Variables, constants, coefficients — algebraic expressions: Exercises

Level 1 — Recognition

(Can you name the parts? No calculation, just correct labelling.)

Exercise 1.1

In the expression $9 m - 4$ name the variable, the coefficient of that variable, and the constant term.

Recall Solution 1.1

Variable: $m$ — it is the letter, the "mystery box" that could hold any number.
Coefficient of $m$ : $9$ — it is the number multiplying $m$ (i.e. $9 m = 9 \times m$ ).
Constant term: $- 4$ — a fixed number standing alone. Keep the minus sign attached: the constant is $- 4$ , not $4$ .

Exercise 1.2

Write down the coefficient of $x$ in each of these: $x, - x, \frac{2}{3} x, - 5 x .$

Recall Solution 1.2

$x$ → coefficient $1$ . We could write $1 \cdot x$ ; the $1$ is invisible but always there.
$- x$ → coefficient $- 1$ . Same idea: $- x = (- 1) \cdot x$ .
$\frac{2}{3} x$ → coefficient $\frac{2}{3}$ . A coefficient does not have to be a whole number.
$- 5 x$ → coefficient $- 5$ . The sign travels with the number.

Exercise 1.3

Classify each symbol below as variable or constant: $y, 7, π, t, - \frac{1}{2} .$

Recall Solution 1.3

$y$ → variable (a chosen letter that can change).
$7$ → constant (fixed number).
$π$ → constant. It looks like a letter but its value is fixed at $\approx 3.14159 \dots$ forever.
$t$ → variable.
$- \frac{1}{2}$ → constant.

Level 2 — Application

(Translate words into expressions, and build expressions correctly.)

Exercise 2.1

A shop sells pens at ₹8 each and notebooks at ₹35 each. Write an expression for the total cost of buying $p$ pens and $n$ notebooks.

Recall Solution 2.1

Pens cost: each of the $p$ pens costs ₹8, so cost $= 8 \times p = 8 p$ .
Notebooks cost: $35 \times n = 35 n$ .
Total: add the two separate costs. $8 p + 35 n$ Variables: $p, n$ . Coefficients: $8$ (for $p$ ), $35$ (for $n$ ). No constant term (no fixed extra charge).

Exercise 2.2

"A number $x$ is tripled, then $7$ is subtracted." Write the expression.

Recall Solution 2.2

"tripled" means multiply by $3$ : that gives $3 x$ .
"then $7$ subtracted" means take away $7$ : $3 x - 7$ .
Order matters — we tripled first, then subtracted. $3 x - 7$

Exercise 2.3

A rectangle has width $w$ and length that is $4$ more than the width. Write an expression for the perimeter (perimeter = twice the length + twice the width).

Recall Solution 2.3

Length $= w + 4$ (four more than the width).
Perimeter $= 2 (length) + 2 (width) = 2 (w + 4) + 2 w$ .
Open the bracket: $2 (w + 4) = 2 w + 8$ .
So perimeter $= 2 w + 8 + 2 w = 4 w + 8$ . $4 w + 8$
Coefficient of $w$ is $4$ ; constant term is $8$ .

Level 3 — Analysis

(Look inside multi-term expressions; account for signs and degrees.)

Exercise 3.1

For $6 x^{2} y - 4 x y + y - 11$ list, term by term, the variables, the coefficient, and the degree (degree = sum of the powers of all variables in that term).

Recall Solution 3.1

Term	Variables	Coefficient	Degree
$6 x^{2} y$	$x, y$	$6$	$2 + 1 = 3$
$- 4 x y$	$x, y$	$- 4$	$1 + 1 = 2$
$y$	$y$	$1$	$1$
$- 11$	none	$- 11$	$0$

Notes: the coefficient of the lone $y$ is the invisible $1$ ; the constant $- 11$ has degree $0$ because it can be seen as $- 11 \cdot x^{0}$ (any variable to the power $0$ equals $1$ ).

Exercise 3.2

In $- \frac{3}{5} x^{4}$ , state the coefficient, the variable, and the power of the variable.

Recall Solution 3.2

Coefficient: $- \frac{3}{5}$ (sign and fraction both belong to the coefficient).
Variable: $x$ .
Power: $4$ . The exponent $4$ is not part of the coefficient — it tells us how many times $x$ multiplies itself.

Two students simplify $5 x - x + 2 - 2$ . Amir writes $4 x$ ; Bela writes $4 x + 0$ ; Carl writes $5 x - x = 4 x$ and "the constants vanish." Are all three answers correct? Explain what each is really claiming.

Recall Solution 3.3

$5 x - x = 5 x - 1 x = 4 x$ (remember $x = 1 x$ ).
$2 - 2 = 0$ .
So the expression equals $4 x + 0 = 4 x$ .
Amir ( $4 x$ ) and Bela ( $4 x + 0$ ) are the same value — writing " $+ 0$ " is legal but redundant.
Carl is also correct: the constants really do cancel to $0$ , so they "vanish." $4 x$

Level 4 — Synthesis

(Combine everything: build, simplify, and reason about structure.)

Exercise 4.1

A taxi charges a fixed ₹50 booking fee plus ₹18 per kilometre. Then a flat ₹20 discount is applied. Write a simplified expression for the fare after $d$ kilometres, and identify the coefficient of $d$ and the constant term.

