Chapter 4 of 9

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Telangana (TSBIE)Class 10 Mathematics★ The textbook's index allots this chapter 15 periods in September - joint largest in the book, tied with Real Numbers - and it runs from textbook page 77 to page 104. That is three numbered exercises (4.1 with nine questions, 4.2 with ten, 4.3 with eleven) plus an Optional Exercise of seven, fourteen worked examples, four Do This boxes, three Try This boxes and two Think and Discuss boxes. No marks distribution is printed anywhere in the volume, so no official weightage can be quoted. The period count is the only number the book itself gives, and it puts this chapter among the two largest in the year.

Pair of Linear Equations in Two Variables

Why one equation in two unknowns is never enough, what the three pictures of a pair of lines mean, how the ratios of the coefficients predict the answer before you draw anything, and the substitution and elimination methods that finish the job when the graph will not. Telangana SCERT Class 10 Mathematics Chapter 4.

⏱ 21 min readMedium difficulty✓ Free · TSBIE syllabus

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By the end of this chapter you'll be able to…

1State what makes an equation linear in two variables, including the condition that a squared plus b squared is not zero
2Explain why a single linear equation in two unknowns cannot determine both, and demonstrate it with two different guesses that both fit
3Solve a pair by the model method, scaling each row until one item appears the same number of times in both
4Draw the graph of a linear equation in two variables by tabulating at least three points, and read a solution off the intersection
5Classify a pair as intersecting, parallel or coincident from the graph, and name it consistent and independent, inconsistent, or consistent and dependent
6Predict the number of solutions from the ratios a1/a2, b1/b2 and c1/c2 without drawing anything
7Find the value of an unknown coefficient that makes a given pair parallel, coincident or uniquely solvable
8Solve a pair by the substitution method following the book's five steps, ending with the check in both original equations
9Solve a pair by the elimination method, choosing correctly between adding and subtracting according to the signs
10Translate word problems on ages, digits, fractions, money, geometry, mixtures and interest into a pair of linear equations
11Reduce a non-linear pair to a linear one by substituting for 1/x, 1/(x+y), 1/root x or a power, and convert back at the end
12Set up and solve speed, upstream-downstream and work-rate problems, in which the natural unknowns appear in denominators

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Why this chapter matters

This is the chapter where a student first meets the idea that information can be insufficient in a way you can prove rather than merely feel. One equation in two unknowns has infinitely many solutions, so Siri's single purchase genuinely cannot fix the price of a pen - not because she was careless, but because the mathematics does not allow it. The chapter then supplies three independent routes to the answer once a second equation arrives: a drawing, a graph, and two algebraic methods. Learning to move between those routes, and to predict from the coefficients alone which of the three outcomes you will get, is the transferable skill. Almost every later modelling problem in school mathematics and in physics reduces to a small system of equations, and the habit of asking 'do I have enough independent equations for the number of unknowns?' starts here.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Linear equations in one variable, and solving them by transposing terms, from class VII and VIII

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Linear equations in two variables and their graphs as straight lines, from class IX

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The Cartesian plane: axes, quadrants, ordered pairs and plotting points

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Ratio and proportion, including recognising when two ratios are equal

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LCM of two small numbers, used when matching coefficients in the elimination method

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Perimeter and area of a rectangle, and the place-value form of a two-digit number, both used in the word problems

Pair of Linear Equations in Two Variables

1. What This Chapter Covers

The chapter opens in a book shop. Siri buys 3 notebooks and 2 pens and her father pays Rs 80. Laxmi buys 4 notebooks and 3 pens of the same kind for Rs 110. Their classmates want to know what one pen costs and what one notebook costs. Siri does not know.

That is the whole chapter in one situation. Two quantities are unknown, and no single purchase pins them down.

The index allots this chapter 15 periods in September — joint largest in the book, tied with Real Numbers. It runs from textbook page 77 to page 104, with three numbered exercises and an Optional Exercise.

2. Why One Equation Is Never Enough

Rubina guesses that a notebook costs Rs 25. Then three notebooks cost Rs 75, the two pens must make up the remaining Rs 5, and each pen is Rs 2.50. Joseph thinks Rs 2.50 is far too little for a pen; he would put it at Rs 16, which forces the notebook to Rs 16 as well.

