1. Introduction
A linear equation in two variables is an equation of the form \(ax + by + c = 0\), where \(x\) and \(y\) are the variables and \(a, b, c\) are real numbers with \(a\) and \(b\) not both zero.
For example, \(3x - 2y + 5 = 0\), \(x + y = 8\) and \(2x = 5y + 1\) are all linear equations in two variables. The word linear means that each variable appears only to the first power, so the graph of such an equation is always a straight line.
Definition
A pair of linear equations in two variables is a set of two linear equations considered together:
\[a_1x + b_1y + c_1 = 0\]
\[a_2x + b_2y + c_2 = 0\]
where \(a_1, b_1, c_1, a_2, b_2, c_2\) are real numbers and in each equation the two coefficients of the variables are not both zero. Throughout the chapter, \(a_1/a_2\), \(b_1/b_2\) and \(c_1/c_2\) (when \(a_2, b_2, c_2 \neq 0\)) are compared to decide the nature of the pair.
Why a pair? A single linear equation in two variables, such as \(x + y = 8\), has infinitely many solutions (\((1,7), (2,6), (3,5), \ldots\)) because we are free to choose one variable and then the other is forced. It is only when two such conditions are given together that the two unknowns can usually be pinned down to one specific value each — that is exactly the situation in most word problems.
Variable
The unknown quantities, usually written \(x\) and \(y\).
Coefficient
The numbers multiplying \(x\) and \(y\), named \(a_1, a_2\) and \(b_1, b_2\).
Constant
The number without a variable, named \(c_1\) or \(c_2\). It may be zero.
Line
Each equation represents a straight line on the \(xy\)-plane.
Common Mistake
Forgetting the condition that \(a\) and \(b\) are not both zero. For example, \(0 \cdot x + 2y - 6 = 0\) is still a linear equation (in fact \(y = 3\)); but an equation like \(0 \cdot x + 0 \cdot y + 5 = 0\) is neither linear nor valid, because no variable is present at all.
Exam Tip
Before solving any pair, rewrite every equation in the clean form \(ax + by + c = 0\) (or \(ax + by = k\)). This makes comparing coefficients and applying the ratio conditions far less error-prone.
2. Solution of a Pair of Linear Equations
Definition
An ordered pair \((x_0, y_0)\) is a solution of the pair of linear equations
\[a_1x + b_1y + c_1 = 0 \quad\text{and}\quad a_2x + b_2y + c_2 = 0\]
if substitution of \(x = x_0\), \(y = y_0\) makes both equations true simultaneously.
Verification of a solution
To verify, substitute the proposed values into each equation separately. Both left-hand sides must equal the respective right-hand sides.
Worked Example — Verification
Check whether \((2, -1)\) is a solution of \(2x + 3y = 1\) and \(x - y = 3\).
Equation 1: \(2(2) + 3(-1) = 4 - 3 = 1\) ✔
Equation 2: \(2 - (-1) = 2 + 1 = 3\) ✔
Because both equations are satisfied, \((2, -1)\) is a solution of the pair. A common check is that even one wrong substitution invalidates the whole answer.
Common Mistake
Checking the pair only in one equation. A pair is satisfied only when the values work in both equations.
Exam Tip
Always finish algebraic solutions with a quick verification in every equation. Examiners give marks for a verified answer and it catches 90% of sign errors.
3. Graphical Representation of a Pair of Linear Equations
Each linear equation \(ax + by + c = 0\) can be rearranged to \(y = -\dfrac{a}{b}x - \dfrac{c}{b}\) (when \(b \neq 0\)), which is the slope–intercept form \(y = mx + k\). Its graph is therefore a straight line. Plotting a line needs only two points: pick two convenient values of \(x\), compute the matching \(y\), and join the two points.
Worked Example — Plotting one line
Plot \(x + y = 6\).
Choose \(x = 0 \Rightarrow y = 6\), \(x = 6 \Rightarrow y = 0\), and \(x = 3 \Rightarrow y = 3\). The points \((0,6)\), \((6,0)\) and \((3,3)\) all lie on the same straight line. Join \((0,6)\) and \((6,0)\).
Interpretation: every point on this line is a solution of \(x + y = 6\) alone (for instance \((4,2)\) too, since \(4 + 2 = 6\)).
When two equations are drawn on the same axes, their two lines interact in exactly three possible ways. This gives us the whole theory of this chapter in one picture.
4. Graphical Method: The Three Cases
Case 1 — Intersecting lines: unique solution
Two lines that cross at one point represent a pair with exactly one solution — the coordinates of the meeting point. Such a pair is called consistent.
Interpretation: the point \((2,3)\) lies on both lines, so it satisfies both equations. Because two straight lines can cross only once, there is exactly one solution.
Case 2 — Parallel lines: no solution
Two lines with the same slope but different intercepts never meet. There is no point common to both lines, hence no solution. Such a pair is called inconsistent.
Interpretation: the lines \(x + 2y = 6\) and \(x + 2y = 12\) have the same slope, so they never meet. No pair of numbers can satisfy both equations.
Case 3 — Coincident lines: infinitely many solutions
Two lines that overlap completely are called coincident. Every point of the single line is common to both equations, giving infinitely many solutions. Such a pair is consistent and dependent.
