NCERT Solutions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials (PDF) – 2027

The expression is the total cost of the garden in Example 2, 200l + 160w + 50lw.

Example 1 gave 4x + 5y + 3. Example 2 gives 200l + 160w + 50lw. Two differences matter.

• The highest power. In 4x + 5y + 3 every variable appears to the power 1 and no term mixes the two letters. In 200l + 160w + 50lw the term 50lw multiplies the two variables together, so that term has degree 1 + 1 = 2.
• The constant. 4x + 5y + 3 has the constant term 3 (the free pens). The garden expression has no constant — if the garden shrinks to nothing, the cost is 0.

The wire is the perimeter, so for every such rectangle

So any pair of positive lengths adding to 10 works. A few more:

The expression is the area of the bent-wire rectangle, 10x − x².

Similarity: both are algebraic expressions — sums of terms, each term a number multiplying letters.

Differences:

The degree is the highest power of the variable that actually appears.

Any expression in one variable whose highest power is the required number will do. For example:

Read the number attached to each power, together with the sign in front of it.

There is no z term written, so its coefficient is

The constant term is the one with no x in it.

The perimeter of a square of side x is the linear polynomial 4x.

Each extra 0.5 cm of side adds 2 cm to the perimeter, every single time.

The amount for m matches is the linear polynomial 200 + 50m. Set it equal to 750 — that turns the polynomial into a linear equation.

Check: 200 + 50 × 11 = 200 + 550 = 750. ✓

Yes. The machine takes in a length x and returns the area of the rectangle that the 20 cm wire then makes.

Check by the shape: if the length is 6 cm, the width is 10 − 6 = 4 cm, and 6 × 4 = 24 cm². ✓

Substitute each value in turn.

Let Salil’s present age be x years. Then his mother’s present age is 3x years.

Salil is 15 years old and his mother is 45 years old.

Check: in 5 years they will be 20 and 50, and 20 + 50 = 70. ✓

The ratio 2 : 5 lets both numbers be written with one letter. Let them be 2k and 5k.

Check: 105 − 42 = 63 ✓  and  42 : 105 = 2 : 5 (divide both by 21). ✓

Let the number of five-rupee coins be f. Then the number of two-rupee coins is 3f.

Ruby has 8 five-rupee coins and 24 two-rupee coins (32 coins in all).

Check: 24 × ₹2 = ₹48 and 8 × ₹5 = ₹40; ₹48 + ₹40 = ₹88. ✓

Let the shorter piece be x feet. Then the longer piece is 4x feet.

The pieces are 60 feet and 240 feet.

Check: 60 + 240 = 300 ✓ and 240 = 4 × 60. ✓

Let the width be w cm. Then the length is (2w + 3) cm.

Width = 3 cm, length = 2(3) + 3 = 9 cm.

Check: perimeter = 2(9 + 3) = 24 cm ✓ and 9 is three more than twice 3. ✓

Stages 1 to 4 of Fig. 2.4 use 1, 3, 5 and 7 tiles. Each new stage adds two tiles — one to each column — so the next three stages use 9, 11 and 13 tiles.

Sequence up to Stage 7: 1, 3, 5, 7, 9, 11, 13.

Forward — stage given, tiles wanted. Substitute in 2n − 1.

Backward — tiles given, stage wanted. Now 2n − 1 becomes a linear equation.

So Stage 11 has 21 tiles and Stage 24 has 47 tiles.

Bela starts with ₹100 and spends ₹5 a day, so after n days she has ₹(100 − 5n).

The money runs out when the amount left is zero.

For a journey of n km (with n ≥ 2) the fare is ₹25 + 15(n − 2) = ₹(15n − 5).

Check the other way: for 9 km the meter charges ₹25 for the first 2 km and ₹15 for each of the remaining 7 km: 25 + 15 × 7 = 25 + 105 = ₹130. ✓

She begins with ₹500 and adds ₹150 at the end of each month.

From the second month onwards the balances are ₹800, ₹950, ₹1100, ₹1250, … — each ₹150 more than the one before.

Check: n = 2 gives 500 + 300 = ₹800. ✓

Check: n = 3 gives 120 − 27 = 93. ✓

With the length fixed at 13 cm, area = 13 × breadth.

Length × breadth = 7 × 11 = 77 cm², and that base is fixed. So volume = 77 × height.

The cost function is C(d) = 100 + 60d.

Now run the machine backwards.

The height after t months is h(t) = 3 − 0.5t.

(i) Seven months of growth at 0.5 ft a month adds 3.5 ft.

(ii)

(iii)

(i)

(ii)

(iii)

(i)

(ii)

(iii)

(i)

It is linear decay because b(x + 1) − b(x) = −15 for every x: a fixed amount is removed over equal intervals, and the coefficient of x is negative. The graph would be a straight line falling from (0, 600).

(ii) The balance runs out when b(x) = 0.

(iii)

Substitute the two observations into y = ax + b.

