NCERT Solutions for Class 9 Maths Chapter 13 Two Variables, One Line (PDF) – 2027

Put the values into the standard form ax + by + c = 0:

So the equation is 3x + 0·y − 1/5 = 0 (which is the same as 3x − 1/5 = 0). It is still an equation in two variables, because the y-term is present with coefficient 0.

Bring every term to the left-hand side so that the right-hand side is 0, then read off a, b and c.

A missing variable has coefficient 0, and a missing number term means the constant is 0.

(i) Charlie’s equation t = 2p is correct. The notebook is the costlier item, so its cost t equals two times the pen’s cost p. Meera’s p = 2t says the pen costs twice the notebook, which is the opposite.

(ii) Yes. The runs of the two batsmen add up to the total, so x + y = 176 is correct. In standard form it is x + y − 176 = 0.

Substitute each pair in the left-hand side 3x + 2y and compare with 12.

(1, 9/2), (0, 6) and (1/2, 21/4) are solutions; (4, 1) is not.

Yes, many more. Choose any value of x, put it in the equation and solve for y:

Every value of x gives one value of y, and there are infinitely many values of x. So the equation has infinitely many solutions.

Yes, I agree. If c = 0 the equation is ax + by = 0. Putting x = 0, y = 0:

This is true for every choice of a and b, so (0, 0) is always a solution and the line always passes through the origin. Toshi’s and Meena’s cases (the y-axis and the x-axis) are just the special cases b = 0 and a = 0.

Put x = 4 and y = 3 in the left-hand side:

The two sides are equal, so (4, 3) is a solution of 5x − 6y = 2. Equivalently, the point (4, 3) lies on the graph of this line.

(i) 7x − 3y = 21

(ii) 2x + 3y = 5

Two solutions: (i) (0, −7) and (3, 0); (ii) (1, 1) and (4, −1).

Substitute x = 2, y = −1 in each equation.

m = 5/2 (= 2.5) and n = 18.

Choose x values that give whole-number y values, then check signs: I (+, +), II (−, +), III (−, −), IV (+, −).

Each pair satisfies its equation; for example in (i), 5(2) + 3(−1) = 7 and 5(−1) + 3(4) = 7. Plotted, both points of each pair lie on that line.

No, C (2, 3) does not lie on the line; the graph shows it well above the line. It does not satisfy the equation either:

No point off the line can satisfy the equation. The graph of an equation is made of exactly its solutions: every solution lies on the line, and every point on the line is a solution.

(i) False. It has infinitely many solutions: each value of x gives a value of y.

(ii) False. Only when c = 0. For x + y = 1: 0 + 0 ≠ 1, so (0, 0) is not on it.

(iii) False. 3x + 2y = 12 has the rational solutions (2, 3) and (1/2, 21/4).

(iv) True. It can be written x + 0·y − 3 = 0, with a = 1, b = 0 (not both zero). Its graph is a vertical line.

(v) True. For every x, y = (7 − 2x)/3, e.g. (2, 1), (5, −1), (0, 7/3), …

(vi) False. 2(1) + 3(2) = 8 ≠ 7.

(i) Let (p, q) be any solution of 3x + 4y = 7, so 3p + 4q = 7. Then

so (p, q) solves the second equation. Conversely, if 6p + 8q = 14, dividing by 2 gives 3p + 4q = 7. So both equations have exactly the same solutions, e.g. (1, 1), (−3, 4), (5, −2).

(ii) If ap + bq = c, multiplying by k gives kap + kbq = kc. If kap + kbq = kc, then, because k ≠ 0, we can divide by k to get ap + bq = c. Each solution of one is a solution of the other, so the solution sets are the same: the two equations draw the same line.

The slope of AB is rise/run = 3/2: every 2 steps right, the line rises 3. From B (4, 3) to x = 8 is a run of 4 = 2 × 2, so the rise is 2 × 3 = 6.

The point is (8, 9). This is the point D in Fig. 13.8.

