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Mathematics

Algebraic Fractions & Laws of Indices

PDF
Matthew Williams
|May 6, 2026|8 min read
AlgebraExponentsFractionsIndicesPaper 01Paper 02

Simplifying and operating on algebraic fractions, plus the rules for working with powers.

Algebraic fractions and indices are simplification tools. They help you rewrite complicated expressions into cleaner forms without changing their value.

In the CSEC exam, these skills often appear inside larger algebra questions rather than as a topic by itself. You may need to simplify before solving, factor before cancelling, or apply index laws before substituting. Each line should preserve equality, so explain what rule you are using when the step is not obvious.

Simplifying Algebraic Fractions

The safest way to simplify an algebraic fraction is to factor first, then cancel common factors. Cancelling terms that are being added is a serious algebra mistake.

Algebraic fractions work exactly like regular fractions, but with variables instead of just numbers.

The Key Principle: You can cancel something from top and bottom if it appears in BOTH places.

Rule 1: Cancel Common Variables

If the same variable appears on top and bottom, you can cross them out.

Example

Simplify 6x9x\frac{6x}{9x}9x6x​:

Method: Break into two fractions 6x9x=69×xx\frac{6x}{9x} = \frac{6}{9} \times \frac{x}{x}9x6x​=96​×xx​

Now: 69=23\frac{6}{9} = \frac{2}{3}96​=32​ and xx=1\frac{x}{x} = 1xx​=1

6x9x=23\frac{6x}{9x} = \frac{2}{3}9x6x​=32​

Why this works: x÷x=1x ÷ x = 1x÷x=1, just like 5÷5=15 ÷ 5 = 15÷5=1

Rule 2: Factor the Numerator, Then Cancel

When the numerator is more complex, factor it first. Then look for common factors with the denominator.

Example

Simplify x2+5xx\frac{x^2 + 5x}{x}xx2+5x​:

Step 1: Factor the numerator (top) x2+5x=x(x+5)x^2 + 5x = x(x + 5)x2+5x=x(x+5)

Why? Because both terms have an xxx: x2=x⋅xx^2 = x \cdot xx2=x⋅x and 5x=5⋅x5x = 5 \cdot x5x=5⋅x

Step 2: Rewrite the fraction x2+5xx=x(x+5)x\frac{x^2 + 5x}{x} = \frac{x(x+5)}{x}xx2+5x​=xx(x+5)​

Step 3: Cancel the xxx from top and bottom x(x+5)x=x+5\frac{x(x+5)}{x} = x + 5xx(x+5)​=x+5

Check: If x=3x = 3x=3: Original = 9+153=243=8\frac{9+15}{3} = \frac{24}{3} = 839+15​=324​=8, Simplified = 3+5=83 + 5 = 83+5=8 ✓

Remember

You can ONLY cancel if:

  • The same thing appears in BOTH numerator and denominator
  • The whole thing is being multiplied (not added)

Wrong: x+2x\frac{x+2}{x}xx+2​, you CANNOT cancel the xxx here! (The xxx is only in one part of a sum)

Right: x(x+2)x\frac{x(x+2)}{x}xx(x+2)​, now you CAN cancel (the xxx is being multiplied)

Adding and Subtracting Algebraic Fractions

Addition and subtraction depend on matching denominators because the pieces must be the same size before they can be combined.

Just like regular fractions, you need a common denominator to add or subtract.

Case 1: Same Denominator (Easy)

If the denominators match, just combine the numerators.

Example

Add 2x+3x\frac{2}{x} + \frac{3}{x}x2​+x3​:

Same denominator: xxx

2x+3x=2+3x=5x\frac{2}{x} + \frac{3}{x} = \frac{2+3}{x} = \frac{5}{x}x2​+x3​=x2+3​=x5​

(This is just like 25+35=55=1\frac{2}{5} + \frac{3}{5} = \frac{5}{5} = 152​+53​=55​=1)

Case 2: Different Denominators (Harder)

Find the Least Common Denominator (LCD), the smallest expression both denominators divide into.

Example

Add 1x+1y\frac{1}{x} + \frac{1}{y}x1​+y1​:

Step 1: What's the LCD? Since we have xxx and yyy, the LCD is xyxyxy

Step 2: Convert each fraction to use the LCD

  • 1x=1×yx×y=yxy\frac{1}{x} = \frac{1 \times y}{x \times y} = \frac{y}{xy}x1​=x×y1×y​=xyy​ (multiply top and bottom by yyy)
  • 1y=1×xy×x=xxy\frac{1}{y} = \frac{1 \times x}{y \times x} = \frac{x}{xy}y1​=y×x1×x​=xyx​ (multiply top and bottom by xxx)