Subtracting (1) from (2) removes b:

Check: x = 10 gives 250 + 150 = ₹400 ✓ and x = 14 gives 350 + 150 = ₹500. ✓

(2) − (1):

Check: 60(15) + 200 = 900 + 200 = ₹1100. ✓

Here °F is the input and °C the output. The two fixed points give:

(2) − (1):

Check: °F = 32 gives (5/9)(0) = 0 °C ✓; °F = 212 gives (5/9)(180) = 100 °C. ✓

The line is y = 2x + 1, so double each x and add 1.

So (1, 3), (2, 5), (5, 11), (7, 15), (9, 19), (12, 25) and (20, 41) all lie on the line.

In every listed point the y-coordinate is −2 times the x-coordinate.

Verifying the hint’s two points on y = ½x:

Both satisfy the equation, so both lie on the line — and two points are all a straight line needs.

Convenient points for the three lines:

Three conclusions, all visible at once when the lines share axes.

1. Every line y = ax passes through the origin (0, 0), whatever a is.
2. All of them rise from left to right, since a > 0.
3. The bigger a is, the steeper the line. Taking y = x as the reference: if a > 1 the line is steeper than y = x; if a < 1 it is less steep.

The picture mirrors the previous one, with one crucial change of sign.

1. Every line y = −ax still passes through the origin.
2. All of them now fall from left to right, because the slope −a is negative — these are pictures of linear decay.
3. Steepness is decided by the size of a, not its sign: if a > 1 the line falls more steeply than y = −x; if a < 1 it falls more gently.

Yes. With a = 2 fixed and b = −1, 1, 5 the three lines are parallel: same tilt, different heights.

Each has b = 0, so plot the origin and one more point.

Role of a and b: here b = 0, so all three lines pass through the origin. Only a changes: the larger a is (4 > 2 > 1), the steeper the line. All the slopes are positive, so all three lines rise from left to right.

Role of a and b: b = 0, so all three pass through the origin. All the slopes are negative, so the lines fall from left to right (linear decay). The bigger the size of a (6 > 3 > 1), the steeper the fall.

Role of a and b: b = 0 for both, so both lines pass through the origin. The slopes 5 and −5 have the same size but opposite signs: the lines are equally steep, one rising (growth) and the other falling (decay).

Role of a and b: a = 3 is the same for all three, so the lines are parallel (equally steep). Changing b (−1, 0, 1) only shifts the line up or down, so they cut the y-axis at (0, −1), (0, 0) and (0, 1).

Role of a and b: y = −2x − 3 and y = −2x have the same a = −2, so they are parallel; b = −3 and 0 decide where they cut the y-axis. y = 2x + 3 has a different slope, +2, so it is not parallel to them and cuts both.

Other correct answers: 4x³ − 7x²,   −x³ − 7x² + 1,   2x³ − 7x² − 3x.

(i) Substitute x = 1.

(ii) Substitute t = a. The input is a letter, so the output is an expression.

Let the number be x. The description translates directly into a linear equation.

Check: (5/2)(−1/2) = −5/4; and −5/4 + 2/3 = −15/12 + 8/12 = −7/12. ✓

Let the smaller number be x. The other is 5x. After adding 21 they become x + 21 and 5x + 21.

Since 5x + 21 is the larger of the two, it must be the one that is twice the other.

The numbers are 7 and 35.

Check: adding 21 gives 28 and 56, and 56 = 2 × 28. ✓

Let n be the number of months. The amount is A = 800 + 250n.

Let the tens digit be t and the units digit be u.

Also the digits differ by 3, so the two digits add to 13 and differ by 3:

The two numbers are 85 and 58.

Check: 85 + 58 = 143 ✓ and 8 − 5 = 3. ✓

First put every equation in the form y = ax + b by dividing through by the coefficient of y. Then a is the slope and b the y-intercept, and the line cuts the y-axis at (0, b).

Two convenient points for each line:

Parallel lines: yes — (ii) and (iv).

(i) Put x = 313.

(ii) Put y = 158 and work backwards.

Check: (9/5)(343 − 273) + 32 = (9/5)(70) + 32 = 126 + 32 = 158 °F. ✓

In words: work = force × distance. With the force fixed at 3 units,

This is a linear equation in the two variables w and d, with slope 3 and constant term 0.

(i) A linear polynomial has the form p(x) = ax + b. The two points give two equations.

(ii) The graph cuts the y-axis where x = 0 and the x-axis where p(x) = 0.

(iii)

Verification: p(1) = 3 + 2 = 5 ✓ and p(3) = 9 + 2 = 11. ✓

Now solve a + c = 6 and a − c = −2.

(i) Stage 4 has four hexagons in the chain and Stage 5 has five.

(ii)

(iii)

(iv)

(v) Set the rule equal to 200.

Since n must be a whole number, no — 200 matchsticks cannot form a stage of this pattern. Stage 39 uses 196 and Stage 40 uses 201.

Finding p(x): the two points give two equations.

Finding q(x): parallel means the same slope, so c = 2. Then use the point (4, −1).

Where they meet the x-axis (set each polynomial to 0):

Factorise, and the answer falls out.

Every such line passes through the point (−1, 0) — they all cut the x-axis at the same place.

They also share:

• slope = y-intercept = a, so each line meets the y-axis at (0, a);
• positive slope (as a > 0), so every line rises from left to right (linear growth).