Chapter 3 · Pair of Linear Equations in Two Variables

3.1 Pairs of Linear Equations and the Graphical Method

Two conditions, two unknowns: write each condition as a straight line, draw both, and the point where they meet is the answer. The coefficients even tell you in advance whether the lines will meet at all.

At the school sports-day stall, $2$ samosas and $3$ cups of tea cost ₹ $80$, while $3$ samosas and $2$ cups of tea cost ₹ $95$. How much is one samosa? One condition alone cannot tell you: "$2$ samosas and $3$ teas cost ₹ $80$" is true for a samosa at ₹ $25$ and tea at ₹ $10$, but also for a samosa at ₹ $10$ and tea at ₹ $20$, and for many other pairs of prices. You need both conditions at once. Two unknowns tied together by two conditions make a pair of linear equations in two variables.

In Class 9 you learnt that the graph of one linear equation in $x$ and $y$ is a straight line. So a pair of linear equations is a pair of straight lines on the same graph paper, and a point that lies on both lines satisfies both conditions. In this section we solve pairs of equations by drawing their lines, and we learn a quick test, using only the coefficients, that tells us whether the lines will cross, run side by side, or lie on top of each other. Sections 3.2 and 3.3 then give exact algebraic methods.

Pairs of linear equations and their solutions

📘 Definition 3.1 — Linear equation in two variables; a pair of them

An equation that can be written in the form $$ax+by+c=0,$$ where $a$, $b$, $c$ are real numbers and $a$, $b$ are not both zero, is called a linear equation in two variables $x$ and $y$. Its graph is a straight line.
Two such equations in the same two variables, considered together, $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2=0,$$ form a pair of linear equations in two variables. This is called the general form of the pair.

An equation such as $3x=7-2y$ is also linear: move everything to one side to get $3x+2y-7=0$, so $a=3$, $b=2$, $c=-7$. Watch the sign of $c$: it is the number on the left once the right-hand side is $0$.

📘 Definition 3.2 — Solution of a pair

A solution of a pair of linear equations is a pair of values $x=x_0$, $y=y_0$ that makes both equations true at the same time. We write it as the point $(x_0, y_0)$. Graphically, it is a point that lies on both lines.

✏️ Example 1 — Is it a solution?

Which of the points $(2, 3)$, $(4, 1)$ and $(1, 4)$ is a solution of the pair $x+y=5$ and $2x-3y=-5$?

💡 Show solution▶

Method Substitute the point into each equation. It is a solution only if both come out true.

(2, 3) $x+y=2+3=5$ ✓ and $2x-3y=4-9=-5$ ✓. Both true, so $(2,3)$ is a solution.

(4, 1) $4+1=5$ ✓ but $2(4)-3(1)=8-3=5\ne-5$ ✗. It lies on the first line only. Not a solution.

(1, 4) $1+4=5$ ✓ but $2(1)-3(4)=2-12=-10\ne-5$ ✗. Not a solution.

Lesson All three points satisfy the first equation; only one satisfies both. Checking only one equation is the most common slip.

✏️ Example 2 — Turning words into a pair of equations

Write the sports-day problem from the start of this section as a pair of linear equations: $2$ samosas and $3$ teas cost ₹ $80$; $3$ samosas and $2$ teas cost ₹ $95$. Then check that a samosa at ₹ $25$ and a tea at ₹ $10$ fits both conditions.

💡 Show solution▶

Name the unknowns Let one samosa cost ₹ $x$ and one cup of tea cost ₹ $y$.

Translate First condition: $2x+3y=80$. Second condition: $3x+2y=95$. In general form: $2x+3y-80=0$ and $3x+2y-95=0$.

Check With $x=25$, $y=10$: $2(25)+3(10)=50+30=80$ ✓ and $3(25)+2(10)=75+20=95$ ✓.

Answer The pair is $2x+3y=80$, $3x+2y=95$, and $(25, 10)$ is a solution: a samosa costs ₹ $25$ and a tea ₹ $10$. (In Section 3.3 you will find these prices without being told them.)

The graphical method

Every solution of $x+y=7$ is a point on one straight line, and every solution of $x-y=1$ is a point on another. A solution of the pair must be on both lines, so it is the point where they cross. To draw a line you need only two points, but always plot a third as a check: if the three points do not lie on one straight line, you have made an arithmetic slip.

