Question 1

Write a linear equation in two variables in which $a = 3$, $b = 0$ and $c = -\dfrac{1}{5}$.

Accepted Answer

The standard form is $ax + by + c = 0$. Put in the given values. $3x + 0 \times y + \left(-\dfrac{1}{5}\right) = 0$ $3x + 0 \times y - \dfrac{1}{5} = 0$ So the required equation is $3x + 0\times y - \dfrac{1}{5} = 0$, which is usually written simply as $3x - \dfrac{1}{5} = 0$.Do not drop the idea of $y$ altogether: because $b = 0$, the term $0 \times y$ is invisible, but the equation is still treated as an equation in two variables, exactly like the book’s example $4x + 0\times y + 7 = 0$. If you clear the fraction by multiplying by 5 you get $15x - 1 = 0$ — the same line, but now with $a = 15$ and $c = -1$, so it no longer matches the values asked for.

Question 2

Complete the following table after expressing the given linear equations in standard form. The equations are: $y - 15 = \sqrt{2}x$; $3y - 2x = 0$; $5x = 3y$; $x = 8$; $3y = 1$. For each, write the standard form, the coefficient of $x$, the coefficient of $y$ and the constant term.

Accepted Answer

Move every term to the left-hand side so the right-hand side is 0, then write the terms in the order $x$-term, $y$-term, constant: $ax + by + c = 0$.

$y - 15 = \sqrt{2}x$: subtract $\sqrt{2}x$ from both sides.

$-\sqrt{2}x + y - 15 = 0$

$3y - 2x = 0$: just reorder the terms.

$-2x + 3y + 0 = 0$

$5x = 3y$: subtract $3y$ from both sides.

$5x - 3y + 0 = 0$

$x = 8$: subtract 8; there is no $y$-term, so its coefficient is 0.

$x + 0 \times y - 8 = 0$

$3y = 1$: subtract 1; there is no $x$-term.

$0 \times x + 3y - 1 = 0$

Standard form	$a$	$b$	$c$
$-\sqrt{2}x + y - 15 = 0$	$-\sqrt{2}$	1	−15
$-2x + 3y + 0 = 0$	−2	3	0
$5x - 3y + 0 = 0$	5	−3	0
$x + 0\times y - 8 = 0$	1	0	−8
$0\times x + 3y - 1 = 0$	0	3	−1

(The rows are in the same order as the equations in the question.)

The standard form is not unique. Multiplying every term by −1 gives an equally correct answer — for example the first row could be $\sqrt{2}x - y + 15 = 0$ with $a = \sqrt{2}$, $b = -1$, $c = 15$. The book itself shows both versions in its Example 1. What matters is that all three numbers change sign together.

Question 3

(i) The cost of a notebook is twice the cost of a pen. Consider the cost of a notebook to be ₹$t$ and that of a pen to be ₹$p$. Charlie wrote $t = 2p$, whereas Meera wrote $p = 2t$. Which of these two representations is correct? (ii) In a one-day International Cricket match between India and

Accepted Answer

(i) “The notebook costs twice the pen” means: notebook cost = 2 × pen cost. $t = 2 \times p = 2p$ So Charlie is correct. Meera’s $p = 2t$ says the pen costs twice the notebook — the reverse of what was given.Quick check with numbers: if a pen costs ₹10, the notebook costs ₹20. Charlie: $t = 2(10) = 20$ ✓. Meera: $10 = 2(20) = 40$ ✗.(ii) The two batsmen’s runs added together give the total of 176. $x + y = 176$ Yes, this is a correct representation. It is a linear equation in two variables ($a = 1$, $b = 1$, $c = -176$ in standard form), and many pairs fit it, for example (100, 76) or (88, 88) — just as we cannot tell from the total alone how the runs were split.

Question 4

Verify if the ordered pair (4, 3) is a solution of $5x - 6y = 2$. Explain your reasoning.

Accepted Answer

An ordered pair is a solution if substituting it makes the two sides equal. Put $x = 4$, $y = 3$.

$\text{LHS} = 5(4) - 6(3)$

$= 20 - 18 = 2 = \text{RHS}$

Both sides equal 2, so yes, (4, 3) is a solution of $5x - 6y = 2$. Geometrically, the point (4, 3) lies on the line $5x - 6y = 2$.

Question 5

Find any two solutions for each of the following equations: (i) $7x - 3y = 21$ (ii) $2x + 3y = 5$

Accepted Answer

Choose a value for one variable, substitute, and solve the resulting one-variable equation for the other.

(i) $7x - 3y = 21$.

Take $x = 0$:

$7(0) - 3y = 21$

$-3y = 21 \Rightarrow y = -7$

Take $y = 0$:

$7x - 3(0) = 21$

$7x = 21 \Rightarrow x = 3$

Two solutions: (0, −7) and (3, 0).

(ii) $2x + 3y = 5$.

Take $x = 1$:

$2(1) + 3y = 5$

$3y = 3 \Rightarrow y = 1$

Take $x = 4$:

$2(4) + 3y = 5$

$3y = -3 \Rightarrow y = -1$

Two solutions: (1, 1) and (4, −1).

These are not the only answers — every linear equation in two variables has infinitely many solutions. Any value of $x$ you choose gives a matching $y$. Values were picked here so that $y$ comes out a whole number, which makes them easy to plot.

Question 6

In the equations given below, $m$ and $n$ are unknown constants: $2mx + 3y = 7$; $4x + ny = -10$. If $(2, -1)$ is the solution of both equations, find the values of $m$ and $n$.

Accepted Answer

Since (2, −1) is a solution of each equation, substituting $x = 2$, $y = -1$ must make each equation true. Each substitution leaves one unknown constant.

