(0 = 9\) indicates the two lines coincide.
(Both false — closest option wording is "A false, R false"). A false statement like \(0=9\) means the pair has NO solution (inconsistent, parallel lines), not infinite. R is false as well — a contradiction means lines are parallel, not coincident.
Frequently Asked Questions
What is the substitution method?
In the substitution method, express one variable in terms of the other from one equation, then substitute into the second equation to get a single-variable equation. Solve, then back-substitute. NCERT Class 10 Chapter 3 details this method.
How does the elimination method work?
Multiply equations so that the coefficients of one variable become equal (or opposite). Add or subtract the equations to eliminate that variable, then solve for the other. Back-substitute to get the first. NCERT Class 10 Maths Chapter 3 uses this.
Which method is better - substitution or elimination?
Use substitution when one equation easily gives a variable in terms of the other. Use elimination when coefficients are easily made equal by small multiplications. Both give the same answer. NCERT Class 10 Chapter 3 recommends judgement.
Example: solve 2x + 3y = 13 and x - y = 1.
From x - y = 1, x = y + 1. Substitute: 2(y+1) + 3y = 13, so 5y + 2 = 13, y = 11/5. Wait, let's check: 2y+2+3y=13, 5y=11, y=11/5=2.2; x=3.2. Alternatively, use integer-friendly examples from NCERT Class 10 Chapter 3.
How do you detect no solution algebraically?
Using the criterion a1/a2 = b1/b2 != c1/c2, the pair has no solution. Alternatively, during elimination, if you get a false statement like 0 = 5, the system is inconsistent. NCERT Class 10 Maths Chapter 3 explains this.
Can you solve word problems with these methods?
Yes. Translate each condition into a linear equation in two variables, then apply substitution or elimination. Many NCERT Class 10 Chapter 3 exercises are word problems from age, distance, digits, or money contexts.
Frequently Asked Questions — Pair of Linear Equations in Two Variables
What is Algebraic Methods in NCERT Class 10 Mathematics?
Algebraic Methods 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 Algebraic Methods step by step?
To solve problems on Algebraic Methods, 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 Algebraic Methods important for the Class 10 board exam?
Algebraic Methods 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 Algebraic Methods?
Common mistakes in Algebraic Methods 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 Algebraic Methods?
End-of-chapter NCERT exercises for Algebraic Methods 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.
AI Tutor
Mathematics Class 10
Ready