Interpretation: \(x + 2y = 6\) and \(2x + 4y = 12\) are really the same condition (the second is just the first multiplied by 2). Both equations describe the same line, so every point of it is a solution.
| Situation | Lines | Number of solutions | Nature of the pair |
|---|---|---|---|
| Intersecting lines | Cross at one point | Exactly one (unique) | Consistent |
| Parallel lines | Never meet | None | Inconsistent |
| Coincident lines | Overlap completely | Infinitely many | Consistent (dependent) |
Common Mistake
Misreading a graph's intersection point. Always read the point by dropping perpendiculars to the two axes, and check that the coordinates satisfy both equations. Also, do not confuse parallel (no solution) with coincident (infinitely many) — parallel lines never touch, coincident lines are the same line.
Exam Tip
In graph questions, label both axes, mark the scales and write the point of intersection clearly. Labelling the two lines near them (for example near their ends) removes any ambiguity for the examiner.
5. Consistent and Inconsistent Pairs
Definitions
Consistent pair: a pair that has at least one solution — either a unique solution (intersecting lines) or infinitely many solutions (coincident lines).
Inconsistent pair: a pair that has no solution (parallel lines).
Key Result
A pair of linear equations is consistent if and only if the two lines intersect or coincide; it is inconsistent if and only if the two lines are parallel.
Unique solution
Intersecting lines. Consistent.
No solution
Parallel lines. Inconsistent.
Infinitely many
Coincident lines. Consistent and dependent.
6. Conditions for the Number of Solutions
Without drawing a graph, we can tell how many solutions a pair has by comparing the ratios of the coefficients of the two equations
\[a_1x + b_1y + c_1 = 0, \qquad a_2x + b_2y + c_2 = 0\]
| Ratio condition | Graphical meaning | Number of solutions | Nature of pair |
|---|---|---|---|
| \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\) | Intersecting lines | Exactly one | Consistent |
| \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\) | Parallel lines | None | Inconsistent |
| \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\) | Coincident lines | Infinitely many | Consistent (dependent) |
Exam Tip
Write the ratios in the fixed order \(a_1/a_2\), \(b_1/b_2\), \(c_1/c_2\) and compare them one at a time. First check whether \(a_1/a_2 = b_1/b_2\). If they differ, it is a unique solution. If they match, then also compare \(c_1/c_2\): equal means infinitely many, unequal means none.
Examples — Identifying the type without solving
Worked Example — Ratio test 1
Identify the type: \(3x + 2y = 8\) and \(6x + 4y = 16\).
\(\dfrac{a_1}{a_2} = \dfrac{3}{6} = \dfrac{1}{2}\), \(\dfrac{b_1}{b_2} = \dfrac{2}{4} = \dfrac{1}{2}\), \(\dfrac{c_1}{c_2} = \dfrac{8}{16} = \dfrac{1}{2}\).
All three ratios are equal, so the lines coincide and the pair has infinitely many solutions (consistent, dependent).
Worked Example — Ratio test 2
Identify the type: \(2x - 3y = 8\) and \(4x - 6y = 9\).
\(\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}\), \(\dfrac{b_1}{b_2} = \dfrac{-3}{-6} = \dfrac{1}{2}\), \(\dfrac{c_1}{c_2} = \dfrac{8}{9}\).
The first two ratios are equal but the third is not, so the lines are parallel and the pair is inconsistent (no solution).
Worked Example — Ratio test 3
Identify the type: \(x + y = 6\) and \(x - y = 2\).
\(\dfrac{a_1}{a_2} = \dfrac{1}{1} = 1\), \(\dfrac{b_1}{b_2} = \dfrac{1}{-1} = -1\).
Since \(1 \neq -1\), the lines intersect, giving a unique solution — noticed instantly, before any solving.
Common Mistake
Comparing ratios carelessly: e.g. comparing \(a_1/a_2\) with \(b_2/b_1\), or forgetting that \(c_1/c_2\) must be compared only when the first two ratios already match. Also, always reduce ratios to simplest form before judging equality.
7. Solving Algebraically
The two algebraic methods you must master are substitution and elimination. Both produce the same unique solution; which one you choose will depend on how the equations look.
7.1 Substitution Method
The idea: use one equation to express one variable in terms of the other, then substitute that expression into the second equation so that only one variable remains.
Worked Example — Substitution 1 (easy)
Solve: \(y = 2x\) and \(3x + y = 10\).
Step 1: \(y = 2x\) is already solved.
Step 2: substitute into \(3x + y = 10\): \(3x + 2x = 10\).
Step 3: \(5x = 10 \Rightarrow x = 2\).
Step 4: \(y = 2(2) = 4\).
Step 5: \(3(2) + 4 = 10\) ✔ and \(4 = 2(2)\) ✔. Solution: \((2, 4)\).
Worked Example — Substitution 2 (easy–moderate)
Solve: \(x = 2y + 1\) and \(3x - y = 8\).
Substitute \(x = 2y + 1\): \(3(2y + 1) - y = 8 \Rightarrow 6y + 3 - y = 8 \Rightarrow 5y = 5 \Rightarrow y = 1\).
Then \(x = 2(1) + 1 = 3\). Check: \(3(3) - 1 = 8\) ✔. Solution: \((3, 1)\).
Worked Example — Substitution 3 (moderate)
Solve: \(x + 2y = 8\) and \(2x - y = 1\).
