An algebraic fraction such as $\dfrac{x^2-9}{x^2+6x+9}$ is handled in exactly the same way as $\dfrac{10}{15}$: to simplify it, add it or divide it, you need to factorise first, just as with numbers you need to know their prime factors. Everything covered in Polynomials and Factorisation comes together here in a single block of calculation.
Simplifying by factorising numerator and denominator
Simplifying an algebraic fraction means cancelling the factors that the numerator and denominator have in common. The essential step, always, is to factorise both first: without factorising there is nothing to cancel.
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Simplify the following algebraic fraction as far as possible by factorising the numerator and the denominator: $\frac{x^2 - 9}{x^2 + 6x + 9}$
Simplify the following algebraic fraction as far as possible by factorising the numerator and the denominator:
Multiplication and division of algebraic fractions
Just as with numerical fractions: to multiply, multiply numerator by numerator and denominator by denominator; to divide, multiply by the reciprocal fraction. Factorising before multiplying almost always lets you simplify along the way, without expanding anything.
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Perform the following multiplication of algebraic fractions and express the result in its simplest form: $\frac{2x}{x-1} \cdot \frac{x^2 - 1}{4x^2}$
Perform the following multiplication of algebraic fractions and express the result in its simplest form:
Addition and subtraction: lowest common multiple of the denominators
With the same denominator, you add the numerators without touching the denominator. With different denominators, you need a common denominator, which is found by factorising each denominator and taking the lowest common multiple.
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Add the following algebraic fractions with common denominators: $\frac{3x - 1}{x^2 + 1} + \frac{x + 5}{x^2 + 1}$
Add the following algebraic fractions with common denominators:
Combined operations: respecting the order of operations
When additions, subtractions, products and brackets are mixed, you first work out what is inside the brackets, then products and quotients, and finally additions and subtractions (each group from left to right), just as with numbers.
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Evaluate the expression inside the brackets and then multiply to simplify the final result: $\frac{x^2 - x}{x+1} \cdot \left( \frac{1}{x} + \frac{1}{x-1} \right)$
Evaluate the expression inside the brackets and then multiply to simplify the final result:
A complex fraction has fractions inside its numerator or its denominator. It is simplified by first reducing the numerator and the denominator separately to a single fraction each, and dividing at the end.
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Simplify the following complex algebraic fraction by first simplifying the numerator and denominator independently: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x}}$
Simplify the following complex algebraic fraction by first simplifying the numerator and denominator independently:
The formulae for resistors in parallel or for thin lenses are, at heart, sums of algebraic fractions that have to be simplified to obtain a single expression in terms of $x$.
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In an electrical circuit, two resistors $R_1$ and $R_2$ are connected in parallel. The equivalent resistance $R_{eq}$ satisfies the relation…
In an electrical circuit, two resistors $R_1$ and $R_2$ are connected in parallel. The equivalent resistance $R_{eq}$ satisfies the relation $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$. If the values of the resistors are given by the polynomials $R_1 = x^2 - 4$ and $R_2 = x^2 + 4x + 4$, obtain the simplified expression for the equivalent resistance $R_{eq}$ in terms of $x$.
The thin lens formula in optics states that $\frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i}$. If the distance from the object to the optical centre is…
The thin lens formula in optics states that $\frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i}$. If the distance from the object to the optical centre is $s_o = \frac{x^2+x}{2}$ and the image distance is $s_i = \frac{x^2-1}{3}$, determine the simplified algebraic expression for the focal length $f$.
The average production cost of an electronic component is given by the expression $C_m(x) = \frac{x^3 + 2x^2 - 5x - 6}{x^2 - 4}$, where $x$ is the production volume in…
The average production cost of an electronic component is given by the expression $C_m(x) = \frac{x^3 + 2x^2 - 5x - 6}{x^2 - 4}$, where $x$ is the production volume in thousands of units. Due to a subsidy adjustment, a fee represented by the fraction $\frac{x^2-1}{x+2}$ must be subtracted. Calculate the new expression for the average cost by first simplifying the first fraction.