First equation, $2mx + 3y = 7$:

$2m(2) + 3(-1) = 7$

$4m - 3 = 7$

$4m = 10 \Rightarrow m = \dfrac{5}{2}$

Second equation, $4x + ny = -10$:

$4(2) + n(-1) = -10$

$8 - n = -10$

$n = 18$

So $m = \dfrac{5}{2}$ and $n = 18$.

Check: the equations become $5x + 3y = 7$ and $4x + 18y = -10$. At (2, −1): $10 - 3 = 7$ ✓ and $8 - 18 = -10$ ✓.

Question 7

Find two solutions which lie in different quadrants for each of the following linear equations. Identify the quadrants in which the points lie. (i) $5x + 3y = 7$ (ii) $5x - 3y = 7$ (iii) $-5x + 3y = 7$ (iv) $-5x - 3y = 7$ Verify your solutions by representing the linear equations on a graph

Accepted Answer

Pick small whole-number values of $x$ (both positive and negative) and solve for $y$. Keep only points where neither coordinate is 0 — a point on an axis is not in any quadrant. Remember: Quadrant I (+, +), II (−, +), III (−, −), IV (+, −).(i) $5x + 3y = 7$, so $3y = 7 - 5x$. $x = 2:\ 3y = 7 - 10 = -3 \Rightarrow y = -1$ $x = -1:\ 3y = 7 + 5 = 12 \Rightarrow y = 4$ Solutions: (2, −1) in Quadrant IV and (−1, 4) in Quadrant II.(ii) $5x - 3y = 7$, so $3y = 5x - 7$. $x = 2:\ 3y = 10 - 7 = 3 \Rightarrow y = 1$ $x = -1:\ 3y = -5 - 7 = -12 \Rightarrow y = -4$ Solutions: (2, 1) in Quadrant I and (−1, −4) in Quadrant III.(iii) $-5x + 3y = 7$, so $3y = 7 + 5x$. $x = 1:\ 3y = 7 + 5 = 12 \Rightarrow y = 4$ $x = -2:\ 3y = 7 - 10 = -3 \Rightarrow y = -1$ Solutions: (1, 4) in Quadrant I and (−2, −1) in Quadrant III.(iv) $-5x - 3y = 7$, so $3y = -7 - 5x$. $x = -2:\ 3y = -7 + 10 = 3 \Rightarrow y = 1$ $x = 1:\ 3y = -7 - 5 = -12 \Rightarrow y = -4$ Solutions: (−2, 1) in Quadrant II and (1, −4) in Quadrant IV.Graphical check: in each graph above, both chosen points sit exactly on the drawn line, and they fall in the quadrants named. Any other point you read off the

Question 8

Consider the graph of the equation $3x - 7y = 21$ shown below (it passes through A(7, 0) and B(0, −3)). Does the point C(2, 3) lie on the line? Does it satisfy the equation? Can points that do not lie on the line satisfy the equation?

Accepted Answer

Does C lie on the line? From the graph, the line passes through A(7, 0) and B(0, −3); at $x = 2$ it is below the $x$-axis, while C(2, 3) is well above it. So C does not lie on the line.Does C satisfy the equation? Substitute $x = 2$, $y = 3$: $\text{LHS} = 3(2) - 7(3)$ $= 6 - 21 = -15$ $-15 \ne 21 = \text{RHS}$ So C does not satisfy the equation either.The point on the line with $x = 2$ actually has $6 - 7y = 21 \Rightarrow y = -\dfrac{15}{7}$ Can a point not on the line satisfy the equation? No. The line is drawn through every pair $(x, y)$ that satisfies $3x - 7y = 21$, and every point on it is such a pair. So the graph and the solution set are the same thing: a point satisfies the equation if and only if it lies on the line.

Question 9

State whether the following sentences are True or False. Justify your answer. (i) A linear equation in two variables has only one solution. (ii) The graph of a linear equation in two variables always passes through the origin. (iii) A linear equation in two variables can never have rational solution

Accepted Answer

(i) False. For any value we give $x$, we can solve for a matching $y$, so there are infinitely many solutions. E.g. $x + y = 4$ has (0, 4), (1, 3), (2, 2), …(ii) False. The origin (0, 0) lies on $ax + by + c = 0$ only when $c = 0$. Counterexample: $x + y = 1$; at (0, 0) the LHS is $0 \ne 1$.(iii) False. $x + y = 2$ has the rational solution (1, 1), and also $\left(\dfrac{1}{2}, \dfrac{3}{2}\right)$.(iv) True. It can be written $1\times x + 0\times y - 3 = 0$, which is the standard form with $a = 1$, $b = 0$, $c = -3$ (and $a$, $b$ are not both zero). Its graph is the vertical line through (3, 0).(v) True. Every linear equation in two variables has infinitely many solutions; for instance (2, 1), $\left(\dfrac{1}{2}, 2\right)$ and (5, −1) all satisfy $2x + 3y = 7$.(vi) False. Substituting: $2(1) + 3(2) = 2 + 6 = 8 \ne 7$

Question 10

(i) Compare the solutions of the equations $3x + 4y = 7$ and $6x + 8y = 14$. Argue that they have the same set of solutions, that is, every solution of one is also a solution of the other. (ii) Show that the equations $ax + by = c$ and $kax + kby = kc$, with $k \ne 0$, have the same set of

Accepted Answer

(i) Notice that the second equation is the first one with every term doubled:

$6x + 8y = 14 \iff 2(3x + 4y) = 2(7)$

Every solution of the first is a solution of the second. If $(p, q)$ satisfies $3p + 4q = 7$, multiply both sides by 2:

$6p + 8q = 14$

Every solution of the second is a solution of the first. If $6p + 8q = 14$, divide both sides by 2 (allowed because 2 is not zero):

$3p + 4q = 7$

So the two equations have exactly the same solutions, and their graphs are the same line. Example: (1, 1) gives $3 + 4 = 7$ and $6 + 8 = 14$.