From \(x + 2y = 8\), \(x = 8 - 2y\). Substitute: \(2(8 - 2y) - y = 1 \Rightarrow 16 - 4y - y = 1 \Rightarrow 5y = 15 \Rightarrow y = 3\).
Then \(x = 8 - 6 = 2\). Check: \(4 - 3 = 1\) ✔, \(2 + 6 = 8\) ✔. Solution: \((2, 3)\).
Worked Example — Substitution 4 (moderate)
Solve: \(4x + 3y = 18\) and \(x = 2y - 1\).
Substitute: \(4(2y - 1) + 3y = 18 \Rightarrow 8y - 4 + 3y = 18 \Rightarrow 11y = 22 \Rightarrow y = 2\).
Then \(x = 2(2) - 1 = 3\). Check: \(4(3) + 3(2) = 12 + 6 = 18\) ✔. Solution: \((3, 2)\).
Worked Example — Substitution 5 (fractions)
Solve: \(3x - 2y = 5\) and \(2x + 3y = 12\).
From \(3x - 2y = 5\), \(x = \dfrac{5 + 2y}{3}\). Substitute:
\[2\left(\frac{5 + 2y}{3}\right) + 3y = 12 \Rightarrow \frac{10 + 4y}{3} + 3y = 12\]
\[10 + 4y + 9y = 36 \Rightarrow 13y = 26 \Rightarrow y = 2\]
Then \(x = \dfrac{5 + 4}{3} = 3\). Check: \(2(3) + 3(2) = 6 + 6 = 12\) ✔. Solution: \((3, 2)\).
Common Mistake — Substitution
Substituting the expression back into the same equation it came from (which only repeats it), forgetting brackets when substituting (e.g. writing \(2x = y - 1\) as \(2x = y - 1\) correctly, but \(x = (y-1)/2\) wrongly as \(x = y - 1/2\)), and making sign errors when a variable had a negative coefficient.
7.2 Elimination Method
The idea: make the coefficients of one variable the same (or opposite) in both equations, then add or subtract to eliminate that variable.
Worked Example — Elimination 1 (easy)
Solve: \(x + y = 9\) and \(x - y = 5\).
The coefficients of \(y\) are \(+1\) and \(-1\), so add: \(2x = 14 \Rightarrow x = 7\).
Then \(7 + y = 9 \Rightarrow y = 2\). Check: \(7 - 2 = 5\) ✔. Solution: \((7, 2)\).
Worked Example — Elimination 2 (easy)
Solve: \(3x + 2y = 8\) and \(5x - 2y = 8\).
Coefficients of \(y\) are opposite, so add: \(8x = 16 \Rightarrow x = 2\).
Then \(3(2) + 2y = 8 \Rightarrow 2y = 2 \Rightarrow y = 1\). Check: \(5(2) - 2(1) = 8\) ✔. Solution: \((2, 1)\).
Worked Example — Elimination 3 (same coefficient)
Solve: \(2x + 3y = 13\) and \(2x + 7y = 25\).
The coefficients of \(x\) are the same, so subtract the first from the second: \(4y = 12 \Rightarrow y = 3\).
Then \(2x + 9 = 13 \Rightarrow x = 2\). Check: \(4 + 21 = 25\) ✔. Solution: \((2, 3)\).
Worked Example — Elimination 4 (scaling needed)
Solve: \(3x + 4y = 18\) and \(2x - 5y = -11\).
Multiply the first by \(5\) and the second by \(4\) to match the \(y\)-coefficients:
\[15x + 20y = 90, \qquad 8x - 20y = -44\]
Add: \(23x = 46 \Rightarrow x = 2\). Then \(3(2) + 4y = 18 \Rightarrow y = 3\).
Check: \(2(2) - 5(3) = 4 - 15 = -11\) ✔. Solution: \((2, 3)\).
Worked Example — Elimination 5 (fractions avoided by scaling)
Solve: \(5x + 2y = 19\) and \(3x - 4y = 1\).
Multiply the first by \(2\): \(10x + 4y = 38\). Now the \(y\)-coefficients are \(+4\) and \(-4\); add:
\[13x = 39 \Rightarrow x = 3\]
Then \(5(3) + 2y = 19 \Rightarrow y = 2\). Check: \(3(3) - 4(2) = 9 - 8 = 1\) ✔. Solution: \((3, 2)\).
Common Mistake — Elimination
Multiplying only one side of an equation, subtracting the wrong way round (sign errors), and forgetting to multiply the constant term when scaling an equation.
Choosing a method
Use substitution when one variable already has coefficient \(1\) or is already expressed in terms of the other (e.g. \(x = 2y + 1\)). Use elimination when coefficients are already matching or easy to match (e.g. \(2x + 3y = 13\), \(2x + 7y = 25\)), or when substitution would produce fractions immediately.
8. Simple Situational (Word) Problems
All word problems follow the same four-step discipline:
- Read the problem carefully and identify the two unknown quantities.
- Define the variables clearly (e.g. "Let the cost of one pen be ₹\(x\) and one pencil be ₹\(y\)").
- Translate each sentence into an equation — one sentence gives one equation.
- Solve the pair, then re-read the question to present the answer in context.
Application 1 — Two numbers
The difference of two numbers is 26, and one number is three times the other. Find the numbers.
