Worked examples — Variables, constants, coefficients — algebraic expressions

This is a practice deep-dive for the parent note Variables, constants, coefficients — algebraic expressions. There we learned the names of the parts. Here we hunt down every kind of situation those names can show up in — positive and negative coefficients, the sneaky invisible $1$, the vanishing coefficient of $0$, fractions, decimals, real word problems, and an exam twist — and we work each one to the bone.

Before we start, one tiny reminder in plain words so no symbol arrives un-introduced:

The scenario matrix

Every question this topic can throw at you falls into one of these cells. Our examples below are chosen so that together they touch every single row.

Cell	What makes it special	Covered by
A. Positive coefficient	plain "$3x$", nothing hidden	Ex 1
B. Invisible coefficient $1$	a lone $x$ or $-x$	Ex 2
C. Negative coefficient	the sign belongs to the number	Ex 2, Ex 3
D. Fractional / decimal coefficient	coefficient need not be a whole number	Ex 4 (fraction), Ex 4b (decimal)
E. Zero coefficient (degenerate)	a term cancels and disappears	Ex 5
F. Constant that is a symbol	$\pi$, $e$ — fixed but written as a letter	Ex 6
G. Real-world word problem	turn a story into an expression	Ex 7
H. Exam twist	coefficient depends on another letter	Ex 8

Read each Forecast and guess before you scroll. Guessing wrong is how the ideas stick.

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Forecast: Which of $3$, $x$, $8$ is the box, which multiplies the box, and which stands alone? Decide now.

1. Spot the variable. The only letter is $x$, so $x$ is the variable. Why this step? Variables are letters — the "mystery boxes." There is exactly one here.
2. Spot the coefficient. The number touching and multiplying $x$ is $3$, so the coefficient is $3$. Why this step? A coefficient must be glued to a variable; $3x$ means $3\times x$.
3. Spot the constant. The number $8$ stands alone with no letter attached, so it is the constant. Why this step? Constants are the terms with no box beside them.
4. Evaluate at $x=4$: replace the box with $4$: $3(4)+8=12+8=20$. Why this step? "Evaluating" means pouring a number into the box (see Evaluating Algebraic Expressions).

Verify: $3\times 4=12$, and $12+8=20$. ✓ Units-free, so nothing to convert — just arithmetic.

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Forecast: There is no number written before $x$. Is its coefficient $0$? Also, $y$ appears twice — do we keep both?

1. Rewrite hidden coefficients openly. $-x$ is really $-1\cdot x$, and the lone $-y$ is $-1\cdot y$: $-1x+7y-1y.$ Why this step? Whenever no number is written, the coefficient is the invisible $1$ (or $-1$ if the term is negative). It is never $0$ — a $0$ coefficient would mean the term isn't there at all.
2. Coefficient of $x$: the number attached to $x$ is $-1$. Why this step? The minus sign belongs to the coefficient, not floating separately.
3. Combine the two $y$ terms. Both are "$y$-boxes," so add their coefficients: $7+(-1)=6$, giving $6y$. Why this step? Like terms (same variable, same power) add through their coefficients — see Algebraic Terms and Like Terms.
4. Coefficient of $y$: $6$.

Verify: Test with $x=2,y=3$. Original: $-2+7(3)-3=-2+21-3=16$. Simplified $-x+6y=-2+18=16$. ✓ Same value ⇒ our coefficients are right.

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Forecast: Is the coefficient of $a$ equal to $4$ or $-4$? The $-$ sits between the two terms — who owns it?

1. Read the expression as a sum of signed terms: $9+(-4a)$. Why this step? Every subtraction is secretly an addition of a negative. Splitting it this way shows each term carries its own sign.
2. Term $9$: no letter, so it is a constant, not a coefficient. Why this step? A number standing alone cannot vary and does not multiply any box, so by definition it is a constant — the word "coefficient" only applies to the number attached to a variable.
3. Term $-4a$: the number glued to $a$, sign included, is $-4$. So the coefficient of $a$ is $\boxed{-4}$. Why this step? The $-$ belongs to the $4$, because the term is $-4a=(-4)\times a$.