(ii) The same argument with a general non-zero $k$. Suppose $(p, q)$ satisfies $ap + bq = c$. Multiplying both sides by $k$:

$k(ap + bq) = kc$

$kap + kbq = kc$

so $(p, q)$ satisfies the second equation. Conversely, if $kap + kbq = kc$, then $k(ap + bq - c) = 0$. Since $k \ne 0$, the other factor must be zero:

$ap + bq - c = 0 \Rightarrow ap + bq = c$

Hence the two equations have the same set of solutions.

Why does the question insist on $k \ne 0$? With $k = 0$ the second equation becomes $0 = 0$, which every pair $(x, y)$ satisfies — so the two solution sets would no longer match. Division by $k$ in the converse step is exactly where $k \ne 0$ is needed.

Question 11

The following diagrams represent ski hills. Rank the hills in order of their steepness, from least to greatest. (Hill A rises 60 m over a horizontal run of 70 m; hill B rises 60 m over 110 m; hill C rises 80 m over 100 m.)

Accepted Answer

Steepness is measured by the slope: $\text{slope} = \dfrac{\text{rise}}{\text{run}}$ $\text{Hill A} = \dfrac{60}{70} = \dfrac{6}{7} \approx 0.857$ $\text{Hill B} = \dfrac{60}{110} = \dfrac{6}{11} \approx 0.545$ $\text{Hill C} = \dfrac{80}{100} = \dfrac{4}{5} = 0.8$ Comparing: $0.545 < 0.8 < 0.857$.Order from least to greatest steepness: Hill B, Hill C, Hill A.Hill C is the tallest (80 m) but it is not the steepest — its long run of 100 m spreads the climb out. Steepness depends on rise and run together, never on height alone.

Question 12

The ramp at a loading dock rises 2.5 metres over a run of 4 metres. Find the slope of the ramp.

Accepted Answer

$\text{slope} = \dfrac{\text{rise}}{\text{run}} = \dfrac{2.5}{4}$ $= \dfrac{25}{40} = \dfrac{5}{8}$ The slope of the ramp is $\dfrac{5}{8}$ (= 0.625): it rises 5 m for every 8 m of horizontal distance.

Question 13

Find the slope of the line $l$ in each of the following diagrams.

Accepted Answer

Pick two points on each line and use $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ Each big square of the grid is taken as 1 unit.(i) The line $l$ is horizontal. Any two of its points have the same $y$-coordinate, so the rise is 0. $m = \dfrac{0}{x_2 - x_1} = 0$ Slope = 0.(ii) The line passes through $(-2, 2)$ and $(2, -2)$ (the two marked points). $m = \dfrac{-2 - 2}{2 - (-2)} = \dfrac{-4}{4} = -1$ Slope = −1 (the line falls 1 unit for every unit to the right).(iii) The line passes through $(-2, -1)$ and $(1, 1)$. $m = \dfrac{1 - (-1)}{1 - (-2)} = \dfrac{2}{3}$ Slope = $\dfrac{2}{3}$ (it rises 2 units for every 3 units to the right).(iv) The line $l$ is vertical. Any two of its points have the same $x$-coordinate, so the run is 0 and the formula would divide by zero.Slope = undefined.In (ii) and (iii) the answer does not depend on which two points you pick or on what one square stands for, as long as both axes use the same scale — rise and run scale together.

Question 14

An accessibility ramp for wheelchairs is to be made alongside the staircase. The guidelines given for the slope of the ramp is 1 cm vertical rise to 12 cm horizontal length. What should be the horizontal length of the ramp if the total height of the stairs is 18 cm?

Accepted Answer

The required slope is $\dfrac{\text{rise}}{\text{run}} = \dfrac{1}{12}$ Let the horizontal length be $L$ cm. The rise is the total height, 18 cm. $\dfrac{18}{L} = \dfrac{1}{12}$ $L = 18 \times 12 = 216$ The ramp must have a horizontal length of 216 cm (2.16 m).

Question 15

Construct a pair of linear equations in two variables for the following situation. On two different days a family buys movie tickets and snack boxes. Let the cost of one movie ticket be ₹$x$ and the cost of one snack box be ₹$y$. (i) On the first day, the family buys 2 movie tickets and 3 snack

Accepted Answer

Cost = (number of tickets) × $x$ + (number of snack boxes) × $y$.(i) First day: 2 tickets and 3 snack boxes cost ₹850. $2x + 3y = 850$ (ii) Second day: 4 tickets and 1 snack box cost ₹1100. $4x + y = 1100$ The pair is $2x + 3y = 850$ and $4x + y = 1100$.Bonus check (solving comes in Section 13.5): the pair has the solution $x = 245$, $y = 120$, i.e. a ticket costs ₹245 and a snack box ₹120: $490 + 360 = 850$ and $980 + 120 = 1100$.

Question 16

Two friends, Sahil and Meena, travelled by taxi. Let the fixed charge for one taxi trip be ₹$x$ and the additional charge per kilometre be ₹$y$. Frame a pair of linear equations in $x$ and $y$ to represent the following situation: (i) Sahil travelled 6 km and paid a total fare of ₹122. (ii)

Accepted Answer

Fare = fixed charge + (distance in km) × charge per km.

(i) Sahil, 6 km:

$x + 6y = 122$

(ii) Meena, 8 km:

$x + 8y = 160$

The pair is $x + 6y = 122$ and $x + 8y = 160$.

Bonus check (solving comes in Section 13.5): subtracting gives $2y = 38$, so $y = 19$ and $x = 8$: a fixed charge of ₹8 and ₹19 per km.

Question 17

In a sports meet, tickets for adults cost ₹150 each and tickets for children cost ₹100 each. A total of 200 people attended the meet, and the total amount collected from ticket sales was ₹25,000. (Let the number of adult tickets sold be $x$ and the number of children’s tickets sold be $y$.) Fram

Accepted Answer

Two facts are given, so there are two equations.