🧭 Problem-solving strategy — solving a pair graphically

For each equation make a small table of three points. Easy choices: put $x=0$ and find $y$; put $y=0$ and find $x$; then pick one more value of $x$ that gives a whole-number $y$.
Draw both axes with the same scale, plot the points, and join each set of three with a ruler. Label each line with its equation.
Read the point where the lines meet. That point $(x_0, y_0)$ is the solution.
Check by substituting $(x_0, y_0)$ into both equations. A graph can only be read approximately, so the check is what makes the answer certain.

✏️ Example 3 — Lines that cross

Solve graphically: $x+y=7$ and $x-y=1$.

💡 Show solution▶

Table for x + y = 7 Write $y=7-x$. $$\begin{array}{c|ccc}x&0&7&3\\\hline y&7&0&4\end{array}$$

Table for x − y = 1 Write $y=x-1$. $$\begin{array}{c|ccc}x&0&1&4\\\hline y&-1&0&3\end{array}$$

Draw and read Plot $(0,7)$, $(7,0)$, $(3,4)$ and join them; plot $(0,-1)$, $(1,0)$, $(4,3)$ and join them. The two lines meet at $(4, 3)$. (Interactive 1 below draws exactly this graph, point by point.)

Check $4+3=7$ ✓ and $4-3=1$ ✓. The solution is $x=4$, $y=3$.

✏️ Example 4 — Lines that never meet, and lines that coincide

Draw the graphs and decide how many solutions each pair has.
(a) $2x+y=6$ and $4x+2y=4$ (b) $x-2y=-2$ and $3x-6y=-6$

💡 Show solution▶

(a) Tables $2x+y=6$: $(0,6)$, $(3,0)$, $(1,4)$. For $4x+2y=4$ divide by $2$ first: $2x+y=2$, giving $(0,2)$, $(1,0)$, $(2,-2)$.

(a) Graph Both lines go down $2$ units for every $1$ unit to the right, so they have the same steepness, but they cut the $y$-axis at different points ($6$ and $2$). They are parallel and never meet: no solution. Indeed $2x+y$ cannot equal $6$ and $2$ at the same time.

(b) Tables $x-2y=-2$: $(0,1)$, $(-2,0)$, $(2,2)$. For $3x-6y=-6$: $(-2,0)$, $(4,3)$, $(6,4)$.

(b) Graph All six points lie on one line. The second equation is just $3$ times the first ($3x-6y=-6$ is $3(x-2y)=3(-2)$), so the two lines coincide. Every point on the line is a solution: infinitely many solutions, for example $(0,1)$, $(2,2)$, $(4,3)$, $(6,4)$, …

So two lines in a plane can behave in only three ways, and each way gives a different number of solutions.

The two lines	Number of solutions	The pair is called
intersect (cross at one point)	exactly one	consistent
coincide (are the same line)	infinitely many	dependent (and consistent)
are parallel (never meet)	none	inconsistent

📘 Definition 3.3 — Consistent, inconsistent and dependent pairs

A pair of linear equations is consistent if it has at least one solution, and inconsistent if it has no solution.
A consistent pair with infinitely many solutions (the two equations describe the same line) is called dependent.

Predicting the answer from the coefficients

Drawing graphs takes time, and a slightly wrong ruler line can hide the truth. Look again at Example 4. In (a), $\dfrac{a_1}{a_2}=\dfrac24=\dfrac12$ and $\dfrac{b_1}{b_2}=\dfrac12$, but the constants $-6$ and $-4$ are in the ratio $\dfrac{3}{2}$. In (b) all three ratios are $\dfrac13$. In Example 3, $\dfrac{a_1}{a_2}=\dfrac11=1$ but $\dfrac{b_1}{b_2}=\dfrac{1}{-1}=-1$. Comparing these three ratios is enough to know the answer before drawing anything.

📐 Theorem 3.1 — The ratio test for a pair of linear equations

Consider the pair $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$, written in general form, where $a_2$, $b_2$, $c_2$ are all non-zero (so that the ratios below make sense).

If $\dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2}$, the lines intersect: the pair has exactly one solution (consistent).
If $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$, the lines coincide: the pair has infinitely many solutions (dependent, consistent).
If $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}$, the lines are parallel: the pair has no solution (inconsistent).