Forming the equations: let the bigger number be \(x\) and the smaller be \(y\). "Difference is 26" gives \(x - y = 26\). "One is three times the other" gives \(x = 3y\).
Solving: substitute \(x = 3y\) into \(x - y = 26\): \(3y - y = 26 \Rightarrow 2y = 26 \Rightarrow y = 13\), and \(x = 39\).
Verify: \(39 - 13 = 26\) ✔ and \(39 = 3 \times 13\) ✔.
Answer: the numbers are 39 and 13.
Application 2 — Age problem
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their present ages.
Forming the equations: let Nuri's present age be \(x\) years and Sonu's be \(y\) years. Five years ago their ages were \(x - 5\) and \(y - 5\), and "Nuri was thrice as old" gives \(x - 5 = 3(y - 5)\). Ten years later their ages are \(x + 10\) and \(y + 10\), so \(x + 10 = 2(y + 10)\).
Solving: the first equation simplifies to \(x = 3y - 10\). The second simplifies to \(x = 2y + 10\). Equating: \(3y - 10 = 2y + 10 \Rightarrow y = 20\), then \(x = 50\).
Verify: 5 years ago: \(45 = 3 \times 15\) ✔. In 10 years: \(60 = 2 \times 30\) ✔.
Answer: Nuri is 50 and Sonu is 20 years old.
Application 3 — Cost of articles
Two pencils and three pens cost ₹54. Four pencils and two pens cost ₹68. Find the price of one pencil and one pen.
Forming the equations: let a pencil cost ₹\(x\) and a pen cost ₹\(y\). Then \(2x + 3y = 54\) and \(4x + 2y = 68\).
Solving (elimination): double the first equation: \(4x + 6y = 108\). Subtract the second: \(4y = 40 \Rightarrow y = 10\). Then \(2x + 30 = 54 \Rightarrow x = 12\).
Verify: \(4(12) + 2(10) = 68\) ✔.
Answer: a pencil costs ₹12 and a pen costs ₹10.
Application 4 — Event tickets
200 tickets were sold for a school event. Adult tickets cost ₹30 and child tickets cost ₹20, and the total collection was ₹4,800. How many of each were sold?
Forming the equations: let adults be \(x\) and children be \(y\). Then \(x + y = 200\) and \(30x + 20y = 4800\).
Solving (elimination): from \(x + y = 200\), \(y = 200 - x\). Then \(30x + 20(200 - x) = 4800 \Rightarrow 10x + 4000 = 4800 \Rightarrow x = 80\), and \(y = 120\).
Verify: \(30(80) + 20(120) = 2400 + 2400 = 4800\) ✔.
Answer: 80 adult and 120 child tickets.
Application 5 — Stationery
Five notebooks and three pens cost ₹190. Three notebooks and five pens cost ₹130. Find the cost of one notebook and one pen.
Forming the equations: \(5n + 3p = 190\) and \(3n + 5p = 130\) where \(n\) is a notebook and \(p\) is a pen (both in ₹).
Solving (elimination): multiply the first by 3 and the second by 5: \(15n + 9p = 570\), \(15n + 25p = 650\). Subtract: \(16p = 80 \Rightarrow p = 5\). Then \(5n + 15 = 190 \Rightarrow n = 35\).
Verify: \(3(35) + 5(5) = 105 + 25 = 130\) ✔.
Answer: a notebook costs ₹35 and a pen costs ₹5.
Application 6 — Perimeter and dimensions
The length of a rectangle exceeds its breadth by 4 m, and its perimeter is 40 m. Find the length and breadth.
Forming the equations: let length be \(l\) m and breadth be \(b\) m. Then \(l - b = 4\) and \(2(l + b) = 40\), i.e. \(l + b = 20\).
Solving (elimination): add the two equations: \(2l = 24 \Rightarrow l = 12\), then \(b = 8\).
Verify: \(2(12 + 8) = 40\) ✔, \(12 - 8 = 4\) ✔.
Answer: length 12 m, breadth 8 m.
Application 7 — Shopping bill
Three kg of apples and two kg of oranges cost ₹360. Two kg of apples and three kg of oranges cost ₹290. Find the price per kg of each fruit.
Forming the equations: let apples be \(x\) ₹/kg and oranges be \(y\) ₹/kg. Then \(3x + 2y = 360\) and \(2x + 3y = 290\).
Solving (elimination): multiplying the first by 2 and the second by 3 gives \(6x + 4y = 720\) and \(6x + 9y = 870\). Subtract: \(5y = 150 \Rightarrow y = 30\). Then \(3x + 60 = 360 \Rightarrow x = 100\).
Verify: \(2(100) + 3(30) = 290\) ✔.
Answer: apples ₹100/kg and oranges ₹30/kg.
Application 8 — Two-digit number
The sum of the digits of a two-digit number is 9. When the digits are interchanged, the new number is 27 more than the original number. Find the number.
Forming the equations: let the tens digit be \(x\) and the units digit be \(y\). The number is \(10x + y\), and after interchange it is \(10y + x\). Then \(x + y = 9\) and \((10y + x) - (10x + y) = 27 \Rightarrow 9y - 9x = 27 \Rightarrow y - x = 3\).
Solving (elimination): \(x + y = 9\) and \(-x + y = 3\). Add: \(2y = 12 \Rightarrow y = 6\), then \(x = 3\).
Verify: the number is 36; digits sum \(3 + 6 = 9\) ✔, and \(63 - 36 = 27\) ✔.