Number of people: adults + children = 200.

$x + y = 200$

Money collected: 150 per adult + 100 per child = 25,000.

$150x + 100y = 25000$

This second equation can be simplified by dividing every term by 50:

$3x + 2y = 500$

The pair is $x + y = 200$ and $150x + 100y = 25000$ (or $3x + 2y = 500$).

Bonus check (solving comes in Section 13.5): the solution is $x = 100$, $y = 100$: 100 adults and 100 children.

Question 18

Form a pair of linear equations for each of the following problems and find their solutions. (i) The sum of two integers is +5 and their difference is −21. Find the two numbers. (ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers. (iii) The coach o

Accepted Answer

(i) Let the integers be $x$ and $y$. $x + y = 5 \ \ \ (1)$ $x - y = -21 \ \ \ (2)$ Elimination: add (1) and (2) — the $y$-terms cancel. $2x = -16 \Rightarrow x = -8$ Put $x = -8$ in (1): $-8 + y = 5 \Rightarrow y = 13$ The integers are −8 and 13. Check: $-8 + 13 = 5$ ✓, $-8 - 13 = -21$ ✓.(ii) Let the larger number be $x$ and the smaller $y$. $x - y = 26$ $x = 3y$ Substitution: put $x = 3y$ into the first equation. $3y - y = 26$ $2y = 26 \Rightarrow y = 13$ $x = 3(13) = 39$ The numbers are 39 and 13. Check: $39 - 13 = 26$ ✓, $39 = 3 \times 13$ ✓.(iii) Let a bat cost ₹$x$ and a ball ₹$y$. $7x + 6y = 8880$ $3x + 5y = 4000$ Elimination: make the $y$-coefficients equal (30) by multiplying the first equation by 5 and the second by 6. $35x + 30y = 44400$ $18x + 30y = 24000$ Subtract: $17x = 20400 \Rightarrow x = 1200$ Put $x = 1200$ in $3x + 5y = 4000$: $3600 + 5y = 4000$ $5y = 400 \Rightarrow y = 80$ A bat costs ₹1200 and a ball ₹80. Check: $7(1200) + 6(80) = 8400 + 480 = 8880$ ✓.(iv) Let the fixed charge be ₹$x$ and the charge per km ₹$y$. $x + 10y = 155$ $x + 15y = 220$ Subtract the first from the second: \(5y =

Question 19

Form a pair of linear equations and find their common solutions graphically. 10 students of Grade 9 took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Accepted Answer

Let the number of boys be $x$ and the number of girls $y$. $x + y = 10$ $y = x + 4 \ \ (\text{i.e. } y - x = 4)$ Table of points for each line:Plot both lines on the same axes. They cross at the point (3, 7).So 3 boys and 7 girls took part. Check: $3 + 7 = 10$ ✓ and $7 = 3 + 4$ ✓.

Question 20

Use the ratios $\dfrac{a_1}{a_2}$, $\dfrac{b_1}{b_2}$, $\dfrac{c_1}{c_2}$ to determine whether the lines representing the following pairs of linear equations intersect at a point, are parallel or are coincident. (i) $5x - 4y + 8 = 0$; $7x + 6y - 9 = 0$ (ii) $9x + 3y + 12 = 0$; \(18x + 6y

Accepted Answer

Rule: $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$ → intersecting; $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$ → parallel; $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ → coincident.(i) $\dfrac{a_1}{a_2} = \dfrac{5}{7}$ $\dfrac{b_1}{b_2} = \dfrac{-4}{6} = -\dfrac{2}{3}$ $\dfrac{5}{7} \ne -\dfrac{2}{3}$, so the lines intersect at a point (one unique solution).(ii) $\dfrac{a_1}{a_2} = \dfrac{9}{18} = \dfrac{1}{2}$ $\dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}$ $\dfrac{c_1}{c_2} = \dfrac{12}{24} = \dfrac{1}{2}$ All three ratios are equal, so the lines are coincident (infinitely many solutions). Indeed the second equation is just the first multiplied by 2.(iii) $\dfrac{a_1}{a_2} = \dfrac{6}{2} = 3$ $\dfrac{b_1}{b_2} = \dfrac{-3}{-1} = 3$ $\dfrac{c_1}{c_2} = \dfrac{10}{9}$ $3 = 3 \ne \dfrac{10}{9}$, so the lines are parallel (no solution).

Question 21

Which of the following pairs of linear equations have solutions? If they have solutions, find them graphically. (i) $x + y = 5$, $2x + 2y = 10$ (ii) $x - y = 8$, $3x - 3y = 16$ (iii) $2x + y - 6 = 0$, $4x - 2y - 4 = 0$ (iv) $2x - 2y - 2 = 0$, $4x - 4y - 5 = 0$

Accepted Answer

First compare the ratios to see what kind of pair it is, then draw.

(i) $x + y - 5 = 0$ and $2x + 2y - 10 = 0$:

$\dfrac{1}{2} = \dfrac{1}{2} = \dfrac{-5}{-10}$

All ratios equal: the lines coincide, so there are infinitely many solutions — every point of $x + y = 5$, such as (0, 5), (2, 3), (5, 0).

(i) x + y = 5 (green) and 2x + 2y = 10 (blue, dashed) are the same line: infinitely many solutions.

(ii) $x - y - 8 = 0$ and $3x - 3y - 16 = 0$:

$\dfrac{1}{3} = \dfrac{-1}{-3}$

$\dfrac{c_1}{c_2} = \dfrac{-8}{-16} = \dfrac{1}{2} \ne \dfrac{1}{3}$

The lines are parallel: no solution.

(ii) x − y = 8 and 3x − 3y = 16 are parallel: no solution.