Recall Solution 4.1

Distance cost: $18 d$ .
Add the booking fee: $18 d + 50$ .
Apply the discount (subtract $20$ ): $18 d + 50 - 20$ .
Combine constants: $50 - 20 = 30$ . $18 d + 30$ Coefficient of $d$ : $18$ . Constant term: $30$ . (Check at $d = 10$ : $18 (10) + 30 = 210$ .)

Exercise 4.2

Write an expression for the perimeter of a triangle with sides $2 a + 1$ , $3 a - 2$ , and $a + 6$ , and simplify it. Then state the coefficient of $a$ and the constant.

Recall Solution 4.2

Perimeter = sum of the three sides.
$P = (2 a + 1) + (3 a - 2) + (a + 6)$ .
Drop brackets (all additions): $P = 2 a + 1 + 3 a - 2 + a + 6$ .
Combine the $a$ -terms: $2 a + 3 a + a = 6 a$ .
Combine constants: $1 - 2 + 6 = 5$ . $P = 6 a + 5$ Coefficient of $a$ : $6$ . Constant: $5$ .

Exercise 4.3

Design an expression in one variable $x$ that satisfies all of these: it has exactly three terms; the coefficient of $x^{2}$ is $- 2$ ; the coefficient of $x$ is the invisible $1$ ; and the constant term is $\frac{1}{2}$ . Then state its value when $x = 2$ .

Recall Solution 4.3

Coefficient of $x^{2}$ is $- 2$ → term $- 2 x^{2}$ .
Coefficient of $x$ is $1$ (invisible) → term $x$ .
Constant term $\frac{1}{2}$ → term $+ \frac{1}{2}$ . $- 2 x^{2} + x + \frac{1}{2}$ Value at $x = 2$ : $- 2 (4) + 2 + \frac{1}{2} = - 8 + 2 + 0.5 = - 5.5$ .

Level 5 — Mastery

(Reason at the meta level: when do the rules break, and can you defend a claim?)

Exercise 5.1

Is $x^{2} + \frac{1}{x}$ a "constant + variable-with-coefficient" style expression like the others on this page? Identify each term's coefficient and power, and explain what is unusual.

Recall Solution 5.1

Rewrite $\frac{1}{x} = x^{- 1}$ .
Terms: $x^{2}$ (coefficient $1$ , power $2$ ) and $x^{- 1}$ (coefficient $1$ , power $- 1$ ).
Unusual: the second term has a negative power — the variable sits in a denominator. It is still an algebraic expression, but it is not a polynomial (polynomials allow only whole-number powers $0, 1, 2, \dots$ ). See Polynomials.
There is no constant term here (no term of power $0$ ).

Exercise 5.2

A student claims: "In $5 x$ , if I let $x$ vary then $5$ is really varying too, because $5 x$ changes." Steel-man this, then decisively refute it.

Recall Solution 5.2

The strongest version of the claim: the product $5 x$ does change as $x$ changes — at $x = 1$ it's $5$ , at $x = 3$ it's $15$ — so "something on the left is moving."
Refutation: yes, the value of the whole term $5 x$ changes, but that change comes entirely from $x$ . The factor $5$ is fixed: every time we multiply by exactly $5$ , never $4.9$ or $6$ . Test it: pick any two values of $x$ and divide the results, $\frac{15}{5} = 3 = \frac{x _{2}}{x _{1}}$ — the ratio is governed by $x$ , and $5$ cancels out because it is constant.
Conclusion: $5$ is the coefficient (fixed), $x$ is the variable (changing). A moving output does not make every factor a variable.

Exercise 5.3

Two expressions are given: $E_{1} = 3 x + 3 x$ and $E_{2} = 3 x^{2}$ . A classmate says "both are just $3$ copies of something, so they're the same." Explain, using coefficients and powers, why $E_{1} \neq = E_{2}$ , and give the correct simplification of each. Verify at $x = 4$ .

Recall Solution 5.3

$E_{1} = 3 x + 3 x$ is adding two like terms: $3 x + 3 x = 6 x$ (coefficient $6$ , power $1$ ).
$E_{2} = 3 x^{2}$ is multiplying $x$ by itself once, then scaling by $3$ (coefficient $3$ , power $2$ ).
" $3$ copies of something" describes $E_{1}$ (three-ish times $x$ , twice), but $E_{2}$ is $x$ squared — a different operation entirely (power, not repeated addition).
Check at $x = 4$ : $E_{1} = 6 (4) = 24$ ; $E_{2} = 3 (16) = 48$ . Different numbers → different expressions. $E_{1} = 6 x, E_{2} = 3 x^{2}$

Recall Quick self-check (cloze)

The coefficient of $x$ in the term $- x$ is == $- 1$ ==. A symbol with a permanently fixed value, like $π$ , is a constant. The degree of the term $6 x^{2} y$ is == $3$ == (sum of the powers $2 + 1$ ). Simplifying $3 x + 3 x$ gives == $6 x$ ==, while $3 x^{2}$ stays == $3 x^{2}$ — addition raises the coefficient, powers raise the exponent==.

Connections

Algebraic Terms and Like Terms — the "combine like terms" step used all over L2–L5.
Evaluating Algebraic Expressions — the substitution checks (e.g. $x = 4$ ) live here.
Equations vs. Expressions — none of these had an equals sign; adding one changes the game.
Polynomials — why Exercise 5.1 is not a polynomial.
Linear Expressions — Exercises 2.1–4.2 are all linear (highest power $1$ ).
Word Problems in Algebra — the translation practice of Level 2.

Exercises — Variables, constants, coefficients — algebraic expressions