Step 3: Now add them yxy+xxy=y+xxy=x+yxy\frac{y}{xy} + \frac{x}{xy} = \frac{y + x}{xy} = \frac{x+y}{xy}xyy​+xyx​=xyy+x​=xyx+y​

Check: If x=2,y=3x = 2, y = 3x=2,y=3: Original = 12+13=36+26=56\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}21​+31​=63​+62​=65​, Simplified = 2+32×3=56\frac{2+3}{2 \times 3} = \frac{5}{6}2×32+3​=65​ ✓

Finding LCD Strategy

Exam Tip

To find the LCD with variable denominators:

  1. List all the factors that appear
  2. Use each factor the maximum number of times it appears in any denominator

Example: denominators x2,xy,yx^2, xy, yx2,xy,y → LCD is x2yx^2yx2y (use xxx twice, yyy once)

Laws of Indices (Exponents)

What Are Indices?

Indices are shorthand for repeated multiplication. They are not decoration; they control how many times the base is used as a factor.

Index (or exponent) is the small number that says "multiply this number by itself how many times."

x3=x×x×x (three times)x^3 = x \times x \times x \text{ (three times)}x3=x×x×x (three times) 24=2×2×2×2=16 (four times)2^4 = 2 \times 2 \times 2 \times 2 = 16 \text{ (four times)}24=2×2×2×2=16 (four times)

The index tells you how many times to use the base.

Rule 1: Multiply

Use this rule only when the bases are the same. x2×x3x^2 \times x^3x2×x3 can be combined, but x2×y3x^2 \times y^3x2×y3 cannot become one power because the bases are different.

When you multiply powers with the SAME base, ADD the exponents:

am×an=am+na^m \times a^n = a^{m+n}am×an=am+n

Why? Because: x2×x3=(x×x)×(x×x×x)=x×x×x×x×x=x5x^2 \times x^3 = (x \times x) \times (x \times x \times x) = x \times x \times x \times x \times x = x^5x2×x3=(x×x)×(x×x×x)=x×x×x×x×x=x5

You're multiplying 2 of them times 3 of them = 5 of them total.

Example

x2×x3x^2 \times x^3x2×x3:

Same base (xxx), so add exponents: 2+3=52 + 3 = 52+3=5

x2×x3=x5x^2 \times x^3 = x^5x2×x3=x5

Example

a4×a2×a3a^4 \times a^2 \times a^3a4×a2×a3:

Add all exponents: 4+2+3=94 + 2 + 3 = 94+2+3=9

a4×a2×a3=a9a^4 \times a^2 \times a^3 = a^9a4×a2×a3=a9

Rule 2: Divide

Division removes repeated factors. Subtracting the exponents counts how many copies of the base remain after cancellation.

When you divide powers with the SAME base, SUBTRACT the exponents:

aman=am−n\frac{a^m}{a^n} = a^{m-n}anam​=am−n

Why? Because: x5x2=x×x×x×x×xx×x=x×x×x=x3\frac{x^5}{x^2} = \frac{x \times x \times x \times x \times x}{x \times x} = x \times x \times x = x^3x2x5​=x×xx×x×x×x×x​=x×x×x=x3

Cancel two of them, and you have three left. 5−2=35 - 2 = 35−2=3.

Example

y5y2\frac{y^5}{y^2}y2y5​:

Same base (yyy), so subtract exponents: 5−2=35 - 2 = 35−2=3

y5y2=y3\frac{y^5}{y^2} = y^3y2y5​=y3

Example

a8a3\frac{a^8}{a^3}a3a8​:

Subtract: 8−3=58 - 3 = 58−3=5

a8a3=a5\frac{a^8}{a^3} = a^5a3a8​=a5

Rule 3: Power of a Power

A power of a power means repeated groups of repeated factors. Multiplying the indices counts the total number of factors.

When you have a power raised to another power, MULTIPLY the exponents:

(am)n=amn(a^m)^n = a^{mn}(am)n=amn

Why? Because: (x2)3=x2×x2×x2=(x×x)×(x×x)×(x×x)=x6(x^2)^3 = x^2 \times x^2 \times x^2 = (x \times x) \times (x \times x) \times (x \times x) = x^6(x2)3=x2×x2×x2=(x×x)×(x×x)×(x×x)=x6

You have 3 groups of x2x^2x2, so 2×3=62 \times 3 = 62×3=6 x's total.

Example

(a2)3(a^2)^3(a2)3:

Multiply exponents: 2×3=62 \times 3 = 62×3=6

(a2)3=a6(a^2)^3 = a^6(a2)3=a6

Example

(y4)2(y^4)^2(y4)2:

Multiply: 4×2=84 \times 2 = 84×2=8

(y4)2=y8(y^4)^2 = y^8(y4)2=y8

Rule 4: Product Rule

When a whole product is raised to a power, every factor inside the brackets is repeated. This includes numerical coefficients, not only variables.