(iii) $2x + y - 6 = 0$ and $4x - 2y - 4 = 0$:

$\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}$

$\dfrac{b_1}{b_2} = \dfrac{1}{-2} = -\dfrac{1}{2}$

The ratios differ, so the lines intersect: exactly one solution. Points for drawing:

$x$	0	3
$2x + y = 6$: $y$	6	0

$x$	0	1
$4x - 2y = 4$: $y$	−2	0

(iii) 2x + y = 6 and 4x − 2y = 4 meet at the single point (2, 2).

The lines meet at (2, 2). Check: $4 + 2 - 6 = 0$ ✓ and $8 - 4 - 4 = 0$ ✓.

(iv) $2x - 2y - 2 = 0$ and $4x - 4y - 5 = 0$:

$\dfrac{2}{4} = \dfrac{-2}{-4} = \dfrac{1}{2}$

$\dfrac{c_1}{c_2} = \dfrac{-2}{-5} = \dfrac{2}{5} \ne \dfrac{1}{2}$

The lines are parallel: no solution.

(iv) 2x − 2y = 2 and 4x − 4y = 5 are parallel (a quarter of a unit apart vertically): no solution.

Summary: pairs (i) and (iii) have solutions — (i) infinitely many, (iii) the single solution $x = 2, y = 2$. Pairs (ii) and (iv) have none.

Question 22

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Accepted Answer

Let the length be $l$ m and the width $w$ m. Half the perimeter is $l + w$. $l + w = 36$ $l = w + 4$ Substitution: $(w + 4) + w = 36$ $2w = 32 \Rightarrow w = 16$ $l = 16 + 4 = 20$ The garden is 20 m long and 16 m wide. Check: $20 + 16 = 36$ ✓, $20 - 16 = 4$ ✓.

Question 23

Given the linear equation $2x + 3y - 8 = 0$, write another linear equation in two variables so that the graphs of the pair formed represent (i) intersecting lines (ii) parallel lines (iii) coincident lines.

Accepted Answer

Here $a_1 = 2$, $b_1 = 3$, $c_1 = -8$. Choose $a_2, b_2, c_2$ to make the ratios behave as needed.(i) Intersecting: need $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$. Take $3x - 2y + 1 = 0$: $\dfrac{2}{3} \ne \dfrac{3}{-2}$ (ii) Parallel: need $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$. Double $a$ and $b$ but not $c$: $4x + 6y - 5 = 0$: $\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{1}{2}$ $\dfrac{-8}{-5} = \dfrac{8}{5} \ne \dfrac{1}{2}$ (iii) Coincident: need all three ratios equal — multiply the whole equation by 2: $4x + 6y - 16 = 0$: $\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{-8}{-16} = \dfrac{1}{2}$ Many other answers are correct. Any non-zero multiple of the given equation gives a coincident line; keep the $x$:$y$ coefficients in the ratio 2 : 3 but change the constant for a parallel line; and break the 2 : 3 ratio for an intersecting line.

Question 24

Here is a problem that was posed by Mahāvīrāchārya in Gaṇita sāra saṅgraha (c. 850 CE). The price of 9 citrons and 7 fragrant wood-apples taken together is 107; and the price of 7 citrons and 9 fragrant wood-apples taken together is 101. O mathematician, tell me quickly the price of each citron and

Accepted Answer

Let a citron cost $x$ and a wood-apple $y$. $9x + 7y = 107 \ \ \ (1)$ $7x + 9y = 101 \ \ \ (2)$ Add (1) and (2): both variables get the same coefficient, 16. $16x + 16y = 208$ $x + y = 13 \ \ \ (3)$ Subtract (2) from (1): both variables get coefficient 2 (with opposite signs). $2x - 2y = 6$ $x - y = 3 \ \ \ (4)$ Now (3) + (4) and (3) − (4): $2x = 16 \Rightarrow x = 8$ $2y = 10 \Rightarrow y = 5$ A citron costs 8 and a fragrant wood-apple costs 5. Check: $72 + 35 = 107$ ✓ and $56 + 45 = 101$ ✓.This is the “quick” method the verse asks for. Whenever the coefficients are swapped between the two equations, adding and subtracting turns the pair into the much simpler pair $x + y = \ldots$, $x - y = \ldots$, which can be solved in your head.

Question 25

5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of each pencil and pen.

Accepted Answer

Let a pencil cost ₹$x$ and a pen ₹$y$. $5x + 7y = 50 \ \ \ (1)$ $7x + 5y = 46 \ \ \ (2)$ The coefficients are swapped, so use the method of the previous question. Add: $12x + 12y = 96 \Rightarrow x + y = 8$ Subtract (1) from (2): $2x - 2y = -4 \Rightarrow x - y = -2$ Adding these two results: $2x = 6 \Rightarrow x = 3$ $y = 8 - 3 = 5$ A pencil costs ₹3 and a pen ₹5. Check: $15 + 35 = 50$ ✓, $21 + 25 = 46$ ✓.

Question 26

Find the height of the stool. (In the first picture a cat sits on the stool and the top of the cat is 85 cm above the floor. In the second picture the cat stands on the floor beside the stool, and the top of the stool is 25 cm higher than the top of the cat.)

Accepted Answer

Let the height of the stool be $s$ cm and the height of the cat $c$ cm.First picture: stool + cat = 85. $s + c = 85$ Second picture: stool − cat = 25. $s - c = 25$ Add the two equations: $2s = 110 \Rightarrow s = 55$ $c = 85 - 55 = 30$ The stool is 55 cm high (and the cat is 30 cm tall). Check: $55 + 30 = 85$ ✓, $55 - 30 = 25$ ✓.