Both guesses fit Siri's purchase exactly. Neither is obviously wrong. The textbook's point is that there are many possible pairs of prices adding to Rs 80, and Siri's situation alone cannot choose between them.

So test the guesses against Laxmi. If Rubina is right, Laxmi's 4 notebooks and 3 pens come to 4 × 25 + 3 × 2.50, which is Rs 107.50. She actually paid Rs 110. If Joseph is right, Laxmi owes 4 × 16 + 3 × 16, which is Rs 112. Also wrong.

The second purchase is what does the work. The book states the rule plainly: when we have two variables, we need at least two independent linear equations to get a unique solution.

3. The Model Method

Before any algebra, the book solves the problem by drawing. Represent a notebook by one rectangle and a pen by another, and lay out both purchases as bars.

The trick is to scale the two rows until one item appears the same number of times in both. Multiply Siri's row by 3 and Laxmi's row by 2, and each row then holds exactly 6 pens.

Nine books and six pens cost Rs 240. Eight books and six pens cost Rs 220. The pens cancel, and the single extra book accounts for the Rs 20 difference. One book is Rs 20.

Now go back to Siri: three books at Rs 20 is Rs 60, so two pens cost Rs 20 and one pen is Rs 10. Checking against Laxmi, 4 × 20 + 3 × 10 = Rs 110, which is what she paid.

The book then gives the general definition. An equation of the form ax + by + c = 0, where a, b and c are real numbers and at least one of a and b is not zero — that is, a² + b² ≠ 0 — is a linear equation in two variables x and y.

4. What a Solution Means, and the Three Pictures

A solution for a pair of linear equations is a pair of values of x and y which together satisfy each one of the equations. Not one of them; both.

The graph of a linear equation in two variables is a straight line. Points on the line are solutions of that equation; points off it are not. So a solution of the pair is a point lying on both lines at once.

That reduces the whole question to geometry. Two lines drawn in the same plane can do only three things.

The textbook attaches names to these. A pair with exactly one solution is consistent and independent. A pair with no solution is inconsistent. A pair whose lines coincide is consistent and dependent, and has infinitely many solutions.

5. The Three Cases, Each From a Real Situation

The book does not present the three cases abstractly. It draws each one from a situation already on the page.

One solution. Writing the book-shop problem with x for the cost of a notebook and y for a pen gives 3x + 2y = 80 and 4x + 3y = 110. Plotting a few points of each produces two lines that cross once.

The crossing point is (20, 10). Substituting confirms it: 3(20) + 2(10) = 80 and 4(20) + 3(10) = 110. Both equations are satisfied, so a notebook is Rs 20 and a pen Rs 10 — exactly what the model method gave.

No solution. The first Think and Discuss situation has 1 kg of potatoes and 2 kg of tomatoes costing Rs 30 one day, and 2 kg of potatoes and 4 kg of tomatoes costing Rs 66 two days later. That is x + 2y = 30 and 2x + 4y = 66.

The lines are parallel, so there is no common solution. The book reads that back into the situation rather than treating it as a failure: it means the vegetables were not the same price on the two days, which is what happens in real markets.

Infinitely many. The cricket coach buys 3 bats and 6 balls for Rs 3900, then one more bat and 2 balls for Rs 1300. These give 3x + 6y = 3900 and x + 2y = 1300. The second is the first divided by 3, so the two lines coincide, every point on one is a point on the other, and the pair has infinitely many solutions. The second purchase told us nothing the first had not.

6. Reading the Answer off the Coefficients

Drawing graphs to find out which of the three cases you are in is slow. The book's section 4.2.3 gives a test that needs no drawing at all. Write the pair as a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, then compare the three ratios a₁/a₂, b₁/b₂ and c₁/c₂.

Check the test against the three situations already solved. The book shop gives 3/4 and 2/3, which differ, so the lines intersect. The vegetables give 1/2, 2/4 and −30/−66; the first two agree and the third does not, so the lines are parallel. The bats and balls give 3/1, 6/2 and 3900/1300, all equal to 3, so the lines coincide.

Worth pausing on why the test works. If a₁/a₂ = b₁/b₂, the two equations have proportional left-hand sides, so the lines have the same slope. Whether they are the same line or merely parallel then depends on whether the constants keep that same proportion.

