From Polynomials to Equations: Solving for the Unknown — Class 9 Mathematics Chapter 2

🖼 Image PendingA wide cinematic depiction of a brass mechanical input-output machine with a number being dropped in at the top, gears spinning inside, and a different number emerging from a slot on the side

AI Generation Prompt

Ultra-wide cinematic banner (16:5 ratio). A vintage brass and copper mechanical input-output machine fills the centre of the frame. At the top, a glowing tag with the number '4' is being dropped into a funnel. Visible through a clear panel in the middle, several brass gears with engraved formulas turn slowly. From a slot on the right side, another glowing tag emerges showing the number '11'. The machine sits on a heavy wooden table in a softly-lit Victorian-style mathematical workshop, with chalk-drawn equations on a blackboard in the background. The image conveys: a polynomial is a machine — input goes in, output comes out, the rule never changes. Painterly cinematic illustration. Dark background. No text labels other than the numbers on the input and output tags.

✦Think About It

**Two numbers add up to 64. One is 10 more than the other. What are the two numbers?** Most students try numbers — "is it 30 and 34? No, 30+34 = 64 but the difference is only 4. Maybe 27 and 37?" — and *eventually* land on the answer. **Is there a way to skip all the trial-and-error?**

Give the smaller number a name — say, $x$. What's the larger one in terms of $x$? Now translate the sentence 'they add up to 64' into algebra.

✦बीजगणित — भास्कराचार्य (११५० CE)

From Bhāskarāchārya's Bījagaṇita — On the Unknown

अव्यक्तं नाम राशिं तं यद् आरम्भे अदृश्यते

क्रियायाः अन्ते एव यत् व्यक्तं रूपेण दृश्यते॥

(avyaktaṃ nāma rāśiṃ taṃ yad ārambhe adṛśyate / kriyāyāḥ ante eva yat vyaktaṃ rūpeṇa dṛśyate)

'अव्यक्त (अनजानी) राशि वो है जो शुरुआत में नहीं दिखती — पर हिसाब के अंत में अपने असली रूप में सामने आ जाती है।'

'The avyakta (unknown) is that quantity which is not visible at the start — but reveals its true form at the end of the calculation.'

Bhāskarāchārya II's Bījagaṇita (1150 CE) is the foundational Sanskrit text on algebra. The very word bījagaṇita means the mathematics of seeds — symbolic seeds (variables) that grow into solved equations. Bhāskara taught the Indian world to treat the unknown $x$ as something you can multiply, add, subtract, and rearrange — just like an ordinary number. Every line of algebra you write today carries his fingerprint.

From a polynomial to an equation

So far in this chapter we've treated a linear polynomial like $50m + 200$ as a rule — "plug in any $m$ , get out a value". But sometimes the question runs the other way: "What value of $m$ would give a particular output?" That's a question about equations, not just polynomials.

When we set a linear polynomial equal to a constant, we get a linear equation:

$50m + 200 = 750 \quad\Longleftrightarrow\quad \text{a linear equation in } m$

A value of the variable that makes the equation true is called a solution (or a root) of the equation. Solving the equation means finding that value.

Standard recipe for solving a linear equation:

Move all the variable terms to one side, all the constants to the other.
Combine like terms.
Divide by the coefficient of the variable.

That's it. Three moves. Always works.

Worked Example 1Sum of two numbers (NCERT Ex 6)

NCERT

The sum of two numbers is 64. One is 10 more than the other. What are the two numbers?

Step 1 — Name the unknown.

Let the smaller number be $x$ . Then the larger is $x + 10$ .

Step 2 — Translate the sentence.

'Their sum is 64' becomes:

$x + (x + 10) = 64$

Simplify the left side: $2x + 10 = 64$ . This is a linear equation in $x$ .

Step 3 — Solve.

Subtract 10 from both sides: $2x = 54$ .
Divide both sides by 2: $x = 27$ .

Step 4 — Answer the original question.

Smaller number: $x = 27$ . Larger number: $x + 10 = 37$ .

Step 5 — Check. $27 + 37 = 64$ ✓ and $37 - 27 = 10$ ✓.