Question 27

The graph of the line $y = 3x$, passing through (0, 0) and (2, 6), is given. (i) Identify the slope of the line from the graph and explain how you calculated it using the two points on the line. (ii) What is the $y$-intercept of the line? Explain its significance. (iii) A water tank is being fil

Accepted Answer

(i) Using $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (2, 6)$: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ $= \dfrac{6 - 0}{2 - 0}$ $= 3$ The slope is 3: going 2 units right, the line rises 6 units, i.e. 3 units up for every unit right. This matches the 3 in $y = 3x$ (form $y = mx + d$ with $m = 3$).(ii) The line crosses the $y$-axis at (0, 0), so the $y$-intercept is 0 ($d = 0$). Significance: when $x = 0$, $y = 0$ — the line passes through the origin. For the water tank, it means the tank was empty at the start (level 0 cm at time 0).(iii) (a) Put $x = 5$: $y = 3(5) = 15$ The water is 15 cm high after 5 minutes.(b) Put $y = 21$: $21 = 3x \Rightarrow x = 7$ It takes 7 minutes.(iv) A point is on the line only if it satisfies $y = 3x$. At $x = 4$: $3(4) = 12 \ne 10$ So (4, 10) does not lie on the line (the point on the line above $x = 4$ is (4, 12)).(v) On the line $x = 3$ every point has $x = 3$. The common point must also satisfy $y = 3x$: $y = 3(3) = 9$ The lines meet at (3, 9), plotted below.

Question 28

In countries like the USA, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Celsius to Fahrenheit: $F = \dfrac{9}{5}C + 32$. (i) Draw the graph of the linear equation above using Celsius on the $x$-axis and

Accepted Answer

(i) Make a table by choosing convenient values of $C$ (multiples of 5 avoid fractions):

$C = 0:\ F = 0 + 32 = 32$

$C = 30:\ F = 54 + 32 = 86$

$C = 100:\ F = 180 + 32 = 212$

$C$	−40	0	30	100
$F$	−40	32	86	212

Plot the points and join them with a straight line. Its slope is $\dfrac{9}{5}$ and its $F$-intercept is 32.

F = (9/5)C + 32. Points (0, 32), (30, 86), (100, 212); the dashed line F = C meets it at (−40, −40).

(ii) $C = 30$:

$F = \dfrac{9}{5}(30) + 32 = 54 + 32 = 86$

30°C = 86°F.

(iii) $F = 95$:

$95 = \dfrac{9}{5}C + 32$

$\dfrac{9}{5}C = 63$

$C = 63 \times \dfrac{5}{9} = 35$

95°F = 35°C.

(iv) $C = 0$: $F = 0 + 32 = 32$, so 0°C = 32°F.

$F = 0$:

$0 = \dfrac{9}{5}C + 32$

$\dfrac{9}{5}C = -32$

$C = -\dfrac{160}{9} \approx -17.8$

0°F = $-\dfrac{160}{9}$°C, about −17.8°C.

(v) We want $F = C$. Put $F = C$ in the equation:

$C = \dfrac{9}{5}C + 32$

$5C = 9C + 160$

$-4C = 160 \Rightarrow C = -40$

Yes: −40°C = −40°F. On the graph this is where the conversion line meets the dashed line $F = C$.

Question 29

Solve the following system of equations graphically: $2x + y = 6$, $2x - y - 2 = 0$.

Accepted Answer

Write the second equation as $2x - y = 2$ and make a table for each line.Plot the points, draw both lines and read off where they cross.The lines intersect at (2, 2), so $x = 2$, $y = 2$.Check: $2(2) + 2 = 6$ ✓ and $2(2) - 2 - 2 = 0$ ✓.

Question 30

Find the point of intersection of the lines shown on the cover page.

Accepted Answer

Method. Read two clearly marked points on each line on the cover, write each line’s equation using the slope $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ and the form $y = mx + d$, then solve the two equations together (by substitution or elimination). The solution $(x, y)$ is the point where the lines cross; check it by substituting into both equations.The two lines for this question are printed on the front cover of Ganita Manjari Part 2, not inside the chapter. Read two clear grid points on each line from your copy of the book and follow the four steps above — the method is exactly the one used in End-of-Chapter Q3 and Exercise Set 13.5 Q2.

Question 31

Give a formula to find the $x$-intercept of the line $y = mx + c$.

Accepted Answer

The $x$-intercept is where the line crosses the $x$-axis, i.e. where $y = 0$. $0 = mx + c$ $mx = -c$ $x = -\dfrac{c}{m} \ \ (m \ne 0)$ So the $x$-intercept is $-\frac{c}{m}$, and the line meets the $x$-axis at $\left(-\frac{c}{m}, 0\right)$.Example: $y = 2x - 6$ has $x$-intercept $-\dfrac{-6}{2} = 3$; indeed $2(3) - 6 = 0$.If $m = 0$ the line $y = c$ is horizontal. It never meets the $x$-axis when $c \ne 0$, and it is the $x$-axis when $c = 0$ — that is why the formula needs $m \ne 0$.

Question 32

A person is choosing between two mobile plans. Plan A: ₹50 monthly fee + ₹0.20 per minute of call time. Plan B: ₹30 monthly fee + ₹0.30 per minute of call time. For how many minutes of calling per month is Plan A cheaper than Plan B? For how many minutes is Plan B cheaper? Also find the number of mi

Accepted Answer

Let $t$ be the minutes of calling in a month and $y$ the monthly bill in ₹. $\text{Plan A: } y = 50 + 0.2t$ $\text{Plan B: } y = 30 + 0.3t$ Equal cost: set the two bills equal. $50 + 0.2t = 30 + 0.3t$ $20 = 0.1t \Rightarrow t = 200$ At 200 minutes both cost $50 + 0.2(200) = 90$, i.e. ₹90.Which is cheaper on each side? Plan B starts lower (₹30 < ₹50) but rises faster (0.30 > 0.20 per minute). $t = 100:\ A = 70,\ B = 60$ $t = 300:\ A = 110,\ B = 120$ Plan A is cheaper for more than 200 minutes a month, Plan B is cheaper for fewer than 200 minutes, and both cost the same (₹90) at exactly 200 minutes.