Example-1 applies the test to 2x + y − 5 = 0 and 3x − 2y − 4 = 0. Here 2/3 is not 1/(−2), so the lines intersect and the pair is consistent. Tabulating points and plotting them gives the unique solution (2, 1).

Example-2 takes 3x + 4y = 2 and 6x + 8y = 4. All three ratios equal 1/2, so the lines are coincident and the pair has infinitely many solutions. The two tables of points in the book turn out identical, which is the arithmetic saying the same thing.

Example-3 takes 2x − 3y = 5 and 4x − 6y = 15. The first two ratios are both 2/1 while the third is 3/1, so the lines are parallel and the pair is inconsistent.

7. Word Problems Answered by Drawing

Example-4 is a puzzle. In a garden there are bees and flowers. If one bee sits on each flower, one bee is left over. If two bees sit on each flower, one flower is left over. How many of each?

Let x be the bees and y the flowers. One bee per flower leaving a bee spare gives x = y + 1, which is x − y − 1 = 0. Two bees per flower leaving a flower spare means the bees fill only x/2 flowers, so y = x/2 + 1, which rearranges to x − 2y + 2 = 0.

The lines cross at (4, 3), so there are 4 bees and 3 flowers. Test it against the words rather than the algebra: one bee on each of 3 flowers uses 3 bees and leaves 1 over, and two bees per flower fills 2 flowers and leaves 1 flower empty. Both conditions hold.

Example-5 is a rectangle whose perimeter is 32 m, so l + b = 16. Increasing the length by 2 m and decreasing the breadth by 1 m leaves the area unchanged, so (l + 2)(b − 1) = lb. Expanding gives lb − l + 2b − 2 = lb, and the lb terms cancel to leave l − 2b + 2 = 0. The lines meet at (10, 6): length 10 m, breadth 6 m.

8. The Substitution Method

Graphs run out of road quickly. The book is explicit about why: the method is not convenient in all cases where the point representing the solution has no integral co-ordinates. It offers √3 and 2√7, or −1.75 and 3.3, or 4/13 and 1/19 as solutions nobody will read off graph paper correctly.

So section 4.3 turns algebraic. The first method is substitution, in five steps.

Example-6 solves 2x − y = 5 and 3x + 2y = 11. The first rearranges to y = 2x − 5. Putting that into the second gives 3x + 2(2x − 5) = 11, so 7x = 21 and x = 3. Back-substituting, 2(3) − y = 5 gives y = 1.

Step 5 is not decoration. Testing (3, 1) in the equation that was not used for back-substitution gives 3(3) + 2(1) = 11, which is correct. A slip in step 1 or 2 survives every later step and only this check catches it.

9. The Elimination Method

Elimination removes a variable by making its coefficients match, then adding or subtracting. The book's step 3 is the one students get wrong: if the variable to be eliminated has the same sign in both equations, subtract; if the signs are opposite, add.

Example-7 takes 3x + 2y = 11 and 2x + 3y = 4. The coefficients of y are 2 and 3, whose LCM is 6, so multiply the first by 3 and the second by 2. That gives 9x + 6y = 33 and 4x + 6y = 8. The y terms have the same sign, so subtract: 5x = 25 and x = 5. Then 3(5) + 2y = 11 gives y = −2.

Example-8 does one problem by both methods so they can be compared. Rubina withdraws Rs 2000 in Rs 50 and Rs 100 notes and receives 25 notes in all. With x fifty-rupee notes and y hundred-rupee notes, x + y = 25 and 50x + 100y = 2000. Both routes give ten Rs 50 notes and fifteen Rs 100 notes.

Example-9 is a competitive exam. Three marks per correct answer and one deducted per wrong answer gave Madhu 40 marks; four per correct and two deducted would have given 50. With x correct and y wrong, 3x − y = 40 and 4x − 2y = 50, which solve to x = 15 and y = 5, so the test had 20 questions.

Example-10 is an age problem. Mary tells her daughter that seven years ago she was seven times as old, and that three years from now she will be three times as old.

From x − 7 = 7(y − 7) and x + 3 = 3(y + 3) come x − 7y + 42 = 0 and x − 3y − 6 = 0, giving Mary 42 and her daughter 12. Seven years back that is 35 and 5; three years on it is 45 and 15.