Why it works. Suppose $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=k$, so $a_1=ka_2$ and $b_1=kb_2$. Then the left side of the first equation is $k(a_2x+b_2y)+c_1$.

If also $c_1=kc_2$, the first equation is exactly $k$ times the second, so the two equations describe the same line.
If $c_1\ne kc_2$: at a common point we would have $a_2x+b_2y=-c_2$ from the second equation, and then the first would give $-kc_2+c_1=0$, i.e. $c_1=kc_2$, which is false. So there is no common point: the lines are parallel.
If $\dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2}$, the two lines have different steepness, so they are not parallel and must cross at exactly one point. Interactive 2 lets you watch all three cases happen.

⚠️ Common mistakes with the ratio test

1. Not writing both equations in the same form first. For $3x=2y+4$ and $6y=9x-12$ you must first rearrange to $3x-2y-4=0$ and $9x-6y-12=0$. Comparing $3$ with $6$ (the coefficient of $y$!) gives nonsense.

2. Losing signs. In $x-4y=6$ the constant in general form is $c=-6$, and $b=-4$. Keep the signs inside the ratios: $\dfrac{-4}{-12}=\dfrac13$.

3. Stopping after two ratios. $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$ only tells you "parallel or coincident". You must compare $\dfrac{c_1}{c_2}$ as well to decide which.

✏️ Example 5 — Classify without drawing

Without drawing the graphs, decide whether the lines intersect, are parallel or coincide, and say whether the pair is consistent.
(a) $4x-3y+7=0$, $5x+2y-6=0$ (b) $6x+4y-10=0$, $9x+6y-15=0$ (c) $2x-5y+3=0$, $6x-15y+4=0$ (d) $3x=2y+4$, $6y=9x-12$

💡 Show solution▶

(a) $\dfrac{a_1}{a_2}=\dfrac45$, $\dfrac{b_1}{b_2}=\dfrac{-3}{2}$. These are different, so the lines intersect: exactly one solution, consistent.

(b) $\dfrac69=\dfrac23$, $\dfrac46=\dfrac23$, $\dfrac{-10}{-15}=\dfrac23$. All equal, so the lines coincide: infinitely many solutions, consistent (dependent). Check: $6x+4y-10=2(3x+2y-5)$ and $9x+6y-15=3(3x+2y-5)$.

(c) $\dfrac26=\dfrac13$, $\dfrac{-5}{-15}=\dfrac13$, but $\dfrac{c_1}{c_2}=\dfrac34$. So the lines are parallel: no solution, inconsistent.

(d) Rearrange first: $3x-2y-4=0$ and $-9x+6y+12=0$. Ratios: $\dfrac{3}{-9}=-\dfrac13$, $\dfrac{-2}{6}=-\dfrac13$, $\dfrac{-4}{12}=-\dfrac13$. All equal: the lines coincide, consistent (dependent).

Check (d) Multiply $3x-2y-4=0$ by $-3$: $-9x+6y+12=0$. Same equation ✓. You could also write the second as $9x-6y-12=0$; then all three ratios are $+\dfrac13$ and the conclusion is the same.

Set the coefficients of the first line $a_1x+b_1y+c_1=0$ and the second line $a_2x+b_2y+c_2=0$. The readouts compare the three ratios, and the graph shows what they predict. Press Play to watch the second line change one coefficient at a time: it crosses the first line, becomes parallel to it, and finally lies right on top of it.

$a_1$1$b_1$2$c_1$−3

$a_2$2$b_2$−1$c_2$−1

$a_1/a_2$—

$b_1/b_2$—

$c_1/c_2$—

the lines—

solution—

✏️ Example 6 — Finding an unknown coefficient

(a) For what value of $k$ does the pair $kx+3y=6$, $4x+6y=5$ have no solution?
(b) Find $c$ so that $2x-3y=5$ and $6x-9y=c$ have infinitely many solutions.
(c) For which values of $k$ does the pair $x+2y=3$, $5x+ky=7$ have a unique solution?

💡 Show solution▶

(a) Condition General form: $kx+3y-6=0$, $4x+6y-5=0$. No solution needs $\dfrac{k}{4}=\dfrac36\ne\dfrac{-6}{-5}$.

(a) Solve $\dfrac k4=\dfrac12$ gives $k=2$. Check the third ratio: $\dfrac12\ne\dfrac65$ ✓. So $k=2$ (the lines $2x+3y=6$ and $4x+6y=5$ are parallel).