Answer: The two numbers are 27 and 37.

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Polynomials as input–output machines (functions)

There is another, equally powerful way to think about a polynomial: as an input-output machine (a function).

Consider the linear polynomial $f(x) = 2x + 3$ . For every value you put in for $x$ , the polynomial gives back a unique value:

Input $x = 4$ → $2(4) + 3 = 11$ . Output: 11.
Input $x = -6$ → $2(-6) + 3 = -9$ . Output: −9.
Input $x = 0$ → $2(0) + 3 = 3$ . Output: 3. (Notice — when the input is zero, the output is the constant term.)

We write $f(4) = 11$ and $f(-6) = -9$ and $f(0) = 3$ . The notation $f(\text{input})$ means "the output of the machine $f$ when fed this input".

Forward direction (substitute): given an input, compute the output.
Reverse direction (solve): given an output, find the input that produces it.

The forward direction is just arithmetic — substitute and simplify. The reverse direction is algebra — set up an equation and solve it. Both are useful, and you'll switch between them constantly.

🖼 Image PendingAn input-output machine illustration. At the top is a small label 'x = 4' attached to a tag dropped into a funnel. The body of the machine, drawn as a metal box with two dials and a small screen, contains the rule 'y = 2x + 3' displayed clearly. At the bottom-right, the machine outputs another tag labelled 'y = 11'.

AI Generation Prompt

An illustration of a stylised input-output machine, drawn in the warm hand-painted style of a children's mathematics textbook. At the top: a small wooden tag labelled 'x = 4' is being dropped through a funnel into the machine. The machine itself is a metallic grey box with two dials, a red button, and a digital display in the middle showing the formula 'y = 2x + 3' in glowing teal letters. At the bottom-right of the machine, an output slot delivers a different wooden tag labelled 'y = 11'. Style: warm illustrative, slightly cartoonish but precise. Dark background, orange accent labels, clean technical illustration style.

📸 Fig. 2.3 — The function $y = 2x + 3$ as an input-output machine. Input $x = 4$ → Output $y = 11$.

Worked Example 2Find the value (NCERT Ex Set 2.2, Q1)

NCERT

Find the value of the linear polynomial $5x - 3$ if (i) $x = 0$ (ii) $x = -1$ (iii) $x = 2$ .

(i) $x = 0$ : $5(0) - 3 = -3$ .

(ii) $x = -1$ : $5(-1) - 3 = -5 - 3 = -8$ .

(iii) $x = 2$ : $5(2) - 3 = 10 - 3 = 7$ .

Side observation: in (i), the value at $x = 0$ is exactly the constant term $-3$ . (Always true.) Notice how the output goes from $-3$ to $-8$ to $7$ — and the per-unit change in $x$ produces a per-unit change of $5$ in the output (the leading coefficient). E.g., from $x = 0$ to $x = 2$ , output went from $-3$ to $7$ , a jump of $10 = 5 \times 2$ . ✓

Worked Example 3Substitute into a quadratic (NCERT Q2)

NCERT

Find the value of the quadratic polynomial $7s^2 - 4s + 6$ if (i) $s = 0$ (ii) $s = -3$ (iii) $s = 4$ .

Substitution works for any polynomial — not just linear ones.

(i) $s = 0$ : $7(0)^2 - 4(0) + 6 = 0 - 0 + 6 = 6$ . (Constant term, as expected.)

(ii) $s = -3$ : $7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81$ .

(iii) $s = 4$ : $7(4)^2 - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 16 + 6 = 102$ .

Important contrast with linear case: the per-unit change here is not constant. From $s = 0$ to $s = 4$ , output jumped by $96$ . From $s = 4$ to $s = 8$ , the jump would be much bigger. This is the signature of a quadratic — the rate of change itself changes.

Worked Example 4Solve a linear equation

SOLVED

Solve $3x - 5 = 16$ .

Step 1 — Move the constant. Add 5 to both sides:

$3x = 21$

Step 2 — Divide by the coefficient. Divide both sides by 3:

$x = 7$

Step 3 — Check. $3(7) - 5 = 21 - 5 = 16$ . ✓

Answer: $x = 7$ .