Verify: At $a=5$: $9-4(5)=9-20=-11$. Using $9+(-4)(5)=9-20=-11$. ✓ Consistent.

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Forecast: The area formula divides by $2$. Does that "$÷2$" become a coefficient, or is it a separate operation?

Figure: a cyan square of side $b$ is drawn on blueprint paper; the amber-shaded triangle is exactly the lower-left half of that square, its two legs (base and height) both labelled $b$, with a small white right-angle marker at the corner. The amber region is literally one-half of the cyan square — that is the picture of the coefficient $\frac{1}{2}$.

1. Recall the area rule: area $=\frac{1}{2}\times \text{base}\times \text{height}$. Why this step? This is the definition of a triangle's area — in the figure, the amber region is exactly half the cyan square of side $b$, so the area is half of $b\times b$.
2. Substitute base $=b$, height $=b$: area $=\frac{1}{2}b\cdot b=\frac{1}{2}{b}^2$. Why this step? Both dimensions are the same box $b$, and $b\cdot b=b^2$.
3. Identify the coefficient. The variable part is $b^2$; the number multiplying it is $\frac{1}{2}$. So the coefficient is $\frac{1}{2}$. Why this step? Coefficients do not have to be whole numbers — a fraction (or decimal) is perfectly allowed (Mistake 3 territory in the parent note).
4. Evaluate at $b=6$: $\frac{1}{2}(6{)}^2=\frac{1}{2}\cdot 36=18$. Why this step? We pour the value $6$ into the box $b$ to test the formula on a concrete case — the same "evaluating" move as Ex 1, now with a squared variable.

Verify: Half of $6\times 6=36$ is $18$ square units. ✓ Units are cm² if $b$ is in cm — an area, as expected.

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Forecast: Is $0.5$ a coefficient, even though it's a decimal and not a whole number?

1. Translate "$0.5$ of the deposit $x$". "Of" means multiply, so the bonus is $0.5\times x=0.5x$. Why this step? "A fraction of a quantity" is multiplication, and the multiplier lands in front of the box as its coefficient.
2. Name the coefficient. The number attached to $x$ is $0.5$, so the coefficient is $0.5$. Why this step? Decimals are just another way to write numbers ($0.5=\frac{1}{2}$); a decimal in front of a variable is a perfectly valid coefficient — the decimal case of Cell D.
3. Evaluate at $x=40$: $0.5\times 40=20$. Why this step? Pour the deposit $40$ into the box to see the actual bonus.

Verify: $0.5\times 40=20$, and since $0.5=\frac{1}{2}$, this is half of $40$, which is indeed $20$. ✓ Same value whether we call the coefficient $0.5$ or $\frac{1}{2}$.

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Forecast: Both $x$-terms have coefficient $5$ but opposite signs. What survives?

1. Group like terms. Collect the $x$-boxes: $5x-5x$; collect constants: $3+2$. Why this step? Only like terms combine, so we sort first.
2. Add the $x$-coefficients: $5+(-5)=0$, so the $x$-part is $0x$. Why this step? $0x=0$ for any value of $x$ — the box multiplied by zero contributes nothing. This is the degenerate case: the variable vanishes.
3. Add the constants: $3+2=5$. Why this step? The bare numbers $3$ and $2$ are like terms too (both attached to no variable), so they combine by ordinary addition — exactly the way variable coefficients combine.
4. Result: $0x+5=5$. The expression no longer contains $x$ at all; the coefficient of $x$ is $0$. Why this step? When a coefficient hits $0$, the term is gone — a genuinely different outcome from the invisible-$1$ case in Ex 2.

Verify: Try $x=100$: $5(100)+3-5(100)+2=500+3-500+2=5$. Try $x=-7$: $-35+3+35+2=5$. Same answer regardless of $x$ ⇒ $x$ truly cancelled. ✓

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Forecast: Two of these — $2$, $\pi$, $r$ — are letters. Are both letters variables?