(b) General form: $2x-3y-5=0$, $6x-9y-c=0$. We need $\dfrac26=\dfrac{-3}{-9}=\dfrac{-5}{-c}$. The first two are both $\dfrac13$, so $\dfrac5c=\dfrac13$, giving $c=15$. Check: $6x-9y=15$ is $3(2x-3y)=3\times5$ ✓.

(c) A unique solution needs $\dfrac15\ne\dfrac2k$, i.e. $k\ne10$. So the pair has a unique solution for every real $k$ except $k=10$. (At $k=10$: $\dfrac15=\dfrac2{10}$, but $\dfrac{-3}{-7}=\dfrac37\ne\dfrac15$, so the lines are parallel.)

Note In (c), $k=0$ is allowed: then $5x=7$ is a vertical line, which still crosses $x+2y=3$. The ratio form needs $b_2\ne0$, so check $k=0$ directly: $x=\tfrac75$, $y=\tfrac45$, a unique solution ✓.

Exercises

Solutions, general form, and turning words into equations.

1 Is $(3, -1)$ a solution of the pair $2x+3y=3$, $x-y=4$?
2 Is $(-2, 5)$ a solution of the pair $3x+2y=4$, $4x+y=-2$?
3 Write $3x=7-2y$ in the form $ax+by+c=0$ and state $a$, $b$ and $c$.
4 Write $y=2x-5$ in the form $ax+by+c=0$ and state $a$, $b$ and $c$.
5 Find $k$ if $x=2$, $y=1$ is a solution of $kx+3y=11$.
6 The sum of two numbers is $42$ and their difference is $8$. Write a pair of equations and show that $25$ and $17$ satisfy both.
7 $5$ pencils and $2$ pens cost ₹ $70$; $3$ pencils and $4$ pens cost ₹ $98$. Write the pair of equations and check that a pencil at ₹ $6$ and a pen at ₹ $20$ fits both.
8 Ravi's father is $26$ years older than Ravi, and the sum of their ages is $54$ years. Form the pair of equations and check that the ages $40$ and $14$ fit.
9 The length of a rectangular kabaddi court is $3$ m more than twice its breadth, and its perimeter is $66$ m. Form a pair of equations in the length $l$ and breadth $b$, and check that $l=23$ m, $b=10$ m satisfies both.

💡 Show solutions▶

1 $2(3)+3(-1)=6-3=3$ ✓ and $3-(-1)=4$ ✓. Yes, it is a solution.

2 $3(-2)+2(5)=-6+10=4$ ✓ but $4(-2)+5=-3\ne-2$ ✗. No.

3 $3x+2y-7=0$: $a=3$, $b=2$, $c=-7$.

4 $2x-y-5=0$: $a=2$, $b=-1$, $c=-5$. (Multiplying by $-1$, $-2x+y+5=0$ is equally correct.)

5 $2k+3(1)=11$, so $2k=8$ and $k=4$.

6 Let the numbers be $x$ and $y$ ($x$ the larger): $x+y=42$, $x-y=8$. Check: $25+17=42$ ✓, $25-17=8$ ✓.

7 Pencil ₹ $x$, pen ₹ $y$: $5x+2y=70$ and $3x+4y=98$. Check: $30+40=70$ ✓, $18+80=98$ ✓.

8 Father $x$ years, Ravi $y$ years: $x-y=26$ and $x+y=54$. Check: $40-14=26$ ✓, $40+14=54$ ✓.

9 $l=2b+3$ and $2l+2b=66$, i.e. $l-2b=3$ and $l+b=33$. Check: $23-20=3$ ✓, $23+10=33$ ✓ (perimeter $2(23+10)=66$ m).

Solve graphically. Use a table of three points for each line, and check your answer by substitution.

1 $x+y=5$ and $x-y=1$
2 $2x+y=10$ and $x-y=2$
3 $x+2y=6$ and $2x-y=2$
4 $y=2x-1$ and $y=-x+5$
5 $3x+y=3$ and $2x-y=7$
6 $x-y=2$ and $2x-2y=8$
7 $x+3y=6$ and $2x+6y=12$
8 Draw the lines $x-y+2=0$ and $2x+y-8=0$. Find the vertices of the triangle formed by the two lines and the $x$-axis, and the area of this triangle.
9 Draw the lines $x+y=6$ and $x-y=2$. Find the vertices of the triangle formed by the two lines and the $y$-axis, and its area.

