Algebraic fractions: simplify, calculate and solve

An algebraic fraction is a fraction containing a variable, such as

$$\frac{3x}{x+2} \quad\text{or}\quad \frac{x^2-9}{x^2+x-6}.$$

The same fraction rules used in arithmetic still apply. The extra challenge is algebraic structure: denominators may need factorising, cancelled factors must multiply the whole numerator and denominator, and some values of the variable are forbidden.

Before you begin

You should be able to:

• find numerical common denominators
• expand and factorise linear and quadratic expressions
• use index laws
• solve linear and quadratic equations

Review algebraic manipulation and factorisation or quadratic equations if those skills are uncertain.

Denominators create restrictions

Division by zero is undefined. Every denominator must therefore be non-zero.

For

$$\frac{5}{x-3},$$

the restriction is $x\ne3$. For

$$\frac{x+1}{x(x-4)},$$

the restrictions are $x\ne0$ and $x\ne4$.

State restrictions from the original expression, before simplifying. Cancellation does not restore an excluded value.

Example 1: a removable factor

Simplify

$$\frac{x^2-4}{x-2}.$$

The original denominator gives $x\ne2$. Factorise the difference of two squares:

$$\frac{x^2-4}{x-2} =\frac{(x-2)(x+2)}{x-2} =x+2,\qquad x\ne2.$$

Although $x+2$ itself can be evaluated at $x=2$, the original fraction cannot. The simplified expression and the original have equal values only on the original domain.

Simplifying by cancelling factors

Cancellation is division of numerator and denominator by the same non-zero factor:

$$\frac{ac}{bc}=\frac ab,\qquad b\ne0,\ c\ne0.$$

You may cancel factors, not terms joined by addition or subtraction.

Example 2: factorise first

Simplify

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

Factorise both parts:

$$x^2+5x+6=(x+2)(x+3),$$ $$x^2-4=(x+2)(x-2).$$

Hence

$$\frac{(x+2)(x+3)}{(x+2)(x-2)} =\boxed{\frac{x+3}{x-2}}, \qquad x\ne-2,2.$$

The restriction $x\ne-2$ remains even though the factor $x+2$ cancels.

Example 3: expose a hidden common factor

Simplify

$$\frac{6-3x}{x^2-4}.$$

Write the numerator in a form that matches a denominator factor:

$$6-3x=-3(x-2),$$

and factorise the denominator:

$$x^2-4=(x-2)(x+2).$$

Therefore

$$\frac{-3(x-2)}{(x-2)(x+2)} =\boxed{-\frac3{x+2}}, \qquad x\ne-2,2.$$

The minus sign belongs to the remaining numerator.

Misconception: cancelling terms

The step

$$\frac{x+3}{x}\ne3$$

is invalid. The $x$ in the numerator is a term in a sum, not a factor of the whole numerator. Instead,

$$\frac{x+3}{x}=\frac xx+\frac3x=1+\frac3x, \qquad x\ne0.$$

A useful test is this: if you cannot place brackets around the whole part being cancelled, it is probably not a common factor.

Multiplying algebraic fractions

Multiply numerators together and denominators together. Factorise first so that common factors can be cancelled before expansion:

$$\frac ab\times\frac cd=\frac{ac}{bd}.$$

Example 4: multiply and simplify

Simplify

$$\frac{x^2-9}{x^2+2x}\times\frac{x}{x-3}.$$

From the original denominators, $x\ne0,-2,3$. Factorise:

$$\frac{(x-3)(x+3)}{x(x+2)}\times\frac{x}{x-3}.$$

Cancel the common factors $x$ and $x-3$:

$$\boxed{\frac{x+3}{x+2}}, \qquad x\ne0,-2,3.$$

Expanding first would hide the factors and create unnecessary work.

Dividing algebraic fractions

To divide by a fraction, multiply by its reciprocal:

$$\frac ab\div\frac cd =\frac ab\times\frac dc.$$

The divisor itself must not equal zero. This may create an additional restriction from its numerator.