Example-11 is a break-even calculation. A publisher spends Rs 320000 in fixed costs plus Rs 31.25 a book, and receives Rs 43.75 a book. Setting 43.75x = 320000 + 31.25x leaves 12.5x = 320000, so the publisher breaks even at 25,600 books.

10. Equations That Are Not Linear Until You Rewrite Them

Section 4.4 handles pairs that are not linear as written but become linear after a substitution. The move is always the same: name the awkward block as a new variable.

Example-12 takes 2/x + 3/y = 13 and 5/x − 4/y = −2. These are not linear in x and y. But putting p = 1/x and q = 1/y turns them into 2p + 3q = 13 and 5p − 4q = −2, which are linear. Eliminating q gives 23p = 46, so p = 2 and q = 3, and therefore x = 1/2 and y = 1/3.

The last step is the one people forget. Solving for p and q is not solving for x and y; you must undo the substitution.

Example-13 is a work-rate problem. Six men and eight women finish a job in 14 days, and eight men and twelve women finish it in 10 days. Let x be the days one man alone would take and y the days one woman would take, so one man does 1/x of the work a day and one woman 1/y.

Ten days of eight men and twelve women completes the job, giving 80/x + 120/y = 1. Fourteen days of six men and eight women gives 84/x + 112/y = 1. Substituting u = 1/x and v = 1/y and eliminating gives 280v = 1, so v = 1/280 and u = 1/140. One man alone takes 140 days and one woman alone 280 days.

Example-14 mixes two speeds. A man covers 370 km partly by train and partly by car. Doing 250 km by train takes 4 hours in all; doing 130 km by train takes 18 minutes longer. With x the train's speed and y the car's, time equals distance over speed, so 250/x + 120/y = 4 and 130/x + 240/y = 43/10.

Substituting a = 1/x and b = 1/y gives 125a + 60b = 2 and 130a + 240b = 43/10. Eliminating b leaves 370a = 37/10, so a = 1/100 and b = 1/80. The train does 100 km/h and the car 80 km/h, which checks: 250/100 + 120/80 = 4 hours exactly.

11. What the Exercises Ask, and Where the Book's Answers Go Wrong

Exercise 4.1 has nine questions: three ratio classifications, nine consistency checks with graphs, and six word problems on pants and skirts, a quiz, pencils and pens, a garden, building a pair with a chosen relationship, and a rectangle's area. Exercise 4.2 has ten word problems. Exercise 4.3 has eight reducible pairs and three word problems, and the Optional Exercise adds six more pairs and a diet-mixture problem.

Answers for all three are printed at textbook pages 375 and 376. Every one of the eight answers to Exercise 4.3 question 1 is correct, as are all ten of Exercise 4.2 except one part. But three printed answers are wrong, and each is worth knowing before you use the key.

Exercise 4.1 question 2(f) gives x + y = 5 and 2x + 2y = 10, and the key answers "Inconsistent". Compare the ratios: 1/2, 1/2 and 5/10, which is 1/2. All three are equal, so by the book's own rule on page 104 the lines coincide. The pair is consistent and dependent with infinitely many solutions. The second equation is simply the first doubled.

Exercise 4.2 question 4(ii) is the Hyderabad taxi. A fixed charge covers the first 3 km and a per-kilometre rate applies beyond it; 10 km costs Rs 166 and 15 km costs Rs 256. The key's own first answer is right: 5 extra kilometres cost Rs 90, so the rate is Rs 18 and the fixed charge is 166 − 7(18) = Rs 40.

But it then prints Rs 490 for a 25 km journey. That is 40 + 25(18), which charges the first 3 km twice. A 25 km trip is 40 + 22(18) = Rs 436.

Exercise 4.3 question 2(iii) asks how long one woman and one man each take on an embroidery job, given that 2 women and 5 men finish in 4 days and 3 women and 6 men finish in 3 days. The key prints "man = 18, woman = 36", and the labels are swapped.

With a woman at 18 days and a man at 36, two women and five men do 2/18 + 5/36 = 1/4 of the job a day, which is the 4 days required. The key's labelling gives 1/3 and fails the condition it was built from.

There is also a wording slip in the key for question 3 of Exercise 4.1, which reports "Number of shirts = 0" for a question about skirts. The number is right.

