Pair of Linear Equations in Two Variables

Pair of Linear Equations in Two Variables

Pair of Linear Equations usually carries 4 to 6 marks in the board paper: 5 in the 2026-27 sample paper, 6 in the 2025 paper and in the second 2026 paper, and 4 in the main 2026 paper.

There is almost always a 1-mark MCQ on the conditions for a unique solution, no solution or infinitely many solutions. The rest comes as a 3-mark word problem, a 5-mark question (either "draw the graphs and find the triangle with an axis" or a word problem), or a 4-mark case study on sales or prices.

Where marks are usually lost:

• mixing up the ratio conditions, especially "no solution" (parallel) and "infinitely many" (coincident);
• in graph questions, not writing a table of values for each line, or reading the vertices of the triangle wrongly;
• defining the variables loosely in word problems (say clearly what $x$ and $y$ stand for);
• stopping after finding $x$ and $y$ when the question asks for something more.

Revise in 5 Minutes

Form: $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. Each equation is a straight line; a solution is a point on both lines.

Nature of solutions

Methods (cross-multiplication is not in the syllabus)

• Substitution: make $x$ or $y$ the subject in one equation and put it in the other.
• Elimination: multiply to make one coefficient equal, then add or subtract.
• Shortcut: if the coefficients are swapped ($ax + by$, $bx + ay$), add and subtract the equations.
• Graphical: make a table of at least 2 (better 3) points for each line, plot both on the same axes, and read the meeting point.

Graph with an axis: a line meets the $x$-axis where $y = 0$ and the $y$-axis where $x = 0$. For the triangle with the $x$-axis, base = distance between the two $x$-intercepts, height = distance of the meeting point from the $x$-axis, that is, $|y|$, area $= \frac{1}{2} \times$ base $\times$ height.

Word problems: state the variables with units, write two equations, solve, and answer exactly what is asked.

• Two-digit number: $10x + y$; reversed: $10y + x$.
• Boat: downstream $u + v$, upstream $u - v$.
• Ages: "$n$ years ago" subtract $n$ from both ages.

Traps

• "Consistent" includes the coincident case.
• A vertical line is $x = a$, a horizontal line is $y = b$.
• Check the answer in both original equations.

How to use this page: try each question on paper first, then read the answer. The marks against each step show what an examiner looks for. The 1-mark MCQs and Assertion-Reason questions are in the quiz at the end, together with questions that test how well you understand the chapter; every quiz answer comes with its explanation.

Short Answer Questions (2 and 3 Marks)

Question 1 (2 marks)

Solve for $x$ and $y$: $\frac{x}{3} + \frac{y}{4} = 4$ and $\frac{x}{2} - \frac{y}{4} = 1$.

Answer.

1. Clearing fractions: $4x + 3y = 48$ and $2x - y = 4$ — 1 mark
2. $y = 2x - 4$ gives $4x + 6x - 12 = 48$, so $x = 6$ and $y = 8$ — 1 mark

Question 2 (2 marks)

Solve for $x$ and $y$: $3x + 2y = 11$ and $2x + 3y = 4$.

Answer.

Model answer:

Look at the coefficients: they are 3, 2 in the first equation and 2, 3 in the second. So adding and subtracting the equations is quicker than the usual method.

Adding: $5x + 5y = 15$, so $x + y = 3$. … (1)

Subtracting the second from the first: $x - y = 7$. … (2)

Adding (1) and (2): $2x = 10$, so $x = 5$. From (1), $y = 3 - 5 = -2$.

Check: $3(5) + 2(-2) = 11$ and $2(5) + 3(-2) = 4$. So $x = 5$, $y = -2$.