Answer: the number is 36.
Exam Tip
Every application problem must end with an interpreted sentence (for example "the shopkeeper sold 80 adult tickets"), never just with \(x = 80\). Always check that the answer makes real-world sense (positive, whole numbers where expected).
9. Competency-Based Examples
Competency questions ask you to reason, choose, interpret or judge rather than simply compute. The next examples cover each type.
Competency 1 — Choosing the correct equations
A shop sells pencils and erasers. Two pencils and three erasers cost ₹24; four pencils and two erasers cost ₹32. Which pair of equations correctly models this if a pencil costs ₹\(x\) and an eraser costs ₹\(y\)?
(a) \(2x + 3y = 24\), \(4x + 2y = 32\) (b) \(2x + 4y = 24\), \(3x + 2y = 32\) (c) \(3x + 2y = 24\), \(2x + 4y = 32\)
Answer: (a). Each sentence translates in order: "two pencils and three erasers cost 24" means \(2x + 3y = 24\) in the same order as the articles are named.
Competency 2 — Interpreting a graph
The graph shows the lines of two equations. Without solving, state the solution of the pair and the number of solutions it has.
Answer: the lines meet at \((2, 3)\), so the unique solution is \(x = 2\), \(y = 3\).
Competency 3 — Identifying consistency
For the pair \(2x + 3y = 7\) and \(4x + 6y = 14\), does the system have a unique solution, no solution, or infinitely many solutions? Justify without solving.
Answer: \(\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{7}{14} = \dfrac{1}{2}\); all ratios equal, so the lines coincide and there are infinitely many solutions.
Competency 4 — Unknown coefficient
If \(x = 2\), \(y = -1\) is a solution of \(3x + ky = 7\), find \(k\).
Answer: substitute: \(3(2) + k(-1) = 7 \Rightarrow 6 - k = 7 \Rightarrow k = -1\). Verify: \(6 - (-1) = 7\) ✔.
Competency 5 — Selecting the appropriate method
The pair \(2x + 5y = 1\) and \(2x - 5y = 19\) can be solved quickly. Which method is most efficient and what suggests it?
Answer: elimination, because the \(y\)-coefficients (\(5\) and \(-5\)) are already opposite. Adding gives \(4x = 20 \Rightarrow x = 5\); then \(2(5) + 5y = 1 \Rightarrow y = -\dfrac{9}{5}\).
Competency 6 — Detecting an incorrect solution
A student writes: "The pair \(x + y = 3\) and \(2x + 2y = 6\) has no solution because the equations are the same kind of line and parallel." Is the conclusion correct? If not, correct it.
Answer: the reasoning is wrong. The second equation is exactly 2 times the first, so the lines coincide, not parallel; the pair has infinitely many solutions.
Competency 7 — Interpreting a solution in context
For a profit-and-loss stall selling adult (₹30) and child (₹20) tickets, the solution of the equations was \(x = 80\) adult and \(y = 120\) child tickets. Why is this solution sensible for the context, and what check confirms the total collected was ₹4,800?
Answer: both values are positive whole numbers that satisfy \(x + y = 200\). The money check is \(30(80) + 20(120) = 4800\), matching the given collection, so the model and its answer are consistent with the real situation.
10. Case-Based Examples
Case-based questions give a small situation followed by two or three questions. Read the passage, extract the data, form the equations, then answer each part.
Case 1 — School canteen
In the school canteen, on Monday a student bought 4 samosas and 5 cold drinks for ₹155. On Tuesday the same student bought 2 samosas and 3 cold drinks for ₹85. The price of a samosa is ₹\(s\) and of a cold drink is ₹\(d\).
(i) Form the pair of linear equations. (ii) Find the price of one samosa and one cold drink. (iii) Verify your answer in both transactions.
Answers
(i) \(4s + 5d = 155\), \(2s + 3d = 85\).
(ii) Doubling the second equation: \(4s + 6d = 170\); subtracting the first gives \(d = 15\), then \(2s + 45 = 85 \Rightarrow s = 20\).
(iii) Monday: \(4(20) + 5(15) = 80 + 75 = 155\) ✔; Tuesday: \(2(20) + 3(15) = 40 + 45 = 85\) ✔.
Case 2 — Stationery shop
Riya buys 3 pens and 2 pencils for ₹64. Om buys 4 pens and 1 pencil for ₹72. The price of a pen is ₹\(p\) and of a pencil is ₹\(c\).
(i) Frame the equations. (ii) Find the cost of a pen and a pencil. (iii) How much do 5 pens and 3 pencils cost?
Answers
(i) \(3p + 2c = 64\), \(4p + c = 72\).
(ii) From the second, \(c = 72 - 4p\). Substitute: \(3p + 2(72 - 4p) = 64 \Rightarrow 3p + 144 - 8p = 64 \Rightarrow -5p = -80 \Rightarrow p = 16\), then \(c = 72 - 64 = 8\).
(iii) \(5(16) + 3(8) = 80 + 24 = ₹104\).
Case 3 — Annual day tickets
For the school annual day, 500 tickets were sold. Adult tickets cost ₹50 and student tickets cost ₹30, and the total collection was ₹21,000. Let adult tickets be \(x\) and student tickets be \(y\).
