Adding/Subtracting Algebraic Fractions (Junior Cert Mathematics): Revision Notes

Adding/Subtracting Algebraic Fractions

Algebraic fractions are fractions where the numerator, the denominator, or both contain algebraic expressions like $x, y$, etc. Just like with regular fractions, you can add and subtract algebraic fractions, but there are a few additional steps because of the variables.

Key Concepts to Remember

1. Common Denominator: Just like with regular fractions, to add or subtract algebraic fractions, they must have the same denominator. If the denominators are different, you need to find a common denominator before you can add or subtract them.
2. Simplifying: After adding or subtracting, always simplify the fraction if possible. This might involve factorising the numerator or denominator and cancelling common factors.

Steps for Adding and Subtracting Algebraic Fractions

Step 1: Find a Common Denominator

If the fractions have different denominators, you must find a common denominator. This could be the lowest common denominator (LCD) or simply multiplying the two denominators together.

Step 2: Adjust the Fractions

Rewrite each fraction so that they have the common denominator. This might involve multiplying the numerator and denominator by the same expression to adjust each fraction.

Step 3: Add or Subtract the Numerators

Once the fractions have the same denominator, you can add or subtract the numerators directly. The denominator remains the same.

Step 4: Simplify the Result

After adding or subtracting, look to simplify the resulting fraction. This might involve factorising and cancelling out common factors.

Worked Examples

Let's go through some detailed examples.

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Review

Adding and subtracting algebraic fractions requires careful attention to the denominators. Always:

• Find a common denominator.
• Adjust the fractions so they have the same denominator.
• Add or subtract the numerators.
• Simplify the result by factoring if necessary.

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Practice Problems: Adding and Subtracting Algebraic Fractions

Below are some practice problems on adding and subtracting algebraic fractions. These problems are designed to gradually increase in complexity, similar to what you might encounter in a Junior Cycle Maths exam. After each problem, you'll find a detailed, step-by-step solution to help you understand how to solve it.

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Solutions

Solution to Problem 1

Step 1: Check the Denominators

• Both fractions have the same denominator, which is $5$. Step 2: Add the Numerators

• Since the denominators are the same, add the numerators directly: $\frac{2x}{5} + \frac{3x}{5} = \frac{2x + 3x}{5}$ Step 3: Simplify the Numerator

• Add the $x$ terms in the numerator: $2x + 3x = 8x$ Final Answer: $\frac{2x + 3x}{5} = \frac{5x}{5}$

• Simplify the fraction: $\frac{5x}{5} = x$

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Explanation:

• We added the numerators because the denominators were the same. The expression simplified to $x$ after cancelling out the common factor of $5$.

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Solution to Problem 2

Step 1: Find a Common Denominator

• The denominators are $9$ and $3$. The smallest common denominator between $9$ and $3$ is $9$. Step 2: Adjust the Fractions to Have the Same Denominator

• The fraction $\frac{2x}{3}$ needs to be rewritten with $9$ as the denominator:

• Multiply both the numerator and denominator of $\frac{2x}{3}$ by $3$: $\frac{2x \times 3}{3 \times 3} = \frac{6x}{9}$

• Now the expression is: $\frac{4x}{9} - \frac{6x}{9}$ Step 3: Subtract the Numerators

• Since the denominators are the same, subtract the numerators: $\frac{4x - 6x}{9}$ Step 4: Simplify the Numerator

• Subtract the $x$ terms in the numerator: $4x - 6x = -2x$

• So the expression simplifies to: $\frac{-2x}{9}$ Final Answer: $\frac{4x}{9} - \frac{6x}{9} = \frac{-2x}{9}$

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Explanation:

• We found the common denominator $9$, adjusted the second fraction by multiplying by $3$, and then subtracted the numerators to get $(-2x)$.

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Solution to Problem 3

Step 1: Find a Common Denominator

• The denominators are $x$ and $2x$. The smallest common denominator is $2x$. Step 2: Adjust the Fractions

• The first fraction $\frac{5}{x}$ needs to be rewritten with $2x$ as the denominator:

• Multiply both the numerator and denominator by $2$: $\frac{5 \times 2}{x \times 2} = \frac{10}{2x}$

• Now the expression looks like this: $\frac{10}{2x} + \frac{4}{2x}$ Step 3: Add the Numerators

• Since the denominators are now the same, add the numerators: $\frac{10 + 4}{2x} = \frac{14}{2x}$ Step 4: Simplify the Fraction

• Simplify the fraction by dividing the numerator by the denominator: $\frac{14}{2x} = \frac{14}{2} \times \frac{1}{x} = 7 \times \frac{1}{x} = \frac{7}{x}$ Final Answer: $\frac{5}{x} + \frac{4}{2x} = \frac{7}{x}$

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Explanation:

• We found the common denominator $2x$, adjusted the first fraction, added the numerators, and then simplified to $\frac{7}{x}$.