Four things in the chapter text itself are worth flagging. The graph on page 87 illustrating Example-3 is labelled 4x − 6y = 9, but the line drawn is the right one for Example-3's 4x − 6y = 15 — it crosses the y-axis at −2.5 and carries the plotted points (0, −2.5), (3, −0.5) and (6, 1.5) straight from the table above it. Only the label is wrong.

Question 5 of the Do This box on page 92 prints 0.2x + 0.3y = 13 alongside 0.4x + 0.5y = 2.3. A decimal point has been dropped: with 1.3 the pair solves cleanly to x = 2 and y = 3, while with 13 it gives x = −290.5 and y = 237.

On page 99, inside Example-13, the line "the portion of work done by 8 women in one day is 8 × 1/x" should say 8 men, since 1/x was defined three lines earlier as one man's daily share. The algebra that follows is correct; only the words are wrong. Finally, the section numbering jumps from 4.2.1 straight to 4.2.3, with no 4.2.2 anywhere in the chapter.

12. Summary

Two unknowns need two independent equations. One equation in two variables has infinitely many solutions, so a single purchase, a single measurement or a single clue can never fix both quantities.

A pair of linear equations is two lines, and a solution is a point on both. Two lines can intersect once, run parallel, or coincide — giving one solution, none, or infinitely many, and named consistent and independent, inconsistent, and consistent and dependent.

The ratios of the coefficients settle which case applies without any drawing. If a₁/a₂ ≠ b₁/b₂ there is one solution; if the first two ratios agree but c₁/c₂ differs there is none; if all three agree there are infinitely many.

Graphs are honest but imprecise, so the chapter supplies two algebraic methods. Substitution writes one variable in terms of the other and reduces the pair to a single equation. Elimination matches a coefficient and then adds or subtracts, subtracting when the signs agree.

Pairs that are not linear can often be made linear by naming the awkward block. Setting 1/x = p, or 1/(x + y) = p, converts reciprocal, speed, work and mixture problems into ordinary pairs — provided you remember to convert back at the end.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Linear equation in two variables

ax + by + c = 0, where a, b, c are real and a^2 + b^2 is not 0

The condition a^2 + b^2 not equal to 0 is just a compact way of saying that a and b are not BOTH zero. If they were, nothing would be left but c = 0.

A pair of linear equations in two variables

a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0

Both equations must be put in this form, with everything on the left, before the ratio test can be used. Signs of c matter.

Condition for a unique solution

a1/a2 is not equal to b1/b2

The lines intersect at one point. The pair is consistent and independent. Nothing needs to be known about c1/c2.

Condition for no solution

a1/a2 = b1/b2, but not equal to c1/c2

The lines are parallel. The pair is inconsistent. Equal a and b ratios fix the slope; a different c ratio moves the line off to one side.

Condition for infinitely many solutions

a1/a2 = b1/b2 = c1/c2

The lines coincide. The pair is consistent and dependent. One equation is a multiple of the other and carries no new information.

What a solution is

a pair (x, y) satisfying BOTH equations at once

Graphically it is a point lying on both lines. A point on only one line solves only that equation.

Substitution method

from one equation write y in terms of x, substitute into the other, solve for x, then back-substitute

Choose the equation where a variable already has coefficient 1 or -1; it avoids fractions. Always finish with the check in both originals.

Elimination method

multiply each equation so one variable has matching coefficients, then subtract if the signs agree and add if they differ

Use the LCM of the two coefficients to decide the multipliers. The sign rule is where most marks are lost.

Reciprocal substitution

put 1/x = p and 1/y = q (or 1/(x+y) = p, 1/(x-y) = q, 1/root x = p)

Turns a reciprocal pair into a linear pair. The step everyone forgets is converting p and q back into x and y at the end.

Work rate

one worker who finishes alone in x days does 1/x of the job per day

Rates of several workers add. n men and m women working d days to finish gives d(n/x + m/y) = 1.

Time from distance and speed

time = distance / speed

In part-train part-car and upstream-downstream problems the unknown speeds land in denominators, so a reciprocal substitution is needed.

Two-digit number in place-value form

10a + b, with the digits reversed giving 10b + a

Their sum is 11(a + b), which is why the sum of a number and its reverse is always a multiple of 11.

Break-even condition

revenue = fixed cost + variable cost per unit times number of units

Example-11 sets 43.75x = 320000 + 31.25x. The break-even point is where the cost line and the revenue line cross.