Worked Example 5Linear equation with a fraction

SOLVED

Solve $\frac{x}{2} + 7 = 12$ .

Step 1 — Subtract 7: $\frac{x}{2} = 5$ .

Step 2 — Multiply by 2: $x = 10$ .

Check: $\frac{10}{2} + 7 = 5 + 7 = 12$ . ✓

Worked Example 6Auto fare word problem

SOLVED

An auto-rickshaw charges a base fare of ₹40 plus ₹15 per kilometre. How many kilometres can you travel for ₹130?

Step 1 — Set up the function. Let $d$ = distance in km. Cost = $40 + 15d$ .

Step 2 — Set the cost equal to ₹130.

$40 + 15d = 130$

Step 3 — Solve.

Subtract 40: $15d = 90$ .
Divide by 15: $d = 6$ .

Step 4 — Check. $40 + 15(6) = 40 + 90 = 130$ ✓.

Answer: You can travel 6 km for ₹130.

Logical ReasoningLevel 4 · Evaluation

Can two **different** inputs to a linear polynomial $f(x) = 2x + 3$ ever produce the **same** output? What about for a quadratic like $g(x) = x^2$?

Practice Yourself — Substitute and Solve

A. Substitute (forward direction):

Find $f(3)$ if $f(x) = 4x - 7$ .
Find $g(-2)$ if $g(y) = -3y + 5$ .
Find $h(0)$ if $h(t) = 9t + 11$ .
Find $p(10)$ if $p(n) = 50n + 200$ .

B. Solve (reverse direction):

$4x + 5 = 21$ .
$7y - 3 = 25$ .
$\frac{m}{4} + 1 = 6$ .
$-2x + 9 = 1$ .

C. Word problems:

The cost of $n$ tickets is given by $T(n) = 80n + 50$ . How many tickets cost ₹530?
A taxi charges ₹100 base + ₹14 per km. How many km can you travel for ₹254?

Answers: 1. 5 · 2. 11 · 3. 11 (the constant term) · 4. 700 · 5. $x = 4$ · 6. $y = 4$ · 7. $m = 20$ · 8. $x = 4$ · 9. 6 tickets · 10. 11 km.

Quick Check

Q1.If $f(x) = 5x - 3$ , what is $f(2)$ ?

AI Generation Prompt

✦Think About It

Give the smaller number a name — say, $x$. What's the larger one in terms of $x$? Now translate the sentence 'they add up to 64' into algebra.

✦बीजगणित — भास्कराचार्य (११५० CE)

From Bhāskarāchārya's Bījagaṇita — On the Unknown

अव्यक्तं नाम राशिं तं यद् आरम्भे अदृश्यते

क्रियायाः अन्ते एव यत् व्यक्तं रूपेण दृश्यते॥

(avyaktaṃ nāma rāśiṃ taṃ yad ārambhe adṛśyate / kriyāyāḥ ante eva yat vyaktaṃ rūpeṇa dṛśyate)

'The avyakta (unknown) is that quantity which is not visible at the start — but reveals its true form at the end of the calculation.'

From a polynomial to an equation

When we set a linear polynomial equal to a constant, we get a linear equation:

$50m + 200 = 750 \quad\Longleftrightarrow\quad \text{a linear equation in } m$

A value of the variable that makes the equation true is called a solution (or a root) of the equation. Solving the equation means finding that value.

Standard recipe for solving a linear equation:

Move all the variable terms to one side, all the constants to the other.
Combine like terms.
Divide by the coefficient of the variable.

That's it. Three moves. Always works.

Worked Example 1Sum of two numbers (NCERT Ex 6)

NCERT

The sum of two numbers is 64. One is 10 more than the other. What are the two numbers?

Step 1 — Name the unknown.

Let the smaller number be $x$ . Then the larger is $x + 10$ .

Step 2 — Translate the sentence.

'Their sum is 64' becomes:

$x + (x + 10) = 64$

Simplify the left side: $2x + 10 = 64$ . This is a linear equation in $x$ .

Step 3 — Solve.