(i) Form the two equations. (ii) How many of each ticket type were sold? (iii) Give one check that proves your numbers match the total collection.
Answers
(i) \(x + y = 500\), \(50x + 30y = 21000\).
(ii) \(30(x + y) + 20x = 21000 \Rightarrow 15000 + 20x = 21000 \Rightarrow x = 300\), \(y = 200\).
(iii) \(50(300) + 30(200) = 15000 + 6000 = 21000\) ✔.
Case 4 — Classroom arrangement
If 3 students sit on each bench in a room, 2 students are left standing. If 4 students sit on each bench, exactly one bench is left empty (no one sits on it). Let \(S\) be the number of students and \(B\) the number of benches.
(i) Frame the two equations. (ii) Find the number of students and benches. (iii) Justify that each equation is satisfied by your values.
Answers
(i) With 3 per bench and 2 left: \(S = 3B + 2\). With 4 per bench and one empty bench, only \(B - 1\) benches are used: \(S = 4(B - 1)\).
(ii) Equate: \(3B + 2 = 4B - 4 \Rightarrow B = 6\), then \(S = 3(6) + 2 = 20\).
(iii) 20 students on 6 benches: \(3 \times 6 + 2 = 20\) ✔ and \(4 \times 5 = 20\) ✔ (6 benches, one empty).
Case 5 — Metro fare
A city metro charges a fixed base amount plus a fixed rate per kilometre. A 5 km journey costs ₹25 and an 8 km journey costs ₹37. If the base fare is \(a\) and the rate per km is \(b\):
(i) Form the two equations. (ii) Find the base fare and the rate per km. (iii) What would a 12 km journey cost?
Answers
(i) \(a + 5b = 25\), \(a + 8b = 37\).
(ii) Subtract: \(3b = 12 \Rightarrow b = 4\), then \(a = 25 - 20 = 5\).
(iii) \(a + 12b = 5 + 48 = ₹53\).
11. More Solved Examples
Basic solved examples
Verify that \((2, -1)\) satisfies both \(2x + 3y = 1\) and \(x - y = 3\).
\(2(2) + 3(-1) = 4 - 3 = 1\) ✔; \(2 - (-1) = 3\) ✔. Hence this ordered pair is a solution.
Without solving, state what the lines \(x + y = 6\) and \(x - y = 2\) do, and find the solution graphically.
Ratios: \(\dfrac{1}{1} \neq \dfrac{1}{-1}\), so the lines intersect — unique solution. Solving gives \(x = 4\), \(y = 2\) (both equations hold).
Solve by substitution: \(x + y = 14\), \(x - y = 4\).
From the second, \(x = y + 4\). Then \((y + 4) + y = 14 \Rightarrow 2y = 10 \Rightarrow y = 5\), \(x = 9\). Check: \(9 - 5 = 4\) ✔.
Solve by elimination: \(2x + 3y = 11\), \(2x - 4y = -24\).
Subtract the second from the first: \(7y = 35 \Rightarrow y = 5\). Then \(2x + 15 = 11 \Rightarrow x = -2\). Check: \(2(-2) - 4(5) = -4 - 20 = -24\) ✔.
Classify: \(3x + 2y = 8\), \(6x + 4y = 16\).
Ratios \(1/2\), \(1/2\), \(1/2\) — coincident lines, infinitely many solutions (consistent, dependent).
Moderate solved examples
Solve: \(3x - 5y = 4\), \(9x - 2y = 7\).
From the first, \(x = \dfrac{4 + 5y}{3}\). Substitute: \(9\cdot\dfrac{4 + 5y}{3} - 2y = 7 \Rightarrow 3(4 + 5y) - 2y = 7 \Rightarrow 12 + 13y = 7 \Rightarrow y = -\dfrac{5}{13}\), then \(x = \dfrac{4 - 25/13}{3} = \dfrac{9}{13}\). Check both equations. Solution: \(\left(\dfrac{9}{13}, -\dfrac{5}{13}\right)\).
Find \(k\) so that \(2x + 3y = 6\) and \(kx + 9y = 18\) have infinitely many solutions.
We need \(\dfrac{2}{k} = \dfrac{3}{9} = \dfrac{6}{18}\). The last two give \(\dfrac{1}{3}\), so \(\dfrac{2}{k} = \dfrac{1}{3} \Rightarrow k = 6\). Verify: \(2/(6) = 3/(9) = 6/18 = 1/3\) ✔.
Solve: \(7x + 15y = 9\), \(2x + 3y = 4\).
Multiply the second by 5: \(10x + 15y = 20\). Subtract from the first: \(-3x = -11 \Rightarrow x = \dfrac{11}{3}\). Then \(2\cdot\dfrac{11}{3} + 3y = 4 \Rightarrow 3y = -\dfrac{10}{3} \Rightarrow y = -\dfrac{10}{9}\). Check: \(7\cdot\dfrac{11}{3} + 15\cdot\left(-\dfrac{10}{9}\right) = \dfrac{77}{3} - \dfrac{50}{3} = 9\) ✔.
Solve: \(x + 6y = 27\), \(2x + 3y = 33\).
From the first, \(x = 27 - 6y\). Substitute: \(2(27 - 6y) + 3y = 33 \Rightarrow 54 - 9y = 33 \Rightarrow y = \dfrac{21}{9} = \dfrac{7}{3}\), then \(x = 27 - 14 = 13\). Check: \(13 + 14 = 27\) ✔; \(26 + 7 = 33\) ✔.