💡 Show solutions▶

1 Points $(0,5)$, $(5,0)$, $(2,3)$ and $(0,-1)$, $(1,0)$, $(3,2)$. Lines meet at $(3,2)$. Check: $3+2=5$, $3-2=1$ ✓.

2 Points $(0,10)$, $(5,0)$, $(3,4)$ and $(0,-2)$, $(2,0)$, $(4,2)$. Solution $(4,2)$. Check: $8+2=10$, $4-2=2$ ✓.

3 Points $(0,3)$, $(6,0)$, $(2,2)$ and $(0,-2)$, $(1,0)$, $(2,2)$. Solution $(2,2)$. Check: $2+4=6$, $4-2=2$ ✓.

4 Points $(0,-1)$, $(1,1)$, $(2,3)$ and $(0,5)$, $(5,0)$, $(2,3)$. Solution $(2,3)$. Check: $2(2)-1=3$, $-2+5=3$ ✓.

5 Points $(0,3)$, $(1,0)$, $(2,-3)$ and $(0,-7)$, $(3,-1)$, $(2,-3)$. Solution $(2,-3)$. Check: $6-3=3$, $4+3=7$ ✓.

6 The second equation is $x-y=4$. Points $(0,-2)$, $(2,0)$ and $(0,-4)$, $(4,0)$: the lines are parallel ($x-y$ cannot be both $2$ and $4$). No solution — inconsistent.

7 The second equation is $2$ times the first; both lines pass through $(0,2)$, $(3,1)$, $(6,0)$. The lines coincide: infinitely many solutions, every point with $x+3y=6$.

8 $x-y+2=0$ meets the $x$-axis ($y=0$) at $(-2,0)$; $2x+y-8=0$ at $(4,0)$. The lines meet at $(2,4)$ (check: $2-4+2=0$, $4+4-8=0$ ✓). Vertices $(-2,0)$, $(4,0)$, $(2,4)$. Base $=6$ units, height $=4$ units, area $=\tfrac12\times6\times4=12$ square units.

9 On the $y$-axis ($x=0$): $(0,6)$ and $(0,-2)$. The lines meet at $(4,2)$. Vertices $(0,6)$, $(0,-2)$, $(4,2)$. The base along the $y$-axis is $8$ units and the height (distance of $(4,2)$ from the $y$-axis) is $4$ units, so the area is $\tfrac12\times8\times4=16$ square units.

III

The ratio test. In 1–5, decide without drawing whether the lines intersect, are parallel or coincide, and whether the pair is consistent.

1 $3x+5y-7=0$ and $2x-4y+1=0$
2 $4x-6y+9=0$ and $10x-15y+2=0$
3 $7x+2y-3=0$ and $21x+6y-9=0$
4 $x-4y=6$ and $3x-12y=15$
5 $2x=3y-8$ and $6y=4x+16$
6 Find $k$ so that $3x+ky=9$ and $6x+8y=5$ have no solution.
7 Find $k$ so that $2x+ky=6$ and $3x+12y=9$ have infinitely many solutions.
8 For which values of $k$ does the pair $kx+2y=5$, $3x+y=1$ have a unique solution?
9 For the pair $x-2y=c$ and $3x-6y=12$, find the value of $c$ for which there are infinitely many solutions. What happens for every other value of $c$?
10 Give a second linear equation which, together with $3x-4y+5=0$, forms (i) intersecting lines, (ii) parallel lines, (iii) coincident lines.

Check: enter a pair in general form and see what the ratio test says

$a_1$$b_1$$c_1$

$a_2$$b_2$$c_2$

ratios $a$, $b$, $c$—

the lines—

solution—

💡 Show solutions▶

1 $\dfrac32$ and $\dfrac{5}{-4}$ are different: intersecting, one solution, consistent.

2 $\dfrac4{10}=\dfrac25$, $\dfrac{-6}{-15}=\dfrac25$, $\dfrac92$: parallel, no solution, inconsistent.

3 $\dfrac7{21}=\dfrac2{6}=\dfrac{-3}{-9}=\dfrac13$: coincident, infinitely many solutions, consistent (dependent).

