Algebraic fractions

Unit 4 Algebraic fractions

Key unit competence: Use operation properties and simplification of rational expressions in solving real life problems.

Key ideas

• An algebraic fraction is a fraction whose numerator and/or denominator contain variables, e.g. $\frac{x+1}{x-2}$.
• Condition of existence: the denominator cannot be zero. For $\frac{x+1}{x-2}$, $x\ne 2$.
• Simplify by factorising the numerator and denominator completely, then cancelling common factors (never terms).
• Add/subtract: find the LCM of the denominators, rewrite each fraction, then combine the numerators.
• Multiply: multiply numerators and denominators (factorise and cancel first). Divide: multiply by the reciprocal.
• Rational equations: multiply both sides by the LCM to clear denominators, solve, then reject values that make a denominator zero.

Formulas to know

Common misconceptions

• "$\frac{x+3}{3}=x$." No: 3 is a term in the numerator, not a factor; you cannot cancel it.
• "$\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y}$." No: $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$.
• "A solution that makes a denominator zero is still valid." No: it must be rejected.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. State the condition of existence and simplify $\frac{2x^2-8}{x^2+2x}$.

   Show answer

   $x\ne 0$, $x\ne -2$. $\frac{2(x-2)(x+2)}{x(x+2)}=\frac{2(x-2)}{x}$.

   Common mistake: Forgetting the condition $x\ne -2$ after cancelling.

2. The sum of a number and its reciprocal is $\frac{10}{3}$. Find the number.

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   $x+\frac{1}{x}=\frac{10}{3}$: $3x^2-10x+3=0$, $(3x-1)(x-3)=0$, $x=3$ or $\frac{1}{3}$.

   Common mistake: Writing the reciprocal of $x$ as $-x$.

3. Simplify $\frac{3}{x+2}+\frac{2}{x-1}$.

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   $\frac{3(x-1)+2(x+2)}{(x+2)(x-1)}=\frac{5x+1}{(x+2)(x-1)}$.

   Common mistake: Adding numerators and denominators separately.

4. Simplify $\frac{1}{a}-\frac{1}{b}$ over $\frac{1}{a}+\frac{1}{b}$, that is $\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}$.

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   Multiply top and bottom by $ab$: $\frac{b-a}{b+a}$.

   Common mistake: Cancelling $\frac{1}{a}$ from top and bottom.

5. Simplify $\frac{x^2-5x+6}{x^2-4}\times \frac{x+2}{x-3}$.

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   $\frac{(x-2)(x-3)}{(x-2)(x+2)}\times \frac{x+2}{x-3}=1$ (for $x\ne \pm 2,3$).

   Common mistake: Cancelling terms such as $x^2$.

6. A tank is filled by pipe A in 4 h and pipe B in 6 h. How long do both take together?

   Show answer

   $\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$ per hour, so $\frac{12}{5}=2.4$ h (2 h 24 min).

   Common mistake: Adding the times (10 h) or averaging them.

7. Solve $\frac{2}{x}-\frac{1}{x+1}=\frac{1}{2}$.

   Show answer

   $4(x+1)-2x=x(x+1)$: $x^2-x-4=0$, $x=\frac{1\pm \sqrt{17}}{2}$.

   Common mistake: Multiplying only some terms by the LCM.

8. Solve $\frac{x+3}{x-1}=2$.

   Show answer

   $x+3=2x-2$, $x=5$ (valid since $x\ne 1$).

   Common mistake: Not checking the condition $x\ne 1$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Simplify $\frac{2}{x+1}-\frac{1}{x}$.	a. $a-b$
   2. Simplify $\frac{x^2-4}{x-2}$.	b. $x=4$
   3. Solve $\frac{2}{x-1}+1=\frac{5}{x-1}$.	c. $\frac{x-1}{x(x+1)}$
   4. Simplify $\frac{\frac{a}{b}-\frac{b}{a}}{\frac{1}{a}+\frac{1}{b}}$.	d. $x+2$

   Show answer

   1 → c ($\frac{x-1}{x(x+1)}$) · 2 → d ($x+2$) · 3 → b ($x=4$) · 4 → a ($a-b$)

   • $\frac{2x-(x+1)}{x(x+1)}=\frac{x-1}{x(x+1)}$.
   • $\frac{(x-2)(x+2)}{x-2}$.
   • Step 1. Subtract $\frac{2}{x-1}$: $1=\frac{3}{x-1}$.
     Step 2. $x-1=3$, so $x=4$ (allowed, since $x\ne 1$).
   • Step 1. Numerator $=\frac{a^2-b^2}{ab}$; denominator $=\frac{a+b}{ab}$.
     Step 2. $\frac{a^2-b^2}{a+b}=\frac{(a-b)(a+b)}{a+b}=a-b$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $2x^2$  ·  $x=2$ or $x=-\frac{4}{3}$  ·  $2x$  ·  $x=2$
   1. The LCM of $x$, $x^2$ and $2x$ is .
   2. Solve $\frac{4}{x}=2$. →
   3. Solve $\frac{1}{x}+\frac{1}{x+2}=\frac{3}{4}$. →
   Show answer

   1: $2x^2$ · 2: $x=2$ · 3: $x=2$ or $x=-\frac{4}{3}$

   • It must contain $2$ and $x^2$.
   • $4=2x$.
   • Step 1. $4(2x+2)=3x(x+2)$: $3x^2-2x-8=0$.
     Step 2. $(3x+4)(x-2)=0$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Simplify $\frac{x^2-1}{x+1}\times \frac{1}{x-1}$.	a. $x\ne 1$ and $x\ne 4$
   2. Find the conditions of existence of $\frac{x+1}{x^2-5x+4}$.	b. $x\ne -1$ and $x\ne 3$
   3. Simplify $\frac{1}{x}+\frac{1}{y}$.	c. 1
   4. Find the conditions of existence of $\frac{x}{(x+1)(x-3)}$.	d. $\frac{x+y}{xy}$

   Show answer

   1 → c (1) · 2 → a ($x\ne 1$ and $x\ne 4$) · 3 → d ($\frac{x+y}{xy}$) · 4 → b ($x\ne -1$ and $x\ne 3$)

   • $\frac{(x-1)(x+1)}{(x+1)(x-1)}$.
   • Step 1. The denominator must not be zero: $x^2-5x+4=(x-1)(x-4)$.
     Step 2. So $x\ne 1$ and $x\ne 4$.
   • LCM $xy$: $\frac{y}{xy}+\frac{x}{xy}$.
   • Each factor of the denominator must be non-zero.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\frac{2}{x}$  ·  $x=1$  ·  $x=6$  ·  $x=-5$
   1. Solve $\frac{3}{x-1}=\frac{2}{x+1}$. →
   2. Simplify $\frac{x}{3}\times \frac{6}{x^2}$. →
   3. Solve $\frac{x}{2}+\frac{x}{3}=5$. →
   Show answer

   1: $x=-5$ · 2: $\frac{2}{x}$ · 3: $x=6$

   • $3x+3=2x-2$.
   • $\frac{6x}{3x^2}=\frac{2}{x}$.
   • Multiply by 6: $3x+2x=30$.

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