Rectangle relations used in the word problems

perimeter = 2(l + b), area = lb

Half the perimeter is l + b, which is already a linear equation. Area conditions expand to something that cancels down to a linear equation.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Trying to solve one equation in two unknowns, and reporting the first pair that fits

✓ A single linear equation in two variables has infinitely many solutions. Rubina's Rs 25 notebook and Joseph's Rs 16 notebook both fit Siri's purchase exactly. You need a second independent equation before any answer is determined.

WATCH OUT

✗ Applying the ratio test before putting both equations in the form ax + by + c = 0

✓ Move every term to the left first. For 2x - 3y = 8 the constant is -8, not 8, and getting that sign wrong flips an inconsistent pair into a dependent one.

WATCH OUT

✗ Concluding 'no solution' whenever a1/a2 equals b1/b2

✓ Equal a and b ratios only mean the slopes match. You must then check c1/c2. If it matches too the lines coincide and there are infinitely many solutions; only if it differs are they parallel.

WATCH OUT

✗ Adding the equations in the elimination method when the signs of the chosen variable are the same

✓ Same sign means subtract; opposite signs mean add. Writing the two equations one above the other and marking the sign of the term to be removed before doing anything takes five seconds and prevents the commonest error in the chapter.

WATCH OUT

✗ Stopping at p and q after a reciprocal substitution

✓ p = 2 does not mean x = 2. It means 1/x = 2, so x = 1/2. In Example-12 the answers are x = 1/2 and y = 1/3 although p = 2 and q = 3.

WATCH OUT

✗ Setting up work problems with the daily output as the unknown, then reporting it as the number of days

✓ If x is the number of days one man takes alone, his daily share is 1/x. Solving gives 1/x, and the answer asked for is its reciprocal. Read the question again before writing the final line.

WATCH OUT

✗ Reading a graphical solution off the picture when the intersection is not at whole-number coordinates

✓ The book warns about this directly, citing solutions such as root 3 and 2 root 7, or 4/13 and 1/19. Use a graph to see which of the three cases you are in, and algebra to get the numbers.

WATCH OUT

✗ Plotting only two points for each line

✓ Two points determine a line, but they cannot reveal an arithmetic slip. Tabulate at least three; if they are not collinear, one of them is wrong.

WATCH OUT

✗ Forgetting that the taxi's fixed charge already covers the first 3 km

✓ Once the fixed charge is Rs 40 and the rate Rs 18, a 25 km trip costs 40 + 18 times 22, not 40 + 18 times 25. The textbook's own printed answer makes exactly this slip.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Pair of Linear Equations in Two Variables?

20 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

20 questions~14 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•One linear equation in two unknowns has infinitely many solutions, so two independent equations are needed to fix both
•A linear equation in two variables is ax + by + c = 0 with a and b not both zero, written compactly as a squared plus b squared not equal to zero
•A solution of a PAIR is a value pair satisfying BOTH equations, not either one
•The graph of a linear equation in two variables is a straight line, and points on it are exactly the solutions of that equation
•Two lines in a plane can only intersect once, be parallel, or coincide - there is no fourth possibility
•Intersecting means one solution, and the pair is called consistent and independent
•Parallel means no solution, and the pair is called inconsistent
•Coincident means infinitely many solutions, and the pair is called consistent and dependent
•If a1/a2 is not equal to b1/b2 there is exactly one solution, whatever c1 and c2 are
•If a1/a2 = b1/b2 but this is not equal to c1/c2, there is no solution
•If a1/a2 = b1/b2 = c1/c2 there are infinitely many solutions
•Both equations must be in the form ax + by + c = 0 before the ratios are compared, signs included
•The model method scales each row until one item appears equally often in both, so that item cancels
•In the book-shop problem a notebook is Rs 20 and a pen Rs 10, found by the model method, by graph and by algebra alike
•Graphs are unreliable when the intersection has non-integer coordinates, which is the book's stated reason for turning to algebra
•Substitution: express one variable from one equation, put it in the other, solve, back-substitute, then check in both originals
•Elimination: match one variable's coefficients using their LCM, then subtract if the signs agree and add if they differ
•Step 5 of substitution, checking in both original equations, is what catches an error made in step 1
•Reciprocal pairs become linear under 1/x = p and 1/y = q, and the answers must be converted back at the end
•A worker finishing alone in x days contributes 1/x of the job per day, and such rates add
•In Example-13 one man alone needs 140 days and one woman alone 280 days
•In Example-14 the train travels at 100 km/h and the car at 80 km/h
•Time equals distance over speed, which is why speed problems produce reciprocal equations
•A two-digit number is 10a + b and its reverse is 10b + a, so their sum is always a multiple of 11
•The textbook's printed answer key contains three wrong answers for this chapter and must be recomputed, not trusted
•Four text errors sit in the chapter itself: a graph label reading 4x - 6y = 9 for a line drawn as 4x - 6y = 15, a dropped decimal point in 0.2x + 0.3y = 13, the phrase '8 women' where the algebra uses 8 men, and a section numbering that skips 4.2.2