Marking scheme:

1. Adding: $5x + 5y = 15$, so $x + y = 3$; subtracting: $x - y = 7$ — 1 mark
2. $2x = 10$, so $x = 5$ and $y = -2$ — 1 mark

Question 3 (2 marks)

Without drawing the graphs, find whether the lines representing each pair intersect, are parallel or coincide: (i) $5x - 3y = 11$ and $-10x + 6y = -22$ (ii) $3x - 4y = 7$ and $6x + 8y = 5$

Answer.

1. (i) $\frac{5}{-10} = \frac{-3}{6} = \frac{11}{-22} = -\frac{1}{2}$, so the lines coincide (infinitely many solutions) — 1 mark
2. (ii) $\frac{3}{6} = \frac{1}{2}$ but $\frac{-4}{8} = -\frac{1}{2}$, so $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ and the lines intersect (unique solution) — 1 mark

Question 4 (2 marks)

In a parallelogram ABCD, $\angle A = (2x + y)^\circ$, $\angle B = (x + 3y)^\circ$ and $\angle C = 70^\circ$. Find the values of $x$ and $y$.

Answer.

1. Opposite angles are equal: $2x + y = 70$; adjacent angles are supplementary: $\angle B = 110^\circ$, so $x + 3y = 110$ — 1 mark
2. $y = 70 - 2x$ gives $x + 210 - 6x = 110$, so $x = 20$ and $y = 30$ — 1 mark

Question 5 (2 marks)

The lines $kx + 2y = 12$ and $3x - y = 9$ meet at a point on the $x$-axis. Find the value of $k$.

Answer.

1. On the $x$-axis $y = 0$; from $3x - y = 9$, $x = 3$, so the point is $(3, 0)$ — 1 mark
2. $(3, 0)$ lies on $kx + 2y = 12$: $3k = 12$, so $k = 4$ — 1 mark

Question 6 (3 marks)

The sum of the digits of a two-digit number is 9. The number obtained by reversing its digits is 27 more than the original number. Find the number.

Answer.

1. Let the tens digit be $x$ and the units digit $y$: number $= 10x + y$, and $x + y = 9$ — 1 mark
2. $(10y + x) - (10x + y) = 27$ gives $9y - 9x = 27$, so $y - x = 3$ — 1 mark
3. $2y = 12$: $y = 6$, $x = 3$; the number is 36 — 1 mark

Question 7 (3 marks)

If 3 is added to the numerator of a fraction and 1 is subtracted from its denominator, the fraction becomes $\frac{5}{4}$. If 1 is subtracted from both the numerator and the denominator, it becomes $\frac{3}{4}$. Find the fraction.

Answer.

1. Let the fraction be $\frac{x}{y}$: $\frac{x + 3}{y - 1} = \frac{5}{4}$ gives $4x - 5y = -17$ — 1 mark
2. $\frac{x - 1}{y - 1} = \frac{3}{4}$ gives $4x - 3y = 1$ — 1 mark
3. Subtracting: $2y = 18$, $y = 9$ and $x = 7$; the fraction is $\frac{7}{9}$ — 1 mark

Question 8 (3 marks)

A motorboat on the Ganga at Varanasi goes 36 km downstream in 2 hours and comes back the same 36 km upstream in 3 hours. Find the speed of the boat in still water and the speed of the current.

Answer.

1. Let the boat's speed in still water be $u$ km/h and the current $v$ km/h; downstream $u + v = \frac{36}{2} = 18$ — 1 mark
2. Upstream $u - v = \frac{36}{3} = 12$ — 1 mark
3. $2u = 30$: $u = 15$ km/h and $v = 3$ km/h — 1 mark

Question 9 (3 marks)

Four years ago, Meena's mother was 5 times as old as Meena. Two years from now, she will be 3 times as old as Meena. Find their present ages.

Answer.

1. Let the present ages be: mother $x$ years, Meena $y$ years; $x - 4 = 5(y - 4)$ gives $x - 5y = -16$ — 1 mark
2. $x + 2 = 3(y + 2)$ gives $x - 3y = 4$ — 1 mark
3. Subtracting: $2y = 20$, $y = 10$, $x = 34$; mother 34 years, Meena 10 years — 1 mark

Question 10 (3 marks)

Solve the following pair of linear equations graphically: $x - y = 1$ and $2x + y = 8$.