From B (4, 3) the line must rise 6 − 3 = 3. A rise of 3 needs a run of 2 (slope 3/2).

The point is (6, 6), which is the point C in Fig. 13.8.

Use slope = (y₂ − y₁)/(x₂ − x₁) for each pair of points:

Every pair gives slope = −5, the same as m in y = −5x + 3. So the slope is verified.

Steepness = slope = rise ÷ run.

From least to greatest steepness: B, C, A.

The slope of the ramp is 5/8 (0.625): it rises 0.625 m for every metre of horizontal distance.

Count squares between two points on l (the marked dots where given): slope = rise/run, with a fall counted as negative.

(i) 0. l is horizontal: the rise between any two points is 0, so slope = 0/run = 0.

(ii) −1. From the upper dot to the lower dot: run = 4 squares right, rise = −4 (down 4). Slope = −4/4 = −1.

(iii) 2/3. From the lower dot to the upper dot: run = 3 squares, rise = 2 squares. Slope = 2/3.

(iv) Undefined. l is vertical: the run is 0, and division by 0 is not defined.

The required slope is rise/run = 1/12. The ramp must rise 18 cm, so

The horizontal length of the ramp should be 216 cm (2.16 m).

Let one paratha cost ₹p and one bowl of dahi ₹d.

A paratha costs ₹110/3 ≈ ₹36.67 and a bowl of dahi costs ₹80/3 ≈ ₹26.67.

Reasoning: two of the first order (2 parathas + 4 dahi) cost ₹180. The second order has the same 2 parathas but 3 more bowls of dahi, and costs ₹80 more. So 3 bowls of dahi cost ₹80.

Why: both groups paid the same ticket prices. So the true prices (x, y) must make Group A’s bill correct and Group B’s bill correct at the same time, i.e. satisfy both equations. One equation alone has infinitely many solutions; only the common solution fits both facts.

Adult ticket ₹180, child ticket ₹80.

Cost = (number of tickets × ₹x) + (number of snack boxes × ₹y).

The pair is 2x + 3y = 850 and 4x + y = 1100.

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

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

One equation counts people, the other counts money.

The pair is x + y = 200 and 150x + 100y = 25000 (the second can be divided by 50 to give 3x + 2y = 500).

Let T, R, H, C be the values of a triangle, rectangle, hexagon and circle.

Triangle = 6, rectangle = 8, hexagon = 7, circle = 4, and the missing column total is 21.

Yes. Substitute y = (3 − x)/2 in (1):

We get x = 49/29, y = 19/29, exactly as before.

Eliminating x:

Substitution: from (2), y = (7x − 2000)/3. Put in (1):

All methods give x = 2000, y = 4000 (incomes ₹18,000 and ₹14,000). Elimination avoids fractions but here needs larger multipliers for x (7 and 9) than for y (3 and 4). Substitution is quick when a coefficient is 1, but here it brings fractions.

No, not always. Both variables vanish together when one equation’s left side is a multiple of the other’s. For example:

Here the equations are the same, so there are infinitely many solutions. But with

there is no solution. So when both variables are eliminated, the result is either “0 = 0” (infinitely many) or “0 = a non-zero number” (none).

Both are 2x − y = 4 in disguise (divide by 5 and by 2), so there are infinitely many solutions, e.g. (0, −4), (2, 0), (3, 2).

Three more examples:

Rule: one equation is a non-zero multiple of the other, i.e. a₁/a₂ = b₁/b₂ = c₁/c₂.

Then that ratio cannot be written (division by 0), so use the rule in its multiple form instead: a₁ = ka₂, b₁ = kb₂, c₁ = kc₂ with k ≠ 0. A zero in one equation must then match a zero in the same place in the other.

So the rule still works; just compare by multiples, or compare only the ratios that exist and check the zero terms directly.

No. A pair can also have no solution. Example:

The same expression x − y cannot equal both 10 and 11, so no pair (x, y) satisfies both. Elimination gives 0 = 1, which is false. Graphically the two lines are parallel (same slope 1, different intercepts) and never meet.