Telangana (TSBIE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: No marks distribution is printed in the textbook for this chapter or anywhere in the volume, so no total is claimed. The index allots 15 periods in September, joint largest in the book. The categories below are the book's own three numbered exercises plus its Optional Exercise and its in-chapter Do This, Try This and Think and Discuss boxes; the marks column indicates question size rather than official weightage. Answers for Exercises 4.1 to 4.3 are printed at textbook pages 375 and 376. All eight answers to Exercise 4.3 question 1 are correct, but THREE printed answers are wrong and should be recomputed rather than trusted: Exercise 4.1 question 2(f) is given as 'Inconsistent' although x + y = 5 and 2x + 2y = 10 have all three ratios equal to 1/2 and so are consistent, dependent and have infinitely many solutions; Exercise 4.2 question 4(ii) gives Rs 490 for a 25 km taxi ride, which double-charges the first 3 km that its own fixed charge of Rs 40 already covers, the correct figure being Rs 436; and Exercise 4.3 question 2(iii) labels the embroidery answer 'man = 18, woman = 36' when the values belong the other way round. The key also writes 'shirts' for 'skirts' in Exercise 4.1 question 3, though the numbers there are right.

Question type	Marks each	Typical count
Exercise 4.1	26	9
Exercise 4.2	32	10
Exercise 4.3	34	11
Optional Exercise	24	7
In-chapter boxes	18	10

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Costing and billing

Costing and billing, where a fixed charge plus a per-unit rate is exactly the pair the Hyderabad taxi question models

Break-even analysis in business

Break-even analysis in business, setting a revenue line against a cost line to find the number of units at which they cross

Blending and dilution in chemistry and pharmacy

Blending and dilution in chemistry and pharmacy, where two stock solutions must be combined to hit a target concentration

Labour planning on a site

Labour planning on a site, working out how long a job takes when teams of different productivity are combined

Journey planning across two modes of transport at differe…

Journey planning across two modes of transport at different speeds, and boat problems where the current adds to or subtracts from the speed

Portfolio splitting

Portfolio splitting, dividing a fixed amount between two instruments with different rates to hit a required overall return

Any calibration with two unknown constants

Any calibration with two unknown constants, such as fitting a straight-line response from two measured readings

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Put both equations into ax + by + c = 0 before touching the ratios; most classification marks are lost to a mishandled constant sign

For 'consistent or inconsistent' questions, the ratio test alone earns the classification marks - draw the graph only if the question says to

When a graph is demanded, tabulate at least three points per line and state the scale on the answer sheet, as every worked example in the book does

Name the outcome with the full phrase the book uses - consistent and independent, inconsistent, or consistent and dependent - rather than just 'one solution'

In word problems, write down what x and y stand for on the first line; unlabelled variables cost marks even when the arithmetic is right

After a reciprocal substitution, write the conversion back to x and y as a separate visible step so it cannot be forgotten under time pressure

Finish every algebraic solution by substituting into both original equations; it is the book's step 5, it is quick, and it catches the errors that method marks will not

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Show that the pair a1x + b1y = c1, a2x + b2y = c2 has a unique solution exactly when a1b2 - a2b1 is not zero, and derive Cramer's formulas x = (c1b2 - c2b1)/(a1b2 - a2b1) and y = (a1c2 - a2c1)/(a1b2 - a2b1)

STRETCH

Find all integer pairs satisfying a single linear equation such as 7x + 11y = 100, and characterise when solutions exist in terms of the gcd of the coefficients