1. Find the true variable. As the circle grows, only the radius $r$ changes; $r$ is the variable. Why this step? A variable is what can take different values in the same problem.
2. Classify $\pi$. $\pi \approx 3.14159\dots$ is always the same number. A fixed value written as a symbol is still a constant, not a variable. Why this step? "Being a letter" does not make something a variable — "being able to change" does (parent note, Mistake 3).
3. Read off the coefficient of $r$. Everything multiplying $r$ is $2\pi$. So the coefficient is $2\pi$ (a constant number, roughly $6.283$). Why this step? A coefficient is the whole numeric factor attached to the box, even when part of it is a symbol like $\pi$.

Verify: At $r=5$: circumference $=2\pi (5)=10\pi \approx 31.4159$. A circle of radius $5$ indeed has circumference $\approx 31.42$ units. ✓

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Forecast: Which number becomes the coefficient and which stays a constant?

1. Assign the variable. The unknown, changing quantity is the distance, so let $d$ = kilometres travelled. Why this step? Model the thing that varies from ride to ride (see Word Problems in Algebra).
2. Translate "₹18 per km." Per-km cost multiplies the number of km: $18\times d=18d$. Why this step? "For every km" is repeated addition of ₹18 — that's multiplication, and $18$ becomes the coefficient.
3. Translate the fixed fee. ₹50 is charged once no matter the distance, so it is a constant term, added on: $18d+50$. Why this step? A charge that doesn't depend on the box is a stand-alone constant.
4. Write the full expression: fare $=18d+50$, with variable $d$, coefficient $18$, constant $50$. Why this step? We combine the distance-dependent part and the fixed part into one "recipe" that gives the fare for any distance — the whole point of using a variable.
5. Evaluate at $d=12$: $18(12)+50=216+50=266$. Why this step? We pour the actual distance $12$ into the box to get a concrete fare, checking the recipe on a real ride.

Verify: $18\times 12=216$ for distance, plus the ₹50 fee $=₹266$. Units are rupees throughout — the ₹18·km "cancels" the km, leaving rupees. ✓

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Forecast: Here the coefficient itself is a letter. Can we still pin it down to one number?

1. Read the roles carefully. $x$ is the variable, $7$ is the constant, and $k$ is the coefficient of $x$ — a fixed number we just haven't been told yet. Why this step? Exam twists blur variable vs coefficient on purpose; sort them first.
2. Substitute the given value $x=3$: $k(3)+7=19$, i.e. $3k+7=19$. Why this step? We now have an equation (it has an equals sign — see Equations vs. Expressions) with one unknown $k$.
3. Isolate the $k$-term: subtract $7$ from both sides: $3k=12$. Why this step? Undo the added constant to leave only the coefficient term.
4. Solve for $k$: divide both sides by $3$: $k=4$. Why this step? Undo the multiplication to free $k$.

Verify: Put $k=4$ back: $4(3)+7=12+7=19$. ✓ Matches the given total.

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Recall Quick self-test (reveal after answering)

Coefficient of $x$ in $-x+6$ ::: $-1$ (the invisible $-1$) Coefficient of $x$ in $7+3-2x-5$ (simplify constants first) ::: $-2$ Coefficient of $b^2$ in the equal-legs triangle area ::: $\frac{1}{2}$ Coefficient of $x$ in "$0.5$ of $x$" ::: $0.5$ (a decimal coefficient, same as $\frac{1}{2}$) What happens to $x$ when its coefficient becomes $0$? ::: The term vanishes; the value no longer depends on $x$. Is $\pi$ a variable or a constant? ::: A constant (fixed value, just written as a symbol).

Connections

• Parent: Variables, constants, coefficients — the definitions these examples exercise
• Algebraic Terms and Like Terms — used in Ex 2 & Ex 5 to combine terms
• Evaluating Algebraic Expressions — the "plug a number in the box" move
• Equations vs. Expressions — Ex 8 crosses into equation territory
• Word Problems in Algebra — Ex 7's translation method
• Linear Expressions — all our one-power examples are linear

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