Question 33

How many lines exist that (i) have a given slope? (ii) have a given slope and pass through a given point?

Accepted Answer

(i) Infinitely many. All lines $y = mx + d$ with the same $m$ and any value of $d$ have that slope. They are all parallel to one another, e.g. $y = 2x$, $y = 2x + 1$, $y = 2x - 5$, …(ii) Exactly one. If the line $y = mx + d$ must pass through $(x_1, y_1)$, then $y_1 = mx_1 + d \Rightarrow d = y_1 - mx_1$ which fixes $d$ completely, so only one line is possible. (Of all the parallel lines in (i), exactly one goes through the given point.)

Question 34

For what values of $p$ does the pair of equations given below have a unique solution? $4x + py + 8 = 0$; $2x + 2y + 2 = 0$.

Accepted Answer

A unique solution needs $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$. $\dfrac{a_1}{a_2} = \dfrac{4}{2} = 2$ $\dfrac{b_1}{b_2} = \dfrac{p}{2}$ $\dfrac{p}{2} \ne 2 \Rightarrow p \ne 4$ The pair has a unique solution for every value of $p$ except $p = 4$.(When $p = 4$: $\dfrac{4}{2} = \dfrac{4}{2} = 2$ but $\dfrac{c_1}{c_2} = \dfrac{8}{2} = 4 \ne 2$, so the lines are parallel and there is no solution at all.)

Question 35

Find the values of $a$ and $b$ for which the following system of equations has infinitely many solutions: $(a + b)x - 2by = 5a + 2b + 1$; $3x - y = 14$.

Accepted Answer

Infinitely many solutions means the lines coincide, so all three ratios are equal. (Comparing the constants on the right-hand side directly gives the same ratio as comparing $c_1/c_2$ after moving them across.)

$\dfrac{a + b}{3} = \dfrac{-2b}{-1} = \dfrac{5a + 2b + 1}{14}$

First = second:

$\dfrac{a + b}{3} = 2b$

$a + b = 6b \Rightarrow a = 5b \ \ \ (1)$

Second = third:

$2b = \dfrac{5a + 2b + 1}{14}$

$28b = 5a + 2b + 1$

$26b - 5a = 1 \ \ \ (2)$

Substitute (1) into (2):

$26b - 25b = 1 \Rightarrow b = 1$

$a = 5(1) = 5$

So $a = 5$, $b = 1$.

Check: the first equation becomes $6x - 2y = 25 + 2 + 1 = 28$, which is exactly $2(3x - y) = 2(14)$ — the same line. ✓

Question 36

Find the value of ‘$k$’ for which the following system of equations represents a pair of coincident lines: $x + 2y = 3$; $(k - 1)x + (k + 1)y = k + 3$.

Accepted Answer

Coincident lines: all three ratios equal.

$\dfrac{1}{k - 1} = \dfrac{2}{k + 1} = \dfrac{3}{k + 3}$

First = second:

$k + 1 = 2(k - 1)$

$k + 1 = 2k - 2 \Rightarrow k = 3$

Check the third ratio with $k = 3$:

$\dfrac{1}{2} = \dfrac{2}{4} = \dfrac{3}{6} = \dfrac{1}{2}$

All three agree, so $k = 3$. The second equation becomes $2x + 4y = 6$, which is the first one doubled.

Question 37

(i) Robot 1 starts from the origin and traces a path by repeatedly moving 3 units to the right and then 4 units upward. Robot 2 starts from the point (10, 0) and repeatedly moves 1 unit to the right and then 2 units upward. Will the paths traced by these two robots intersect and if so, where? (ii) R

Accepted Answer

After each complete move a robot has gone a fixed run and a fixed rise, so the points it reaches lie on a straight line whose slope is rise ÷ run. We find each line and then solve the pair.(i) Robot 1: starts at (0, 0), slope $\dfrac{4}{3}$. $y = \dfrac{4}{3}x$ Robot 2: starts at (10, 0), slope $\dfrac{2}{1} = 2$. $y - 0 = 2(x - 10) \Rightarrow y = 2x - 20$ Substitution: $\dfrac{4}{3}x = 2x - 20$ $4x = 6x - 60 \Rightarrow x = 30$ $y = \dfrac{4}{3}(30) = 40$ Yes, the paths meet at (30, 40). It is a point both robots actually reach: Robot 1 after 10 moves $(3 \times 10, 4 \times 10)$ and Robot 2 after 20 moves $(10 + 20, 2 \times 20)$.(ii) Robot 1: starts at (3, 0), slope $\dfrac{5}{5} = 1$. $y = x - 3 \ \ (x \ge 3)$ Robot 2: starts at (7, 0), slope $\dfrac{-3}{5}$. $y = -\dfrac{3}{5}(x - 7) \ \ (x \ge 7)$ Solve the two equations together: $x - 3 = -\dfrac{3}{5}x + \dfrac{21}{5}$ $5x - 15 = -3x + 21$ $8x = 36 \Rightarrow x = 4.5,\ y = 1.5$ The lines meet at (4.5, 1.5) — but Robot 2 starts at $x = 7$ and only moves to the right, so it never gets back to $x = 4.5$. For every $x \ge 7$, Robot 1 is above the $x$-axis ($y = x - 3 \ge 4$) and Robot

Question 38

* At a certain time, Jacob notices that his digital watch reads ‘$a$’ minutes after two o’clock. Fifteen minutes later, it reads ‘$b$’ minutes after three o’clock. He noticed that $a$ is six times greater than $b$. What time was it when he looked at his watch for the second time?