STRETCH

For what values of k does the pair kx + 3y = k - 3 and 12x + ky = k have no solution? Explain why one root of the resulting quadratic in k must be rejected

STRETCH

Extend the three-case classification to three equations in two unknowns, and describe geometrically what makes such a system consistent

STRETCH

Prove that the sum of a two-digit number and its reverse is always divisible by 11, and find the corresponding statement for three-digit numbers

STRETCH

Generalise the work-rate method to three workers and three conditions, and say what goes wrong when the three equations are not independent

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

Telangana SSC public examination - Mathematics Paper I, where a graphical solution of a pair and a reducible-to-linear word problem are both standing items

Navodaya and Telangana residential school entrance tests, which lean heavily on the age, digit and fraction word problems from Exercise 4.2

NTSE and state-level mathematics talent tests, where the coefficient-ratio conditions appear as quick multiple-choice classification questions

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Because 3 notebooks and 2 pens costing Rs 80 is one equation in two unknowns, and such an equation has infinitely many solutions. The book demonstrates this rather than asserting it: Rubina's guess of Rs 25 a notebook and Rs 2.50 a pen fits perfectly, and so does Joseph's Rs 16 for both. Nothing in Siri's purchase can choose between them. Only when Laxmi's purchase supplies a second, independent equation do the two together pin the prices down to Rs 20 and Rs 10.

No. As soon as a1/a2 and b1/b2 differ, the two lines have different slopes, so they must cross somewhere and there is exactly one solution regardless of the constants. The c ratio only matters in the case where the first two ratios agree, because then the lines are certainly parallel or certainly identical, and c1/c2 decides which.

Yes, and the textbook asks this directly in a Think and Discuss box on page 83. Dependent means one equation is a multiple of the other, so both describe the same line. Every point on that line satisfies both equations, so solutions certainly exist - infinitely many of them. Consistent simply means at least one solution exists, so a dependent pair cannot fail to be consistent.

Substitution is easier when one variable already has coefficient 1 or -1, because expressing it costs nothing and creates no fractions. Elimination is easier when neither coefficient is 1 but the two coefficients of one variable share a small LCM, as with 2 and 3 in Example-7. Neither is more correct; the book solves Example-8 both ways to make that point. If asked for a specific method in an examination, use the one named.

Look only at the signs of the variable you are removing, after the multiplications. If it carries the same sign in both equations - say +6y and +6y - subtract one equation from the other. If the signs are opposite - say +12q and -12q - add them. The book states this as step 3 and then applies it visibly in Examples 7, 8, 10, 12 and 14, labelling each one 'same sign, so subtract' or 'opposite sign, so we add'.

Because the quantity you are asked for sits in a denominator. If one man alone takes x days, his daily contribution is 1/x, and combining several workers produces an equation in 1/x and 1/y rather than in x and y. The same happens with time equals distance over speed. Setting u = 1/x and v = 1/y makes the pair linear, and the final step converts back. Forgetting to convert back is the single commonest error in this section.

Mostly, but not entirely. All eight answers to Exercise 4.3 question 1 are right, and so are all of Exercise 4.2 except one part. But three printed answers are wrong: Exercise 4.1 question 2(f) is called inconsistent when its three ratios are all 1/2 and the lines coincide; Exercise 4.2 question 4(ii) gives Rs 490 for a 25 km taxi ride when its own fixed charge and rate give Rs 436; and Exercise 4.3 question 2(iii) has the man and woman labels swapped. Recompute before concluding your own work is wrong.

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Sets

What makes a collection a set, the two ways of writing one down, and then the whole vocabulary — empty set, universal set, subset, union, intersection, difference, equal sets, finite and infinite, and cardinality — with Venn diagrams doing the explaining throughout. Telangana SCERT Class 10 Mathematics Chapter 2.

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What a polynomial is and what its degree means, what a zero is and where it sits on the graph, the relationships that tie zeroes to coefficients for quadratics and cubics, and the division algorithm that lets one known zero unlock the rest. Telangana SCERT Class 10 Mathematics Chapter 3.

Quadratic Equations

What makes an equation quadratic and why you must simplify before deciding, the three ways to find its roots, how completing the square turns into the quadratic formula, and how the discriminant tells you the answer before you compute it. Telangana SCERT Class 10 Mathematics Chapter 5.

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