In each, multiplying the first equation makes the left sides identical but the right sides differ (e.g. 4x − 6y = 8 against 4x − 6y = 7), and subtracting gives a false statement such as 0 = 1. So there is no solution.

Rule: a₁/a₂ = b₁/b₂ ≠ c₁/c₂: the variable parts are in the same ratio but the constants are not.

Yes, all three converses are true (with a₂, b₂, c₂ non-zero):

1. If the pair has a unique solution, then a₁/a₂ ≠ b₁/b₂.
2. If it has infinitely many solutions, then all three ratios are equal.
3. If it has no solution, then a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Reason: the three conditions cover every possible pair, and no pair satisfies two of them. The three outcomes also cannot happen together. So, for example, if a pair has a unique solution it cannot be in case 2 or 3 (those give infinitely many or none); it must be in case 1.

Yes. Two examples:

1. Comparison (equating) method. Make the same variable the subject of both equations and set the two expressions equal. For x + y = 25 and x − y = 11:

2. Adding and subtracting. When the coefficients are swapped, as in 2x + 3y = 600, 3x + 2y = 700, adding gives x + y = 260 and subtracting gives x − y = 100, two very simple equations.

Pair 1: the lines cross at one point, (8/3, 2/3). Pairs 2 and 3: the two lines coincide (one line). Pair 4: the lines are parallel and never meet.

A solution of the pair must satisfy both equations, so it is a point that lies on both lines. We read off the common point(s) of the two lines.

(i) Intersect at a point: exactly one common point, so a unique solution, e.g. (8/3, 2/3) for Pair 1.
(ii) Parallel: no common point, so no solution (Pair 4).
(iii) Coincident: every point of the line is common, so infinitely many solutions (Pairs 2 and 3).

Take two different lines with the same slope: y = mx + d₁ and y = mx + d₂, with d₁ ≠ d₂. Suppose they had a common point (p, q). Then

This contradicts d₁ ≠ d₂. So the lines have no common point: they are parallel. (If d₁ = d₂, they are the same line.)

Let the integers be x and y, with difference x − y.

The numbers are −8 and 13.

Let the larger number be x and the smaller y.

The numbers are 39 and 13.

Let a bat cost ₹x and a ball ₹y.

A bat costs ₹1200 and a ball costs ₹80.

Let the fixed charge be ₹x and the charge per km ₹y.

Fixed charge ₹25, ₹13 per km; a 25 km journey costs ₹350.

Let the fraction be x/y.

The fraction is 7/9.

Let the fraction be x/y.

The fraction is 3/5.

Let Nuri’s present age be x years and Sonu’s y years.

Nuri is 50 years old and Sonu is 20 years old.

Let the tens digit be x and the units digit y. The number is 10x + y; reversed it is 10y + x.

The number is 18.

Let her get x notes of ₹50 and y notes of ₹100.

10 notes of ₹50 and 15 notes of ₹100.

Let the fixed charge be ₹x and the charge per extra day ₹y. Seven days = 3 + 4 extra days; five days = 3 + 2 extra days.

Fixed charge ₹15; ₹3 for each extra day.

Let the number of boys be x and of girls y.

The lines meet at (3, 7). So 3 boys and 7 girls took part.

(i) Intersect at a point (a₁/a₂ ≠ b₁/b₂).
(ii) Coincident (all three ratios equal 1/2).
(iii) Parallel (a₁/a₂ = b₁/b₂ = 3, but c₁/c₂ = 10/9 ≠ 3).

(i) has infinitely many solutions: every point of the line x + y = 5, e.g. (0, 5), (2, 3), (5, 0).
(ii) No solution.
(iii) Unique solution x = 2, y = 2, where the lines through (0, 6), (3, 0) and (0, −2), (1, 0) meet.
(iv) No solution.

Let the length be l m and the width w m. Half the perimeter is l + w.

The garden is 20 m long and 16 m wide.