Accepted Answer

The first reading is 2:$a$. Fifteen minutes later the hour has changed to 3, so $a + 15$ minutes past 2 is the same as $b$ minutes past 3, i.e. $60 + b$ minutes past 2: $a + 15 = 60 + b$ $a - b = 45 \ \ \ (1)$ “$a$ is six times $b$”: $a = 6b \ \ \ (2)$ Substitute (2) into (1): $6b - b = 45 \Rightarrow b = 9$ $a = 54$ The first time was 2:54 and the second, 15 minutes later, was 3:09.“Six times greater than” is read here as $a = 6b$, which gives whole-minute readings. Reading it as $a = b + 6b = 7b$ would give $6b = 45$, $b = 7.5$ — not a possible digital-watch reading, which confirms the intended meaning.

Question 39

* The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.

Accepted Answer

Let the tens digit be $x$ and the units digit $y$. The number is $10x + y$; reversed, $10y + x$. $x + y = 15$ $(10y + x) - (10x + y) = 9$ $9y - 9x = 9 \Rightarrow y - x = 1$ Add $x + y = 15$ and $y - x = 1$: $2y = 16 \Rightarrow y = 8$ $x = 15 - 8 = 7$ The number is 78. Check: $7 + 8 = 15$ ✓ and $87 - 78 = 9$ ✓.

Question 40

* In a cyclic quadrilateral ABCD, $\angle A = (x + 7)^\circ$, $\angle B = (y + 8)^\circ$, $\angle C = (3y + 23)^\circ$ and $\angle D = (4x + 12)^\circ$. Find all four angles of the cyclic quadrilateral.

Accepted Answer

Opposite angles of a cyclic quadrilateral add up to 180°.

$\angle A + \angle C = 180^\circ$:

$(x + 7) + (3y + 23) = 180$

$x + 3y = 150 \ \ \ (1)$

$\angle B + \angle D = 180^\circ$:

$(y + 8) + (4x + 12) = 180$

$4x + y = 160 \ \ \ (2)$

From (2), $y = 160 - 4x$. Substitute into (1):

$x + 3(160 - 4x) = 150$

$x + 480 - 12x = 150$

$-11x = -330 \Rightarrow x = 30$

$y = 160 - 120 = 40$

$\angle A = 30 + 7 = 37^\circ$

$\angle B = 40 + 8 = 48^\circ$

$\angle C = 3(40) + 23 = 143^\circ$

$\angle D = 4(30) + 12 = 132^\circ$

$\angle A = 37^\circ$, $\angle B = 48^\circ$, $\angle C = 143^\circ$, $\angle D = 132^\circ$. Check: $37 + 143 = 180$ ✓, $48 + 132 = 180$ ✓, and all four add to 360°.

Question 41

* A train moving with uniform speed for a certain distance takes 6 hours less if its speed is increased by 6 km/hour. It would have taken 6 hours more had its speed been decreased by 4 km/hour. Find the distance travelled and the speed of the train.

Accepted Answer

Let the speed be $v$ km/h and the time taken $t$ hours. Distance = $vt$, and the distance is the same in all three cases.Faster by 6 km/h, 6 hours less: $(v + 6)(t - 6) = vt$ $vt - 6v + 6t - 36 = vt$ $-6v + 6t = 36 \Rightarrow t - v = 6 \ \ \ (1)$ Slower by 4 km/h, 6 hours more: $(v - 4)(t + 6) = vt$ $vt + 6v - 4t - 24 = vt$ $6v - 4t = 24 \Rightarrow 3v - 2t = 12 \ \ \ (2)$ The $vt$ terms cancel, leaving a linear pair. From (1), $t = v + 6$. Substitute into (2): $3v - 2(v + 6) = 12$ $v - 12 = 12 \Rightarrow v = 24$ $t = 30$ $\text{distance} = 24 \times 30 = 720$ Speed 24 km/h, distance 720 km. Check: $720 \div 30 = 24$ h (6 h less) ✓; $720 \div 20 = 36$ h (6 h more) ✓.

Question 42

* The age of a father is equal to the sum of the ages of his four children. After 20 years, the sum of the ages of the children will be twice the age of the father. Find the age of the father.

Accepted Answer

Let the father’s present age be $x$ years and the sum of the four children’s ages $y$ years. $x = y \ \ \ (1)$ In 20 years each of the four children is 20 years older, so their total grows by $4 \times 20 = 80$; the father grows by 20. $y + 80 = 2(x + 20) \ \ \ (2)$ Substitute $y = x$ from (1): $x + 80 = 2x + 40$ $x = 40$ The father is 40 years old. Check: in 20 years the children total $40 + 80 = 120$ and the father is 60; $120 = 2 \times 60$ ✓.The trap is adding only 20 to the children’s total. Each of the four children ages by 20 years, so their sum grows by 80.

Standard form	\(a\)	\(b\)	\(c\)
\(-\sqrt{2}x + y - 15 = 0\)	\(-\sqrt{2}\)	1	−15
\(-2x + 3y + 0 = 0\)	−2	3	0
\(5x - 3y + 0 = 0\)	5	−3	0
\(x + 0\times y - 8 = 0\)	1	0	−8
\(0\times x + 3y - 1 = 0\)	0	3	−1

Equation	Point 1	Point 2
\(5x + 3y = 7\)	(2, −1): IV	(−1, 4): II
\(5x - 3y = 7\)	(2, 1): I	(−1, −4): III
\(-5x + 3y = 7\)	(1, 4): I	(−2, −1): III
\(-5x - 3y = 7\)	(−2, 1): II	(1, −4): IV

Two Variables, One Line

Exercise Set 13.1

Exercise Set 13.2

Exercise Set 13.3

Exercise Set 13.4

Exercise Set 13.5

End-of-Chapter Exercises

Pair	Ratios	Lines
(i)	\(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\)	intersecting
(ii)	all three equal	coincident
(iii)	\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\)	parallel

Statement	Verdict
(i)	False
(ii)	False
(iii)	False
(iv)	True
(v)	True
(vi)	False