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Algebra - Algebraic Fractions and their Operations

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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An algebraic fraction is a fraction where the numerator and/or the denominator contain algebraic expressions. Examples include x2\frac{x}{2}, 5y\frac{5}{y}, and x+2xβˆ’3\frac{x+2}{x-3}.

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To simplify an algebraic fraction, factorize both the numerator and the denominator completely, then cancel out any common factors. Remember: you can only cancel factors, not individual terms added or subtracted.

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When multiplying algebraic fractions, multiply the numerators together and the denominators together: abΓ—cd=acbd\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}. It is often easier to simplify by canceling common factors before performing the multiplication.

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To divide one algebraic fraction by another, multiply the first fraction by the reciprocal of the second (flip the second fraction): abΓ·cd=abΓ—dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}.

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For addition and subtraction, fractions must have a common denominator. Find the Lowest Common Multiple (LCM) of the denominators, convert each fraction to an equivalent fraction with this denominator, and then combine the numerators: acΒ±bd=adΒ±bccd\frac{a}{c} \pm \frac{b}{d} = \frac{ad \pm bc}{cd}.

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Restrictions: Since division by zero is undefined, any value of the variable that makes the denominator equal to zero must be excluded from the domain.

πŸ“Formulae

kakb=ab, where k,b≠0\frac{ka}{kb} = \frac{a}{b}, \text{ where } k, b \neq 0

abΓ—cd=acbd\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}

abΓ·cd=abΓ—dc=adbc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}

ac+bc=a+bc\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}

abβˆ’cd=adβˆ’bcbd\frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd}

πŸ’‘Examples

Problem 1:

Simplify the expression: x2βˆ’9x2+5x+6\frac{x^2 - 9}{x^2 + 5x + 6}

Solution:

x2βˆ’9x2+5x+6=(xβˆ’3)(x+3)(x+2)(x+3)=xβˆ’3x+2\frac{x^2 - 9}{x^2 + 5x + 6} = \frac{(x-3)(x+3)}{(x+2)(x+3)} = \frac{x-3}{x+2}

Explanation:

First, factorize the numerator using the difference of squares identity a2βˆ’b2=(aβˆ’b)(a+b)a^2 - b^2 = (a-b)(a+b). Then, factorize the quadratic trinomial in the denominator. Finally, cancel the common factor (x+3)(x+3) from both the top and bottom.

Problem 2:

Perform the operation and simplify: 4xβˆ’1+3x+2\frac{4}{x-1} + \frac{3}{x+2}

Solution:

4(x+2)(xβˆ’1)(x+2)+3(xβˆ’1)(xβˆ’1)(x+2)=4x+8+3xβˆ’3(xβˆ’1)(x+2)=7x+5(xβˆ’1)(x+2)\frac{4(x+2)}{(x-1)(x+2)} + \frac{3(x-1)}{(x-1)(x+2)} = \frac{4x + 8 + 3x - 3}{(x-1)(x+2)} = \frac{7x + 5}{(x-1)(x+2)}

Explanation:

To add these fractions, we find the Lowest Common Denominator, which is (xβˆ’1)(x+2)(x-1)(x+2). We multiply the numerator of the first fraction by (x+2)(x+2) and the numerator of the second fraction by (xβˆ’1)(x-1), then expand and combine like terms in the numerator.

Problem 3:

Divide: 5y2xΓ·10yx2\frac{5y^2}{x} \div \frac{10y}{x^2}

Solution:

5y2xΓ—x210y=5Γ—yΓ—yΓ—xΓ—xxΓ—2Γ—5Γ—y=xy2\frac{5y^2}{x} \times \frac{x^2}{10y} = \frac{5 \times y \times y \times x \times x}{x \times 2 \times 5 \times y} = \frac{xy}{2}

Explanation:

Convert the division into multiplication by taking the reciprocal of the second fraction. Then, cancel common factors (55, one xx, and one yy) to simplify the result.