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Solution to Problem 4

Step 1: Find a Common Denominator

• The denominators are $(x - 1)$ and $(x + 2)$. The common denominator will be $(x - 1)(x + 2)$. Step 2: Adjust the Fractions

• Rewrite each fraction with the common denominator $(x - 1)(x + 2)$:

• For $\frac{7}{x - 1}$, multiply both the numerator and denominator by $(x + 2)$: $\frac{7(x + 2)}{(x - 1)(x + 2)}$

• For $\frac{3}{x + 2}$, multiply both the numerator and denominator by $(x - 1)$: $\frac{3(x - 1)}{(x - 1)(x + 2)}$

• Now the expression looks like this: $\frac{7(x + 2)}{(x - 1)(x + 2)} - \frac{3(x - 1)}{(x - 1)(x + 2)}$ Step 3: Subtract the Numerators

• Since the denominators are now the same, subtract the numerators: $\frac{7(x + 2) - 3(x - 1)}{(x - 1)(x + 2)}$ Step 4: Expand and Simplify the Numerator

• Expand the expressions in the numerator: $[ 7(x + 2) = 7x + 14 ] [ 3(x - 1) = 3x - 3 ]$

• Subtract the two expressions: $7x + 14 - (3x - 3) = 7x + 14 - 3x + 3 = 4x + 17$ Final Answer: $\frac{7(x + 2) - 3(x - 1)}{(x - 1)(x + 2)} = \frac{4x + 17}{(x - 1)(x + 2)}$

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Explanation:

• We found the common denominator $(x - 1)(x + 2)$, adjusted each fraction, expanded and subtracted the numerators, and then simplified the expression to $( \frac{4x + 17}{(x - 1)(x + 2)} )$.

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Solution to Problem 5

Step 1: Factorise the Denominator

• The denominator $x^2 - 9$ is a difference of squares, which can be factorised as: $x^2 - 9 = (x - 3)(x + 3)$

Step 2: Find a Common Denominator

• The common denominator between $(x - 3)(x + 3)$ and $(x + 3)$ is $(x - 3)(x + 3)$. Step 3: Adjust the Fractions

• The first fraction already has the denominator $(x - 3)(x + 3)$.

• For the second fraction $\frac{2}{x + 3}$, multiply both the numerator and denominator by $(x - 3)$: $\frac{2(x - 3)}{(x + 3)(x - 3)}$

• Now the expression looks like this: $\frac{3x}{(x - 3)(x + 3)} + \frac{2(x - 3)}{(x + 3)(x - 3)}$ Step 4: Add the Numerators

• Since the denominators are now the same, add the numerators: $\frac{3x + 2(x - 3)}{(x - 3)(x + 3)}$ Step 5: Expand and Simplify the Numerator

• Expand the expression: $2(x - 3) = 2x - 6$

• Add $3x$ and $(2x - 6)$: $3x + 2x - 6 = 5x - 6$ Final Answer: $\frac{3x + 2(x - 3)}{(x - 3)(x + 3)} = \frac{5x - 6}{(x - 3)(x + 3)}$

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Explanation:

• We factorised the denominator, found the common denominator, adjusted the second fraction, expanded the numerator, and then simplified the expression.

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Solution to Problem 6

Step 1: Find a Common Denominator

• The denominators are $(x + 2)$ and $(x - 2)$. The common denominator is $(x + 2)(x - 2)$. Step 2: Adjust the Fractions

• For $\frac{4}{x + 2}$, multiply both the numerator and denominator by $(x - 2)$: $\frac{4(x - 2)}{(x + 2)(x - 2)}$

• For $\frac{5}{x - 2}$, multiply both the numerator and denominator by $(x + 2)$: $\frac{5(x + 2)}{(x - 2)(x + 2)}$

• Now the expression is: $\frac{4(x - 2)}{(x + 2)(x - 2)} + \frac{5(x + 2)}{(x - 2)(x + 2)}$ Step 3: Add the Numerators

• Since the denominators are now the same, add the numerators: $\frac{4(x - 2) + 5(x + 2)}{(x + 2)(x - 2)}$ Step 4: Expand the Numerators

• Expand each term in the numerator: $[ 4(x - 2) = 4x - 8 ] [ 5(x + 2) = 5x + 10 ]$

• Now combine them: $4x - 8 + 5x + 10 = 9x + 2$ Final Answer: $\frac{4(x - 2) + 5(x + 2)}{(x + 2)(x - 2)} = \frac{9x + 2}{(x + 2)(x - 2)}$

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Explanation:

• We found the common denominator $(x + 2)(x - 2)$, adjusted each fraction, expanded the numerators, added them, and simplified the expression.

