Fractions and Exponents: The Algebra Behind Every Problem

Teaching resource for Intermediate Microeconomics and Mathematics, Sciences Po Paris.

Most of the time a problem set goes wrong it is not the economics. You understand that the consumer sets the MRS equal to the price ratio, you write the right condition, and then the algebra stops you: a fraction inside a fraction, an exponent of −1/2, an equation with x in the denominator. This note collects the rules for fractions and exponents in one place, shows how they combine, and then uses them where you need them in the course: solving equations, taking derivatives, and finding a demand function from start to finish. There are three practice sheets at the end, from numbers only to full economics problems.

1. Fractions with numbers

A fraction a/b is a divided by b. Four operations, four rules:

\text{simplify:}\ \ \frac{6 \cdot 3}{6 \cdot 4} = \frac{3}{4} \] \[ \text{multiply:}\ \ \frac{a}{b}\cdot\frac{c}{d} = \frac{ac}{bd}

\text{divide:}\ \ \frac{a/b}{c/d} = \frac{a}{b}\cdot\frac{d}{c} = \frac{ad}{bc} \] \[ \text{add:}\ \ \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}

Simplifying means dividing the top and the bottom by the same number. 18/24 = 3/4 because 6 divides both.
Multiplying needs no common denominator: tops together, bottoms together. 3/4 · 8/9 = 24/36 = 2/3.
Dividing by a fraction is multiplying by its reciprocal: flip the second fraction and multiply. (2/3) ÷ (4/5) = 2/3 · 5/4 = 5/6. This also explains why dividing by a number smaller than 1 makes things bigger: 6 ÷ (3/4) = 6 · 4/3 = 8.
Adding is the only operation that needs a common denominator. Rewrite both fractions over the same bottom, then add the tops only:

\frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}

A whole number is a fraction with denominator 1, so 3 − 2/5 = 15/5 − 2/5 = 13/5. And “of” means multiply: two fifths of 30 is 2/5 · 30 = 12.

Fractions of fractions. When a fraction has fractions inside it, simplify the top and the bottom separately until each is a single fraction, then divide by flipping:

\frac{1 - \frac13}{1 + \frac13} = \frac{2/3}{4/3} = \frac{2}{3}\cdot\frac{3}{4} = \frac{1}{2}

2. Fractions with letters

The rules are exactly the same. The letters just mean you cannot compute a number at the end.

\frac{1}{x} + \frac{1}{y} = \frac{y}{xy} + \frac{x}{xy} = \frac{x + y}{xy} \] \[ \frac{x}{y}\cdot\frac{y^2}{x^3} = \frac{x y^2}{x^3 y} = \frac{y}{x^2}

The one rule that causes most mistakes: cancel factors, never terms. You can cancel something that multiplies the whole top and the whole bottom. You cannot cancel a piece of a sum.

\frac{3x + 6}{3} = \frac{3(x + 2)}{3} = x + 2 \] \[ \text{but} \] \[ \frac{x + 3}{x} \neq 3

When in doubt, factor first. (x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) = x + 2: once the top is written as a product, the bracket (x − 2) is a factor and it cancels.

Splitting a fraction works when the sum is on top, never when it is on the bottom:

\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c} \] \[ \text{but} \] \[ \frac{c}{a + b} \neq \frac{c}{a} + \frac{c}{b}

Check any rule you are unsure about with numbers. With a = b = c = 1, the left one gives 2 = 1 + 1, true; the right one gives 1/2 = 2, false.

3. Exponents

xⁿ means x multiplied by itself n times. Everything else follows from that:

Rule	Example
x^a · x^b = x^a+b (same base, multiply: add exponents)	x³ · x⁵ = x⁸
x^a / x^b = x^a−b (same base, divide: subtract)	x⁷ / x³ = x⁴
(x^a)^b = x^ab (power of a power: multiply)	(x²)⁴ = x⁸
(xy)^a = x^ay^a (the power hits every factor)	(2x³)² = 4x⁶
x⁰ = 1	5⁰ = 1
x^−a = 1/x^a (negative means “one over”)	2⁻³ = 1/8
x^1/n = ⁿ√x (fraction means root)	16^1/2 = 4, 27^1/3 = 3
x^m/n = (ⁿ√x)^m (root first, then power)	8^2/3 = 2² = 4

Where the strange ones come from:

x⁰ = 1 because x²/x² = 1 and also x²/x² = x²⁻² = x⁰.
x⁻¹ = 1/x for the same reason: x²/x³ = 1/x and also x²⁻³ = x⁻¹. A negative exponent does not make a number negative. 2⁻³ is 1/8, a positive number.
x^1/2 = √x because (x^1/2)² = x¹ = x, which is what a square root does.

Negative exponents on a fraction flip it, and fractional exponents on a fraction act on the top and bottom separately:

\left(\frac{2}{3}\right)^{-1} = \frac{3}{2} \] \[ \left(\frac{1}{2}\right)^{-2} = 2^2 = 4

\left(\frac{8}{27}\right)^{2/3} = \left(\frac{2}{3}\right)^{2} = \frac{4}{9} \] \[ 9^{-1/2} = \frac{1}{\sqrt 9} = \frac13

There are no rules for sums. (x + y)² ≠ x² + y² (try x = y = 1: 4 ≠ 2), and xᵃ + xᵇ does not simplify. Similarly 2⁻¹ + 2⁻² is just 1/2 + 1/4 = 3/4: compute each power, then add.