Subtract 10 from both sides: $2x = 54$ .
Divide both sides by 2: $x = 27$ .

Step 4 — Answer the original question.

Smaller number: $x = 27$ . Larger number: $x + 10 = 37$ .

Step 5 — Check. $27 + 37 = 64$ ✓ and $37 - 27 = 10$ ✓.

Answer: The two numbers are 27 and 37.

Loading simulator…

Polynomials as input–output machines (functions)

There is another, equally powerful way to think about a polynomial: as an input-output machine (a function).

Consider the linear polynomial $f(x) = 2x + 3$ . For every value you put in for $x$ , the polynomial gives back a unique value:

Input $x = 4$ → $2(4) + 3 = 11$ . Output: 11.
Input $x = -6$ → $2(-6) + 3 = -9$ . Output: −9.
Input $x = 0$ → $2(0) + 3 = 3$ . Output: 3. (Notice — when the input is zero, the output is the constant term.)

We write $f(4) = 11$ and $f(-6) = -9$ and $f(0) = 3$ . The notation $f(\text{input})$ means "the output of the machine $f$ when fed this input".

Forward direction (substitute): given an input, compute the output.
Reverse direction (solve): given an output, find the input that produces it.

AI Generation Prompt

📸 Fig. 2.3 — The function $y = 2x + 3$ as an input-output machine. Input $x = 4$ → Output $y = 11$.

Worked Example 2Find the value (NCERT Ex Set 2.2, Q1)

NCERT

Find the value of the linear polynomial $5x - 3$ if (i) $x = 0$ (ii) $x = -1$ (iii) $x = 2$ .

(i) $x = 0$ : $5(0) - 3 = -3$ .

(ii) $x = -1$ : $5(-1) - 3 = -5 - 3 = -8$ .

(iii) $x = 2$ : $5(2) - 3 = 10 - 3 = 7$ .

Worked Example 3Substitute into a quadratic (NCERT Q2)

NCERT

Find the value of the quadratic polynomial $7s^2 - 4s + 6$ if (i) $s = 0$ (ii) $s = -3$ (iii) $s = 4$ .

Substitution works for any polynomial — not just linear ones.

(i) $s = 0$ : $7(0)^2 - 4(0) + 6 = 0 - 0 + 6 = 6$ . (Constant term, as expected.)

(ii) $s = -3$ : $7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81$ .

(iii) $s = 4$ : $7(4)^2 - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 16 + 6 = 102$ .

Worked Example 4Solve a linear equation

SOLVED

Solve $3x - 5 = 16$ .

Step 1 — Move the constant. Add 5 to both sides:

$3x = 21$

Step 2 — Divide by the coefficient. Divide both sides by 3:

$x = 7$

Step 3 — Check. $3(7) - 5 = 21 - 5 = 16$ . ✓

Answer: $x = 7$ .

Worked Example 5Linear equation with a fraction

SOLVED

Solve $\frac{x}{2} + 7 = 12$ .

Step 1 — Subtract 7: $\frac{x}{2} = 5$ .

Step 2 — Multiply by 2: $x = 10$ .

Check: $\frac{10}{2} + 7 = 5 + 7 = 12$ . ✓

Worked Example 6Auto fare word problem

SOLVED

An auto-rickshaw charges a base fare of ₹40 plus ₹15 per kilometre. How many kilometres can you travel for ₹130?

Step 1 — Set up the function. Let $d$ = distance in km. Cost = $40 + 15d$ .

Step 2 — Set the cost equal to ₹130.

$40 + 15d = 130$

Step 3 — Solve.

Subtract 40: $15d = 90$ .
Divide by 15: $d = 6$ .

Step 4 — Check. $40 + 15(6) = 40 + 90 = 130$ ✓.

Answer: You can travel 6 km for ₹130.

Logical ReasoningLevel 4 · Evaluation

Can two **different** inputs to a linear polynomial $f(x) = 2x + 3$ ever produce the **same** output? What about for a quadratic like $g(x) = x^2$?

Quick Check

Q1.If $f(x) = 5x - 3$ , what is $f(2)$ ?