Answer.

Model answer:

For $x - y = 1$, that is, $y = x - 1$:

For $2x + y = 8$, that is, $y = 8 - 2x$:

Plot these points and join them to get two straight lines on the same graph. The lines meet at the point $(3, 2)$.

So the solution is $x = 3$, $y = 2$. Check: $3 - 2 = 1$ and $2(3) + 2 = 8$.

Marking scheme:

1. Tables: $x - y = 1$ through $(1, 0)$, $(0, -1)$, $(3, 2)$; $2x + y = 8$ through $(4, 0)$, $(0, 8)$, $(3, 2)$ — 1 mark
2. Both lines plotted correctly on the same axes — 1 mark
3. The lines meet at $(3, 2)$, so $x = 3$, $y = 2$ — 1 mark

Long Answer and Case-Based Questions

Question 11 (5 marks)

Draw the graphs of the equations $2x + y = 6$ and $x - 2y + 2 = 0$. Find the coordinates of the vertices of the triangle formed by these two lines and the $x$-axis, shade the triangular region, and find its area.

Answer.

Model answer:

For $2x + y = 6$, that is, $y = 6 - 2x$:

For $x - 2y + 2 = 0$, that is, $y = \frac{x + 2}{2}$:

Plot the points and draw both lines on the same axes. They meet at $(2, 2)$.

The line $2x + y = 6$ meets the $x$-axis at $(3, 0)$ and the line $x - 2y + 2 = 0$ meets it at $(-2, 0)$. So the triangle has vertices A$(2, 2)$, B$(-2, 0)$ and C$(3, 0)$.

Its base BC lies on the $x$-axis: BC $= 3 - (-2) = 5$ units. The height is the distance of A from the $x$-axis, which is 2 units.

Area $= \frac{1}{2} \times 5 \times 2 = 5$ square units.

Marking scheme:

1. Table for $2x + y = 6$: $(0, 6)$, $(3, 0)$, $(2, 2)$ — 1 mark
2. Table for $x - 2y + 2 = 0$: $(-2, 0)$, $(0, 1)$, $(2, 2)$ — 1 mark
3. Both lines drawn correctly and the triangle shaded — 1 mark
4. Vertices: $(2, 2)$, $(3, 0)$ and $(-2, 0)$ — 1 mark
5. Base $= 3 - (-2) = 5$ units, height $= 2$ units; area $= \frac{1}{2} \times 5 \times 2 = 5$ square units — 1 mark

Question 12 (5 marks)

Kiran, a school teacher in Bhubaneswar, invested ₹ 25,000 in two schemes. The first pays simple interest at 8% per annum and the second at 10% per annum. Her total interest at the end of one year was ₹ 2,280. How much did she invest in each scheme? If she now moves ₹ 5,000 from the first scheme to the second, what will her interest for one year be?

Answer.

Model answer:

Let Kiran invest ₹ $x$ in the 8% scheme and ₹ $y$ in the 10% scheme.

Total investment: $x + y = 25000$. … (1)

Interest for one year: $\frac{8x}{100} + \frac{10y}{100} = 2280$. Multiplying by 100 and dividing by 2: $4x + 5y = 114000$. … (2)

Multiplying (1) by 4: $4x + 4y = 100000$. Subtracting this from (2): $y = 14000$. Then $x = 25000 - 14000 = 11000$.

Check: $8\%$ of 11000 is 880 and $10\%$ of 14000 is 1400, and $880 + 1400 = 2280$.

So she invested ₹ 11,000 at 8% and ₹ 14,000 at 10%.