4. Fractions and exponents together

The two topics meet constantly, because a fraction can always be written as a power and the other way round. These four translations are worth knowing by heart:

\frac{1}{x} = x^{-1} \] \[ \frac{3}{x^2} = 3x^{-2} \] \[ \frac{1}{\sqrt x} = x^{-1/2} \] \[ x\sqrt{x} = x^{3/2}

Only what is in the denominator changes sign when it moves up. A number in front stays where it is: 3/x² = 3x⁻², and 1/(2x) = (1/2)x⁻¹, not 2x⁻¹.

Simplifying a ratio of powers is where all of this comes together, and it is exactly what computing an MRS asks for. The rule is: numbers with numbers, x with x, y with y. For U = x^1/3 y^2/3:

MRS = \frac{U_x}{U_y} = \frac{\tfrac13\, x^{-2/3} y^{2/3}}{\tfrac23\, x^{1/3} y^{-1/3}} = \frac{1/3}{2/3}\cdot x^{-\frac23 - \frac13}\cdot y^{\frac23 + \frac13} = \frac12\, x^{-1} y = \frac{y}{2x}

Three separate skills in one line: dividing fractions ((1/3)/(2/3) = 1/3 · 3/2 = 1/2), subtracting exponents (−2/3 − 1/3 = −1), and subtracting a negative exponent (2/3 − (−1/3) = 1). If any one of them is shaky, the MRS comes out wrong even though the economics was right.

5. Solving equations

The principle never changes: do the same thing to both sides until the unknown is alone. Fractions and exponents just need the right “same thing”.

Fractions: clear them. Multiply every term by the common denominator, or cross-multiply when there is one fraction on each side.

\frac{x}{3} + \frac{x}{6} = 5 \;\overset{\times 6}{\Longrightarrow}\; 2x + x = 30 \;\Longrightarrow\; x = 10 \] \[ \frac{x - 1}{2} = \frac{x + 2}{5} \;\Longrightarrow\; 5(x - 1) = 2(x + 2) \;\Longrightarrow\; x = 3

Keep the brackets when you cross-multiply: the whole numerator is multiplied, not just its first term.

Exponents: raise both sides to the reciprocal power. The reciprocal of 1/2 is 2, of 2/3 is 3/2, of −1 is −1:

x^{1/2} = 6 \;\Longrightarrow\; x = 6^2 = 36 \] \[ x^{2/3} = 9 \;\Longrightarrow\; x = 9^{3/2} = 27 \] \[ x^{-1/2} = \frac15 \;\Longrightarrow\; x^{1/2} = 5 \;\Longrightarrow\; x = 25

If the unknown is in the exponent, as in 2ˣ = 32, write both sides as powers of the same base (2ˣ = 2⁵, so x = 5) or take logs.

Solving for one letter in terms of another. In economics you often solve for y “as a function of x”. Treat x exactly as if it were a number:

\frac{y}{2x} = \frac{3}{4} \;\Longrightarrow\; y = \frac{3}{4}\cdot 2x = \frac{3x}{2}

6. Derivatives of fractions and roots: rewrite as powers first

Most derivative mistakes with fractions and roots disappear with one habit: rewrite as a power, differentiate with (xⁿ)′ = nxⁿ⁻¹, then rewrite back. The power rule works for every exponent, negative and fractional included.

f(x) = \frac{3}{x^2} = 3x^{-2} \quad\Longrightarrow\quad f'(x) = 3\cdot(-2)\,x^{-3} = -\frac{6}{x^3} \] \[ f(x) = \sqrt{x} = x^{1/2} \quad\Longrightarrow\quad f'(x) = \tfrac12\, x^{-1/2} = \frac{1}{2\sqrt x} \] \[ f(x) = \frac{4}{\sqrt x} = 4x^{-1/2} \quad\Longrightarrow\quad f'(x) = 4\cdot\left(-\tfrac12\right) x^{-3/2} = -\frac{2}{x^{3/2}}

The step that goes wrong is “exponent minus one” when the exponent is already a fraction or negative: −2 − 1 = −3, 1/2 − 1 = −1/2, −1/2 − 1 = −3/2.

If the bottom is a single term, split the fraction first and you will not need the quotient rule:

f(x) = \frac{x^2 + 1}{x} = x + x^{-1} \quad\Longrightarrow\quad f'(x) = 1 - x^{-2} = 1 - \frac{1}{x^2}

If the bottom is a sum, it cannot be split (section 2). Then use the quotient rule, or write the bracket as a power and use the chain rule: 1/(1 + x) = (1 + x)⁻¹, whose derivative is −(1 + x)⁻². The derivatives refresher has all the rules.

7. Everything at once: a Cobb-Douglas demand

A consumer with U = x^1/3 y^2/3, income 60, and prices p_x = 2, p_y = 4. Each step below uses one of the sections above.