(i) Intersecting: 3x + 2y − 7 = 0, since 2/3 ≠ 3/2.

(ii) Parallel: 4x + 6y − 9 = 0, since 2/4 = 3/6 = 1/2 but −8/−9 = 8/9 ≠ 1/2.

(iii) Coincident: 4x + 6y − 16 = 0, since 2/4 = 3/6 = −8/−16 = 1/2.

The coefficients 9 and 7 are swapped, so adding and subtracting give very simple equations.

A citron costs 8 and a fragrant wood-apple costs 5 (in the units of money of the problem).

Let a pencil cost ₹x and a pen ₹y.

A pencil costs ₹3 and a pen costs ₹5.

Let the height of the stool be h cm and the height of the sitting cat be c cm.

The stool is 55 cm high (and the cat is 30 cm tall).

(i) Slope = 3: from (0, 0) to (2, 6) the run is 2 and the rise is 6.

(ii) y-intercept = 0: the line meets the y-axis at the origin (0, 0). For the tank, the water level is 0 cm at time 0, i.e. the tank starts empty.

(iii)(a) y = 3 × 5 = 15 cm. (b) 21 = 3x ⇒ x = 7 minutes. (The level rises 3 cm every minute.)

(iv) No. At x = 4, y = 3 × 4 = 12, not 10, so (4, 10) does not satisfy y = 3x.

(v) (3, 9). Every point of x = 3 has x-coordinate 3; on y = 3x that gives y = 9.

(i) Table: C = 0, 30, 100, −40 gives F = 32, 86, 212, −40. Plot and join:

(ii) 86 °F. (iii) 35 °C. (iv) 0 °C = 32 °F; 0 °F = −160/9 °C ≈ −17.8 °C. (v) Yes: −40 °C = −40 °F.

The two lines meet at (2, 2). So the solution is x = 2, y = 2.

The cover picture is not printed inside the chapter, so read it from the cover of your own copy of the book and work as follows.

1. On each line, pick two points that sit exactly on grid corners. 2. Find each line’s slope m = rise/run and its y-intercept d, and write y = mx + d. 3. Solve the two equations by elimination or substitution. 4. Check that the point agrees with the picture.

Sample (not the actual cover): if one line passes through (0, 4) and (4, 0) and the other through (0, 0) and (2, 2):

The line meets the x-axis where y = 0:

x-intercept = −c/m; the line cuts the x-axis at (−c/m, 0).

For t minutes, the monthly cost in ₹ is:

Both plans cost the same (₹90) at 200 minutes. Plan A is cheaper for more than 200 minutes; Plan B is cheaper for fewer than 200 minutes.

(i) Infinitely many. All lines y = mx + d with the given m and any d have that slope; they are all parallel to each other.

(ii) Exactly one. If the line passes through (p, q), then q = mp + d, so d = q − mp is fixed. Only one value of d, so only one line.

A unique solution needs a₁/a₂ ≠ b₁/b₂:

The pair has a unique solution for all real values of p except p = 4.

Infinitely many solutions need all three ratios equal:

a = 5, b = 1.

Coincident lines need all three ratios equal:

k = 3.

Each robot repeats one step, so its path follows a straight line of slope rise/run from its starting point.

(i) Yes, at (30, 40).

(ii) No. The lines would cross at (4.5, 1.5), but Robot 2 starts at x = 7 and only moves right, so it never gets there.

The first time is 2 : a. Fifteen minutes later the hour has changed to 3, so a + 15 = 60 + b.

He first looked at 2:54, and the second time it was 9 minutes past 3, i.e. 3:09.

Let the tens digit be x and the units digit y; the number is 10x + y.

The number is 78.

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

∠A = 37°, ∠B = 48°, ∠C = 143°, ∠D = 132°.

Let the speed be v km/h and the time t h; the distance is vt.

Speed 24 km/h; distance 720 km.

Let the father’s age be x years and the sum of the children’s ages y years.

The father is 40 years old.