Do \(2x - 3y = 8\) and \(4x - 6y = 9\) have a solution? Justify.
\(\dfrac{2}{4} = \dfrac{-3}{-6} = \dfrac{1}{2}\) but \(\dfrac{8}{9} \neq \dfrac{1}{2}\). The lines are parallel, so the pair has no solution.
The sum of two numbers is 15 and their difference is 3. Find them.
Let the numbers be \(x\) and \(y\). Then \(x + y = 15\), \(x - y = 3\). Add: \(2x = 18 \Rightarrow x = 9\), \(y = 6\). Verify: \(9 + 6 = 15\) ✔, \(9 - 6 = 3\) ✔.
Solve: \(3x + 2y = 12\), \(5x - 2y = 4\).
Add: \(8x = 16 \Rightarrow x = 2\). Then \(6 + 2y = 12 \Rightarrow y = 3\). Check: \(5(2) - 2(3) = 10 - 6 = 4\) ✔.
12. Important Formulas and Results
Quick Formula Sheet
- General form: \(a_1x + b_1y + c_1 = 0\), \(a_2x + b_2y + c_2 = 0\).
- Unique solution: \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\).
- No solution: \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\).
- Infinitely many: \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\).
- Substitution: express one variable, substitute into the other equation.
- Elimination: equalise a coefficient, then add or subtract.
- Every solution must satisfy both equations; always verify.
13. Common Examination Mistakes
- Sign errors when moving terms across the equals sign, e.g. writing \(x - y = 6\) as \(y = x + 6\).
- Incorrect substitution — substituting back into the same equation instead of the other one, or losing brackets (writing \(2x = y - 1\) leads to \(x = \dfrac{y - 1}{2}\), not \(x = y - \dfrac{1}{2}\)).
- Incorrect elimination — multiplying a two-term equation and forgetting the constant, or subtracting in the wrong order.
- Forgetting to verify — an unverified answer can hide a sign slip.
- Misreading graph intersections — reading the wrong coordinate or not dropping perpendiculars to both axes.
- Confusing no solution with infinitely many — parallel lines give zero solutions; coincident lines give infinitely many.
- Incorrect ratio conditions — comparing ratios in the wrong order, or failing to reduce them to simplest form.
- Forming incorrect equations from word problems — for example adding the collections instead of equating them, or mixing up which quantity is bigger in "exceeds by".
14. Exam Tips
- Define variables clearly in word problems, stating units ("Let the cost of a pencil be ₹\(x\)").
- Write equations carefully before rushing to solve; one sentence of the problem gives one equation.
- Show algebraic steps — examiners award stepwise marks.
- Verify answers in both equations before writing the final answer.
- Interpret the answer in context with a concluding sentence.
- Label graph axes and scales and mark the intersection point when asked to solve graphically.
- Check consistency conditions carefully (ratio test) even when solving algebraically — it confirms the number of solutions you should get.
15. Quick Revision Sheet
Scan this before the exam
- Standard form: \(a_1x + b_1y + c_1 = 0\), \(a_2x + b_2y + c_2 = 0\).
- Each equation is a straight line.
- Intersecting \(\Rightarrow\) unique solution \(\Rightarrow\) consistent.
- Parallel \(\Rightarrow\) no solution \(\Rightarrow\) inconsistent.
- Coincident \(\Rightarrow\) infinitely many \(\Rightarrow\) consistent, dependent.
- Unique: \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\).
- None: \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\).
- Infinite: \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\).
- Substitution: express one variable \(\to\) substitute \(\to\) solve \(\to\) back-substitute \(\to\) verify.
- Elimination: equalise a coefficient \(\to\) add/subtract \(\to\) solve \(\to\) back-substitute \(\to\) verify.
- Word problems: define \(\to\) translate \(\to\) solve \(\to\) interpret \(\to\) verify.
16. Self-Practice Questions
25 original questions to test your understanding. Do not treat these as board PYQs — they are designed as self-assessment, with answers at the end.
Conceptual (Q1 – Q5)
Q1.
Write the general form of a pair of linear equations in two variables, and state when it represents a valid pair.
Q2.
What does the solution of a pair of linear equations represent graphically?
Q3.
A pair of linear equations is consistent and dependent. What does this tell you about the lines?
Q4.
If \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\), how many solutions does the pair have?
Q5.
Define an inconsistent pair of linear equations.
Graphical (Q6 – Q10)
Q6.
For \(x + y = 4\) and \(x - y = 2\), identify the type of graph and find the solution.
Q7.
Classify \(2x + 3y = 6\) and \(4x + 6y = 9\) without drawing.
Q8.
Classify \(3x - y = 6\) and \(6x - 2y = 12\) without drawing.
Q9.
If two lines intersect at the point \((2,1)\), what is the solution of the pair? Give one example of a pair of equations whose lines intersect there.
Q10.
State the ratio condition under which a pair of linear equations has no solution.
Algebraic (Q11 – Q15)
Q11.
Solve by substitution: \(x + y = 5\), \(2x - 3y = 4\).
Q12.
Solve: \(3x + 4y = 10\), \(2x - 2y = 2\).
Q13.
Solve: \(2u + v = 7\), \(u - 2v = 1\).
Q14.