4 General form $x-4y-6=0$, $3x-12y-15=0$: $\dfrac13$, $\dfrac{-4}{-12}=\dfrac13$, $\dfrac{-6}{-15}=\dfrac25$. Parallel, inconsistent.

5 General form $2x-3y+8=0$ and $-4x+6y-16=0$: $\dfrac{2}{-4}=\dfrac{-3}{6}=\dfrac{8}{-16}=-\dfrac12$. Coincident, consistent (dependent). The second equation is $-2$ times the first.

6 $\dfrac36=\dfrac k8$ gives $k=4$; then $\dfrac{c_1}{c_2}=\dfrac{-9}{-5}=\dfrac95\ne\dfrac12$ ✓. $k=4$.

7 $\dfrac23=\dfrac k{12}=\dfrac{-6}{-9}$. The last ratio is $\dfrac23$ ✓, and $\dfrac k{12}=\dfrac23$ gives $k=8$.

8 Unique solution when $\dfrac k3\ne\dfrac21$, i.e. $k\ne6$.

9 General form $x-2y-c=0$, $3x-6y-12=0$. The first two ratios are $\dfrac13$. Infinitely many solutions when $\dfrac{-c}{-12}=\dfrac13$, i.e. $c=4$. For every other $c$ the lines are parallel and there is no solution. (The pair can never have exactly one solution.)

10 Many answers are possible. (i) Change the ratio of the $x$- and $y$-coefficients, e.g. $x+y-2=0$ (ratios $3$ and $-4$). (ii) Keep $3x-4y$ but change the constant, e.g. $6x-8y+3=0$ (ratios $\tfrac12,\tfrac12,\tfrac53$). (iii) Multiply the whole equation by a non-zero number, e.g. $6x-8y+10=0$.

Modelling. Form a pair of linear equations, decide how many solutions it has, and solve it graphically where possible. Give units in your answers.

1 Auto fares. Auto service A charges ₹ $25$ plus ₹ $12$ per km; service B charges ₹ $40$ plus ₹ $9$ per km. For what distance do they cost the same, and what is that cost? Which is cheaper for an $8$ km ride, and by how much?
2 Piggy bank. Meena's piggy bank holds $25$ coins, all ₹ $2$ or ₹ $5$ coins, worth ₹ $80$ in total. How many of each coin does she have?
3 A suspicious bill. One shop bill says $2$ notebooks and $3$ pens cost ₹ $90$; another bill from the same shop says $4$ notebooks and $6$ pens cost ₹ $200$. Can both bills be correct if the prices did not change? Explain using the ratio test.
4 School garden. The length of a rectangular garden is $4$ m more than its width, and half its perimeter is $36$ m. Find the length and the width.
5 Water tanks. Tank A holds $600$ L and is being emptied at $25$ L per minute. Tank B holds $200$ L and is being filled at $15$ L per minute. After how many minutes do they hold the same amount of water, and how much is it?
6 Cricket. In a T20 match two openers scored $156$ runs between them, and one scored $24$ runs more than the other. Find each opener's score.

💡 Show solutions▶

1 Cost ₹ $y$ for $x$ km: $y=25+12x$ and $y=40+9x$. Ratios of the $x$-coefficients differ ($12\ne9$), so one solution. Table: A $(0,25)$, $(5,85)$, $(10,145)$; B $(0,40)$, $(5,85)$, $(10,130)$. Lines meet at $(5, 85)$: equal cost ₹ $85$ at $5$ km. For $8$ km: A costs $25+96=$ ₹ $121$, B costs $40+72=$ ₹ $112$, so B is cheaper by ₹ $9$.

2 $x$ coins of ₹ $2$, $y$ coins of ₹ $5$: $x+y=25$ and $2x+5y=80$. Ratios $\tfrac12\ne\tfrac15$: one solution. Points $(25,0)$, $(0,25)$, $(15,10)$ and $(40,0)$, $(0,16)$, $(15,10)$: they meet at $(15,10)$. $15$ two-rupee coins and $10$ five-rupee coins. Check: $30+50=80$ ✓.

3 $2x+3y-90=0$ and $4x+6y-200=0$: $\dfrac24=\dfrac36=\dfrac12$ but $\dfrac{90}{200}=\dfrac{9}{20}$. The lines are parallel, so no prices fit both bills: at least one bill is wrong (if $2$ notebooks and $3$ pens cost ₹ $90$, then twice as many must cost ₹ $180$, not ₹ $200$).