When you raise a product to a power, apply the power to EACH part:

(ab)n=anbn(ab)^n = a^n b^n(ab)n=anbn

Why? Because: (xy)3=(xy)×(xy)×(xy)=(x×x×x)×(y×y×y)=x3y3(xy)^3 = (xy) \times (xy) \times (xy) = (x \times x \times x) \times (y \times y \times y) = x^3 y^3(xy)3=(xy)×(xy)×(xy)=(x×x×x)×(y×y×y)=x3y3

Example

(2x)3(2x)^3(2x)3:

Apply the power to both: 232^323 and x3x^3x3

(2x)3=23×x3=8x3(2x)^3 = 2^3 \times x^3 = 8x^3(2x)3=23×x3=8x3

Example

(ab)2(ab)^2(ab)2:

Apply to each: a2a^2a2 and b2b^2b2

(ab)2=a2b2(ab)^2 = a^2 b^2(ab)2=a2b2

Rule 5: Zero Power

The zero power rule often surprises students because it does not mean "nothing". It comes from a pattern of division where equal powers cancel completely.

Any number (except 0) raised to the power 0 equals 1:

a0=1a^0 = 1a0=1

Why? Think about the divide rule: anan=1 (something divided by itself)\frac{a^n}{a^n} = 1 \text{ (something divided by itself)}anan​=1 (something divided by itself)

But also: anan=an−n=a0\frac{a^n}{a^n} = a^{n-n} = a^0anan​=an−n=a0

So a0=1a^0 = 1a0=1.

Example

50=15^0 = 150=1

x0=1x^0 = 1x0=1 (for any x≠0x \neq 0x=0)

(7xyz)0=1(7xyz)^0 = 1(7xyz)0=1

No matter how complicated the base, if the power is 0, the answer is always 1.

Rule 6: Negative Powers

A negative exponent shows position, not negativity. It moves the factor to the other side of a fraction bar and makes the exponent positive.

A negative power means "put it in a fraction":

a−n=1ana^{-n} = \frac{1}{a^n}a−n=an1​

Why? Using the divide rule: a0an=a0−n=a−n\frac{a^0}{a^n} = a^{0-n} = a^{-n}ana0​=a0−n=a−n

But a0=1a^0 = 1a0=1, so: a−n=1ana^{-n} = \frac{1}{a^n}a−n=an1​

Example

2−22^{-2}2−2:

Flip to a fraction with positive power: 2−2=122=142^{-2} = \frac{1}{2^2} = \frac{1}{4}2−2=221​=41​

Example

x−3x^{-3}x−3:

Flip: x−3=1x3x^{-3} = \frac{1}{x^3}x−3=x31​

Example

(2x)−1(2x)^{-1}(2x)−1:

Flip: (2x)−1=12x(2x)^{-1} = \frac{1}{2x}(2x)−1=2x1​

Remember

Negative power ≠ negative answer!

2−2=142^{-2} = \frac{1}{4}2−2=41​ (positive!)

−22=−4-2^2 = -4−22=−4 (this is different, the negative is NOT an exponent)

All Six Rules Summary

RuleFormulaExampleWhy
Multiplyam×an=am+na^m \times a^n = a^{m+n}am×an=am+nx2×x3=x5x^2 \times x^3 = x^5x2×x3=x5Counting total uses
Divideaman=am−n\frac{a^m}{a^n} = a^{m-n}anam​=am−ny5y2=y3\frac{y^5}{y^2} = y^3y2y5​=y3Canceling
Power of Power(am)n=amn(a^m)^n = a^{mn}(am)n=amn(a2)3=a6(a^2)^3 = a^6(a2)3=a6Nested multiplication
Product(ab)n=anbn(ab)^n = a^n b^n(ab)n=anbn(2x)3=8x3(2x)^3 = 8x^3(2x)3=8x3Distributing power
Zeroa0=1a^0 = 1a0=150=15^0 = 150=1Self-division
Negativea−n=1ana^{-n} = \frac{1}{a^n}a−n=an1​2−2=142^{-2} = \frac{1}{4}2−2=41​Reciprocal
Exam Tip

Strategy for index problems:

  1. Identify which rule applies (same base? power of power? product?)
  2. Apply that rule
  3. Simplify
  4. Check whether multiple rules need to be combined

Example: (2x2)3×x−4(2x^2)^3 \times x^{-4}(2x2)3×x−4

  • First: Apply product rule → 23(x2)3=8x62^3(x^2)^3 = 8x^623(x2)3=8x6
  • Then: Multiply with x−4x^{-4}x−4 → 8x6×x−4=8x6−4=8x28x^6 \times x^{-4} = 8x^{6-4} = 8x^28x6×x−4=8x6−4=8x2
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