Solve: \(2x + 3y = 8\), \(x - y = -1\).
Q15.
Solve by elimination: \(3x - y = 3\), \(2x + y = 7\).
Application (Q16 – Q20)
Q16.
The sum of two numbers is 40 and one number is 8 more than the other. Find the numbers.
Q17.
A father is 30 years older than his son. In 12 years, the father will be twice as old as the son. Find their present ages.
Q18.
4 kg of rice and 3 kg of dal cost ₹300; 2 kg of rice and 5 kg of dal cost ₹290. Find the price per kg of each.
Q19.
The perimeter of a rectangle is 44 m and its length is 6 m more than its breadth. Find the dimensions.
Q20.
In a test of 50 questions, a student scores 4 marks for each correct answer and loses 1 mark for each wrong answer. She scores 190. How many did she get correct?
Competency / Reasoning (Q21 – Q23)
Q21.
Ravi says the pair \(2x + 3y = 6\) and \(4x + 6y = 12\) has infinitely many solutions because the two equations describe the same line. Is Ravi correct?
Q22.
For what value of \(k\) does the pair \(3x - y = 1\) and \(kx + 2y = 5\) have a unique solution? State the condition on \(k\).
Q23.
A graph shows two lines meeting at \((1, 2)\). Write the solution of the corresponding pair and one example of two equations with that intersection.
Case-based (Q24 – Q25)
Q24.
An auto-rickshaw charges a fixed amount plus a rate per kilometre. A 2 km trip costs ₹40 and a 5 km trip costs ₹85. If the base fare is \(a\) and the per-km rate is \(b\): (i) form the equations; (ii) find \(a\) and \(b\); (iii) predict the fare for an 8 km trip.
Q25.
At a school fair, 4 samosas and 5 drinks cost ₹155, while 2 samosas and 3 drinks cost ₹85. (i) Frame the equations. (ii) Find the price of a samosa and a drink. (iii) What would 3 samosas and 2 drinks cost?
Answers
A1.
\(a_1x + b_1y + c_1 = 0\), \(a_2x + b_2y + c_2 = 0\), with \(a_1, b_1\) not both zero and \(a_2, b_2\) not both zero.
A2.
The coordinates of the point (or points) common to the two lines — i.e., their intersection when the lines meet.
A3.
The lines coincide, so the pair has infinitely many solutions.
A4.
No solution — the system is inconsistent (parallel lines).
A5.
A pair of linear equations that has no common solution — its lines are parallel.
A6.
Intersecting lines (unique solution). Solving: \(2x = 6 \Rightarrow x = 3\), \(y = 1\). Solution \((3, 1)\).
A7.
\(\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{1}{2}\) but \(\dfrac{6}{9} = \dfrac{2}{3}\), so parallel lines — no solution.
A8.
\(\dfrac{3}{6} = \dfrac{-1}{-2} = \dfrac{1}{2}\) and \(\dfrac{6}{12} = \dfrac{1}{2}\), all equal — coincident lines, infinitely many solutions.
A9.
Solution is \(x = 2\), \(y = 1\). One example: \(x + y = 3\) and \(x - y = 1\) (both hold at \((2,1)\)).
A10.
\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\).
A11.
\(x = \dfrac{19}{5}\), \(y = \dfrac{6}{5}\). Check: \(2(19/5) - 3(6/5) = (38 - 18)/5 = 4\) ✔.
A12.
\(x = 2\), \(y = 1\). Check: \(6 + 4 = 10\) ✔, \(4 - 2 = 2\) ✔.
A13.
\(u = 3\), \(v = 1\). Check: \(6 + 1 = 7\) ✔, \(3 - 2 = 1\) ✔.
A14.
\(x = 1\), \(y = 2\). Check: \(2 + 6 = 8\) ✔, \(1 - 2 = -1\) ✔.
A15.
Add: \(5x = 10 \Rightarrow x = 2\), \(y = 3\). Check: \(4 + 3 = 7\) ✔.
A16.
24 and 16. Check: \(24 + 16 = 40\), \(24 - 16 = 8\).
A17.
Father 48, son 18. Check: \(48 + 12 = 2 \times 30\) ✔.
A18.
Rice ₹45/kg, dal ₹40/kg. Check: \(4(45) + 3(40) = 300\) ✔, \(2(45) + 5(40) = 290\) ✔.
A19.
Length 14 m, breadth 8 m. Check: \(2(14 + 8) = 44\) ✔, \(14 - 8 = 6\) ✔.
A20.
48 correct, 2 wrong. Check: \(4(48) - 2 = 190\) ✔.
A21.
Yes. The second equation is the first multiplied by 2, so the lines coincide and there are infinitely many solutions.
A22.
Unique solution needs \(\dfrac{3}{k} \neq -\dfrac{1}{2}\), i.e. \(k \neq -6\).
A23.
Solution \(x = 1\), \(y = 2\). Example: \(x + y = 3\), \(x - y = -1\) (both true at \((1,2)\)).
A24.
(i) \(a + 2b = 40\), \(a + 5b = 85\). (ii) \(b = 15\), \(a = 10\). (iii) \(10 + 8 \times 15 = ₹130\).
A25.
(i) \(4s + 5d = 155\), \(2s + 3d = 85\). (ii) Samosa ₹20, drink ₹15. (iii) \(3(20) + 2(15) = ₹90\).