4 Length $l$ m, width $w$ m: $l-w=4$ and $l+w=36$. Points $(4,0)$, $(10,6)$, $(20,16)$ and $(36,0)$, $(0,36)$, $(20,16)$. Solution $l=20$ m, $w=16$ m. Check: $20-16=4$, $20+16=36$ ✓.

5 After $t$ minutes the volume $V$ litres is $V=600-25t$ in A and $V=200+15t$ in B. Points A: $(0,600)$, $(8,400)$, $(10,350)$; B: $(0,200)$, $(10,350)$, $(20,500)$. They meet at $(10, 350)$: after $10$ minutes, $350$ L each. Check: $600-250=350=200+150$ ✓.

6 Scores $x$ and $y$ runs ($x$ larger): $x+y=156$, $x-y=24$. Ratios $1$ and $-1$ differ: one solution. The lines meet at $(90, 66)$: $90$ runs and $66$ runs. Check: $90+66=156$, $90-66=24$ ✓.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study

How this is examined: board papers test the ratio test in 1-mark MCQs and Assertion–Reason items, ask for a value of $k$ in 2-mark questions, and set a 3-mark "solve graphically and find the triangle" question. Write the three ratios explicitly — that line earns the marks.

E1. Board · 1 mark The pair of equations $3x-2y=5$ and $9x-6y=10$ has (A) a unique solution (B) no solution (C) infinitely many solutions (D) exactly two solutions

💡 Show solution▶

Ratios $\dfrac{a_1}{a_2}=\dfrac39=\dfrac13$, $\dfrac{b_1}{b_2}=\dfrac{-2}{-6}=\dfrac13$, $\dfrac{c_1}{c_2}=\dfrac{-5}{-10}=\dfrac12$.

Trap Option (C) is what you get if you stop after the first two equal ratios. The third ratio is different, so the lines are parallel, not coincident. (D) is impossible: two different straight lines meet at most once.

Answer (B) no solution.

E2. Board · 1 mark Assertion (A): The lines $x+2y=4$ and $2x+4y=8$ coincide. Reason (R): If $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$, then the lines $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ coincide.
(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

Check A $\dfrac12=\dfrac24=\dfrac{-4}{-8}=\dfrac12$: all three ratios are equal, so the lines coincide. A is true.

Check R Equal $a$- and $b$-ratios alone only give "parallel or coincident". For example $x+2y=4$ and $2x+4y=5$ have $\dfrac12=\dfrac24$ but are parallel. R is false.

Answer (C). Choosing (A) is the trap: R sounds like the rule, but it leaves out the condition on $\dfrac{c_1}{c_2}$.

E3. Board · 2 marks Find the value of $k$ for which the pair $kx-4y=3$ and $6x-12y=9$ has infinitely many solutions.

💡 Show solution▶

Condition Infinitely many solutions: $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$, i.e. $\dfrac k6=\dfrac{-4}{-12}=\dfrac{-3}{-9}$. (1 mark)

Solve $\dfrac{-4}{-12}=\dfrac13$ and $\dfrac{-3}{-9}=\dfrac13$ ✓, so $\dfrac k6=\dfrac13$ and $k=2$. (1 mark)

Check With $k=2$: $6x-12y=9$ is $3(2x-4y)=3\times3$ ✓.

E4. Board · 2 marks Check whether the pair $x+y=4$, $2x-3y=3$ is consistent. If it is, solve it graphically.

💡 Show solution▶

Test $\dfrac{a_1}{a_2}=\dfrac12$, $\dfrac{b_1}{b_2}=\dfrac{1}{-3}=-\dfrac13$. They differ, so the lines intersect: consistent, with a unique solution.

Tables $x+y=4$: $(0,4)$, $(4,0)$, $(3,1)$. $2x-3y=3$: $(0,-1)$, $(3,1)$, $(6,3)$.

Answer The lines meet at $(3,1)$: $x=3$, $y=1$. Check: $3+1=4$, $6-3=3$ ✓. (Choose "pair 2" in Interactive 1 to see this graph.)

E5. Board · 3 marks Solve graphically: $2x-y=1$ and $x+y=5$. Also find the coordinates of the points where the lines meet the $y$-axis, and the area of the triangle formed by the two lines and the $y$-axis.