Marginal utilities (power rule, section 6): U_x = (1/3) x^−2/3 y^2/3 and U_y = (2/3) x^1/3 y^−1/3.
MRS (ratio of powers, section 4): U_x / U_y = y/(2x).
Tangency (solving for y in terms of x, section 5):

\frac{y}{2x} = \frac{p_x}{p_y} = \frac{2}{4} = \frac12 \quad\Longrightarrow\quad y = \frac12\cdot 2x = x

Budget (substitute and solve): 2x + 4y = 60 with y = x gives 6x = 60, so x = 10 and y = 10.
Check: spending on x is 2 · 10 = 20, one third of income, which is the exponent on x. For any Cobb-Douglas x^a y^1−a, the consumer spends the share a of income on x.

None of the steps is hard on its own. The difficulty is that five of them come in a row, and one slip anywhere breaks the answer. That is why the algebra is worth drilling separately.

8. Common mistakes

Adding fractions by adding tops and bottoms: 1/2 + 1/3 is 5/6, not 2/5.
Cancelling a term instead of a factor: (x + 3)/x is 1 + 3/x, not 3.
Splitting a sum in the denominator: 1/(x + y) is not 1/x + 1/y.
Reading a negative exponent as a negative number: 2⁻³ = 1/8, not −8.
Moving the coefficient with the power: 1/(2x) = (1/2)x⁻¹, not 2x⁻¹.
Distributing a power over a sum: (x + y)² is x² + 2xy + y², not x² + y².
Getting “exponent minus one” wrong with fractions: the derivative of x^−1/2 has exponent −3/2, not 1/2.
Dropping brackets when you cross-multiply or subtract: x/2 − (x − 1)/3 = (3x − 2x + 2)/6. The minus applies to the whole numerator.

Practice

Six to check yourself. Then the three sheets, in order:

Starter: 50 calculations with numbers only, fractions then powers and roots.
Intermediate: 58 exercises with letters, simplifying, single powers and solving equations.
Upper intermediate: 20 derivatives of fractions and roots, then 11 economics problems where every algebra step is shown.

1.
Compute \( \dfrac{\frac12 + \frac13}{\frac14} \).
Solution
\[ \frac12 + \frac13 = \frac36 + \frac26 = \frac56 \] \[ \frac{5/6}{1/4} = \frac56 \cdot 4 = \frac{20}{6} = \frac{10}{3} \]
Top first, as a single fraction, then divide by flipping.
2.
Simplify \( \dfrac{1}{x} - \dfrac{1}{x + 1} \).
Solution
\[ \frac{x + 1}{x(x + 1)} - \frac{x}{x(x + 1)} = \frac{x + 1 - x}{x(x + 1)} = \frac{1}{x(x + 1)} \]
The common denominator is the product \( x(x+1) \). Each fraction is multiplied top and bottom by what it is missing.
3.
Compute \( 32^{-2/5} \) and \( \left(\tfrac{8}{27}\right)^{2/3} \) without a calculator.
Solution
\[ 32^{-2/5} = \frac{1}{\left(32^{1/5}\right)^2} = \frac{1}{2^2} = \frac14 \] \[ \left(\frac{8}{27}\right)^{2/3} = \left(\frac{8^{1/3}}{27^{1/3}}\right)^2 = \left(\frac23\right)^2 = \frac49 \]
Minus sign: one over. Denominator of the exponent: which root. Numerator: which power. Root first keeps the numbers small.
4.
Write \( \dfrac{x^2 y^5}{x^4 y^2} \) with positive exponents, and \( \dfrac{\sqrt[3]{x^2}}{x} \) as a single power of \( x \).
Solution
\[ \frac{x^2 y^5}{x^4 y^2} = x^{2-4}\, y^{5-2} = x^{-2} y^3 = \frac{y^3}{x^2} \] \[ \frac{x^{2/3}}{x^{1}} = x^{\frac23 - 1} = x^{-1/3} \]
5.
Solve \( \tfrac12\, x^{-1/2} = \tfrac14 \) for \( x > 0 \).
Solution
\[ x^{-1/2} = \frac12 \;\Longrightarrow\; \frac{1}{x^{1/2}} = \frac12 \;\Longrightarrow\; x^{1/2} = 2 \;\Longrightarrow\; x = 4 \]
This is a first-order condition: the derivative of \( \sqrt x \) set equal to a price of \( \frac14 \).
6.
Differentiate \( f(x) = \dfrac{2x + 3}{\sqrt{x}} \).
Solution
The bottom is a single term, so split and rewrite as powers:
\[ f(x) = \frac{2x}{x^{1/2}} + \frac{3}{x^{1/2}} = 2x^{1/2} + 3x^{-1/2} \] \[ f'(x) = x^{-1/2} - \tfrac32\, x^{-3/2} = \frac{2x - 3}{2x^{3/2}} \]
For the single fraction, put both terms over \( 2x^{3/2} \): \( x^{-1/2} = \frac{2x}{2x^{3/2}} \).

Once these feel comfortable, the next step is the derivatives refresher and then the maths companion for Intermediate Micro. All notes and practice sheets are on the resources page.

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