After moving ₹ 5,000, the amounts are ₹ 6,000 at 8% and ₹ 19,000 at 10%. Interest $= 480 + 1900 = 2380$, so her yearly interest will be ₹ 2,380, that is, ₹ 100 more.

Marking scheme:

1. Let the amounts be ₹ $x$ (8%) and ₹ $y$ (10%): $x + y = 25000$ — 1 mark
2. $\frac{8x}{100} + \frac{10y}{100} = 2280$, that is, $4x + 5y = 114000$ — 1 mark
3. $4x + 4y = 100000$; subtracting, $y = 14000$ — 1 mark
4. $x = 11000$: ₹ 11,000 at 8% and ₹ 14,000 at 10% — 1 mark
5. New amounts ₹ 6,000 and ₹ 19,000; interest $= 480 + 1900 = 2380$, that is, ₹ 2,380 — 1 mark

Question 13 (4 marks)

A self-help group of women in a village near Nashik makes jute bags and cloth purses. Each jute bag is sold at ₹ $x$ and each cloth purse at ₹ $y$. In the first week they sold 30 bags and 20 purses and earned ₹ 4400. In the second week they sold 20 bags and 40 purses and earned ₹ 4000.

(i) Show that the second week's sales give $x + 2y = 200$. (1 mark)

Answer.

1. $20x + 40y = 4000$; dividing by 20, $x + 2y = 200$ — 1 mark

(ii) A member guesses that a bag costs ₹ 100 and a purse ₹ 70. Check her guess against both weeks. (1 mark)

Answer.

1. Week 1: $3000 + 1400 = 4400$ fits, but week 2: $2000 + 2800 = 4800 \neq 4000$; so the guess is wrong — 1 mark

(iii) Find the selling price of a jute bag and of a cloth purse. (2 marks)

Answer.

1. Week 1: $3x + 2y = 440$; subtracting $x + 2y = 200$: $2x = 240$, $x = 120$ — 1 mark
2. $2y = 80$, $y = 40$; bag ₹ 120, purse ₹ 40 — 1 mark

OR

(iii) In the third week the group sold 50 items in all (bags and purses) at the same prices, a bag for ₹ 120 and a purse for ₹ 40, and earned ₹ 4400. Form a pair of linear equations and find how many bags and how many purses they sold. (2 marks)

Answer.

1. Let $b$ bags and $p$ purses be sold: $b + p = 50$ and $120b + 40p = 4400$, that is, $3b + p = 110$ — 1 mark
2. Subtracting: $2b = 60$; 30 bags and 20 purses — 1 mark

Question 14 (4 marks)

Two cab services run in Lucknow. Service P charges a fixed ₹ 50 plus ₹ 10 per km. Service Q has no fixed charge and charges ₹ 20 per km. The figure shows the fare $y$ (in ₹) against the distance $x$ (in km) for both services; the two lines meet at E.

(i) Write the pair of linear equations for the two services. (1 mark)

Answer.

1. P: $y = 10x + 50$; Q: $y = 20x$ — 1 mark

(ii) For what distance do the two services charge the same fare? What is that fare? (1 mark)

Answer.

1. $20x = 10x + 50$ gives $x = 5$ km; fare ₹ 100 (point E is $(5, 100)$) — 1 mark

(iii) Riya has to travel 12 km. Which service is cheaper for her, and by how much? (2 marks)

Answer.

1. P: $10 \times 12 + 50 = 170$; Q: $20 \times 12 = 240$ — 1 mark
2. Service P is cheaper, by ₹ 70 — 1 mark

OR

(iii) Service Q changes its fare to a fixed ₹ 10 plus ₹ 15 per km. Find the distance at which the two services now charge the same fare, and that fare. (2 marks)

Answer.

1. New Q: $y = 15x + 10$; $15x + 10 = 10x + 50$ — 1 mark
2. $5x = 40$, $x = 8$ km; fare $= 10 \times 8 + 50 = 130$, that is, ₹ 130 — 1 mark