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Solution to Problem 7

Step 1: Factorise the Denominator

• The denominator $(x^2 - 4)$ is a difference of squares, which factorises to: $x^2 - 4 = (x - 2)(x + 2)$ Step 2: Find a Common Denominator

• The common denominator between $(x - 2)(x + 2)$ and $(x + 2)$ is $(x - 2)(x + 2)$. Step 3: Adjust the Fractions

• The first fraction already has the denominator $(x - 2)(x + 2)$.

• For the second fraction $\frac{x}{x + 2}$, multiply both the numerator and denominator by $(x - 2)$: $\frac{x(x - 2)}{(x + 2)(x - 2)}$

• Now the expression looks like this: $\frac{3x}{(x - 2)(x + 2)} - \frac{x(x - 2)}{(x - 2)(x + 2)}$ Step 4: Subtract the Numerators

• Subtract the numerators: $\frac{3x - x(x - 2)}{(x - 2)(x + 2)}$ Step 5: Expand and Simplify the Numerator

• Expand $x(x - 2)$ in the numerator: $x(x - 2) = x^2 - 2x$

• Now subtract: $3x - (x^2 - 2x) = 3x - x^2 + 2x = -x^2 + 5x$ Final Answer: $\frac{3x - x(x - 2)}{(x - 2)(x + 2)} = \frac{-x^2 + 5x}{(x - 2)(x + 2)}$

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Explanation:

• We factorised the denominator, found the common denominator, adjusted the second fraction, expanded and subtracted the numerators, and simplified the expression.

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Solution to Problem 8

Step 1: Factorise the Denominator

• The denominator $x^2 - 4x + 4$ is a perfect square trinomial, which factorises to: $x^2 - 4x + 4 = (x - 2)(x - 2) = (x - 2)^2$ Step 2: Find a Common Denominator

• The common denominator between $(x - 2)^2$ and $(x - 2)$ is $(x - 2)^2$. Step 3: Adjust the Fractions

• The first fraction already has the denominator $(x - 2)^2$.

• For the second fraction $\frac{2}{x - 2}$, multiply both the numerator and denominator by $(x - 2)$: $\frac{2(x - 2)}{(x - 2)(x - 2)} = \frac{2(x - 2)}{(x - 2)^2}$

• Now the expression is: $\frac{x}{(x - 2)^2} + \frac{2(x - 2)}{(x - 2)^2}$ Step 4: Add the Numerators

• Add the numerators: $\frac{x + 2(x - 2)}{(x - 2)^2}$ Step 5: Expand and Simplify the Numerator

• Expand $2(x - 2)$ in the numerator: $2(x - 2) = 2x - 4$

• Now add $(x)$ and$(2x - 4)$: $[ x + 2x - 4 = 3x - 4 ]$ Final Answer: $\frac{x + 2(x - 2)}{(x - 2)^2} = \frac{3x - 4}{(x - 2)^2}$

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Explanation:

• We factorised the quadratic denominator, found the common denominator, adjusted the second fraction, expanded and added the numerators, and simplified the expression.

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Solution to Problem 9

Problem: Simplify $( \frac{2x}{x^2 - 9} - \frac{3}{x + 3} ).$

Step 1: Factorise the Denominator

• The denominator $(x^2 - 9)$ is a difference of squares, which factorises to: $x^2 - 9 = (x - 3)(x + 3)$ Step 2: Find a Common Denominator

• The common denominator between $(x - 3)(x + 3)$ and $(x + 3)$ is $(x - 3)(x + 3)$. Step 3: Adjust the Fractions

• The first fraction already has the denominator $(x - 3)(x + 3)$.

• For the second fraction $\frac{3}{x + 3}$, multiply both the numerator and denominator by $(x - 3)$: $\frac{3(x - 3)}{(x + 3)(x - 3)} = \frac{3(x - 3)}{(x - 3)(x + 3)}$

• Now the expression is: $\frac{2x}{(x - 3)(x + 3)} - \frac{3(x - 3)}{(x - 3)(x + 3)}$ Step 4: Subtract the Numerators

• Subtract the numerators: $\frac{2x - 3(x - 3)}{(x - 3)(x + 3)}$ Step 5: Expand and Simplify the Numerator

• Expand $3(x - 3)$ in the numerator: $3(x - 3) = 3x - 9$

• Now subtract: $2x - (3x - 9) = 2x - 3x + 9 = -x + 9$ Final Answer: $\frac{2x - 3(x - 3)}{(x - 3)(x + 3)} = \frac{-x + 9}{(x - 3)(x + 3)}$

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Explanation:

• We factorised the quadratic denominator, found the common denominator, adjusted the second fraction, expanded and subtracted the numerators, and simplified the expression.

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