💡 Show solution▶

Tables $2x-y=1$, i.e. $y=2x-1$: $(0,-1)$, $(1,1)$, $(2,3)$. $x+y=5$: $(0,5)$, $(5,0)$, $(2,3)$. (1 mark)

Solution The lines meet at $(2,3)$. Check: $4-3=1$, $2+3=5$ ✓. (1 mark)

Triangle On the $y$-axis the lines pass through $(0,-1)$ and $(0,5)$. The triangle has base $5-(-1)=6$ units along the $y$-axis and height $2$ units (the $x$-coordinate of $(2,3)$). Area $=\tfrac12\times6\times2=6$ square units. (1 mark)

E6. Board · Case study Scooter rentals in Goa. Company P rents a scooter for ₹ $150$ plus ₹ $4$ per km. Company Q charges ₹ $90$ plus ₹ $6$ per km. Let ₹ $y$ be the cost of riding $x$ km.
(i) Write the pair of linear equations, and use a graph to find the distance at which both companies cost the same and what that cost is. (2 marks)
(ii) Is the pair consistent? Justify using the coefficients. (1 mark)
(iii) A tourist plans a $50$ km ride. Which company is cheaper, and by how much? (1 mark)

💡 Show solution▶

(i) P: $y=150+4x$, i.e. $4x-y+150=0$. Q: $y=90+6x$, i.e. $6x-y+90=0$. Tables: P $(0,150)$, $(30,270)$, $(50,350)$; Q $(0,90)$, $(30,270)$, $(50,390)$. The lines meet at $(30,270)$: both cost ₹ $270$ for $30$ km.

(ii) $\dfrac{a_1}{a_2}=\dfrac46=\dfrac23$ and $\dfrac{b_1}{b_2}=\dfrac{-1}{-1}=1$. They differ, so the lines intersect: the pair is consistent with a unique solution.

(iii) P: $150+4(50)=$ ₹ $350$; Q: $90+6(50)=$ ₹ $390$. P is cheaper by ₹ $40$. (Beyond $30$ km, the line with the smaller per-km rate stays lower.)

🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. For two real price plans that people around you choose between — two mobile recharge packs, two cab or auto apps, two gym or swimming-pool memberships — at what amount of use does the cheaper plan switch?

Why it matters. Many prices have the form "fixed charge + rate × amount used". Each such plan is a straight line, and the break-even point where two lines cross tells a customer which plan to choose. Electricity companies, airlines and city transport planners all over the world compare tariffs in exactly this way, and businesses use the same idea to find how many items they must sell before they make a profit.

What to do.

Choose two plans that charge a fixed amount plus a rate (ask family members, or look at a rate card). Write down the fixed charge and the rate for each, with units (₹, ₹ per km, ₹ per visit…).
Write each plan as an equation $y=(\text{fixed})+(\text{rate})\,x$ and then in the general form $ax+by+c=0$. Use the ratio test to predict whether the lines cross.
Make a table of at least five values of $x$ for each plan in a spreadsheet (or by hand) and draw both lines on one graph.
Read off the break-even point. Check it by substituting into both equations.
Ask five people which plan they use and how much they use it. According to your graph, are they on the cheaper plan?

Tools: Graph paper or a spreadsheet (Google Sheets / Excel — insert a line chart with both columns), a calculator.

Hand in: The two equations, the table, the graph with the crossing point marked, your check by substitution, and a two-sentence recommendation: "Choose plan … if you use less than …".

Going further. What if the two plans have the same rate but different fixed charges? Sketch the lines and explain what the ratio test says. Then look for a plan with a "free first $k$ units" rule — is its graph still a single straight line?

Teacher/parent note: real tariffs often have slabs, so a single straight line is a model that is valid only over a range of use. Encourage the student to state that range. The algebraic methods of Sections 3.2–3.3 give the exact break-even point to compare with the graph reading.

🧪 Self-check quiz

0/5 correct

Q1. The graph of a pair of linear equations that has no solution is a pair of

Q2. If $\dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2}$, the pair $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$

Q3. For which value of $k$ are the lines $2x+3y=7$ and $4x+ky=14$ coincident?

Q4. The lines $x+y=6$ and $x-y=2$ meet at

Q5. The pair $x-3y=5$ and $2x-6y=10$ is

← Chapter 3 — Pair of Linear Equations in Two Variables 3.2 The Substitution Method →