Example 5: divide and track the domain

Simplify

$$\frac{x^2-4}{x^2-x-6}\div\frac{x+2}{x-3}.$$

The displayed denominators require

$$x^2-x-6=(x-3)(x+2)\ne0,$$

so $x\ne3,-2$. The divisor must also be non-zero. Since its numerator is $x+2$, this confirms $x\ne-2$.

Now invert the second fraction and multiply:

$$\frac{(x-2)(x+2)}{(x-3)(x+2)} \times\frac{x-3}{x+2}.$$

Cancelling gives

$$\boxed{\frac{x-2}{x+2}}, \qquad x\ne-2,3.$$

Do not invert the first fraction, and do not invert both fractions.

Adding and subtracting algebraic fractions

Fractions can be added or subtracted only after they have a common denominator:

$$\frac ab+\frac cd=\frac{ad+bc}{bd}.$$

This formula is reliable, but using the lowest common denominator often gives shorter algebra.

Example 6: different linear denominators

Simplify

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

The common denominator is $x(x+1)$:

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

Therefore

$$\boxed{\frac{5x+2}{x(x+1)}}, \qquad x\ne0,-1.$$

Notice that the numerator is $2(x+1)+3x$, not merely $2+3$.

Example 7: subtraction and brackets

Simplify

$$\frac{4}{x-2}-\frac{1}{x+3}.$$

Use the common denominator $(x-2)(x+3)$:

$$\frac{4(x+3)-(x-2)}{(x-2)(x+3)}.$$

The subtraction applies to the whole second numerator:

$$4(x+3)-(x-2)=4x+12-x+2=3x+14.$$

Thus

$$\boxed{\frac{3x+14}{(x-2)(x+3)}}, \qquad x\ne2,-3.$$

Example 8: denominators with a shared factor

Simplify

$$\frac{3}{x^2-4}+\frac{1}{x+2}.$$

Since $x^2-4=(x-2)(x+2)$, the lowest common denominator is already $x^2-4$. Rewrite only the second fraction:

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

Hence

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

So the result is

$$\boxed{\frac{x+1}{x^2-4}}, \qquad x\ne-2,2.$$

Using the product of the original denominators would work, but it would introduce a repeated factor that later has to be cancelled.

Fractions inside fractions

A fraction whose numerator or denominator contains fractions is sometimes called a complex fraction. Multiply every term in the large numerator and denominator by the lowest common denominator of the smaller fractions.

Example 9: simplify a complex fraction

Simplify

$$\frac{\dfrac2x+1}{\dfrac3x-1}.$$

The small denominators have common denominator $x$, so multiply the entire top and bottom by $x$:

$$\frac{x\left(\dfrac2x+1\right)}{x\left(\dfrac3x-1\right)} =\frac{2+x}{3-x}.$$

The original expression requires $x\ne0$. Its large denominator must also be non-zero:

$$\frac3x-1\ne0 \quad\Longrightarrow\quad x\ne3.$$

Therefore

$$\boxed{\frac{x+2}{3-x}}, \qquad x\ne0,3.$$

Multiplying only one term by $x$ changes the value. The multiplier must apply to every term in both parts of the large fraction.

Solving equations containing algebraic fractions

The safest method is:

1. Record values excluded by the original denominators.
2. Find the lowest common denominator.
3. Multiply every term of the equation by it.
4. Solve the resulting equation.
5. Reject any excluded value and check valid solutions in the original equation.

Example 10: a linear equation

Solve

$$\frac{3}{x}+2=\frac{5}{x}, \qquad x\ne0.$$

Multiply every term by $x$:

$$3+2x=5.$$

Thus $2x=2$, so

$$\boxed{x=1}.$$

Checking gives $3+2=5$, so the solution is valid.

Example 11: an equation producing a quadratic

Solve

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

The restrictions are $x\ne1,-1$. Multiply every term by $(x-1)(x+1)$:

$$2(x+1)+(x-1)=(x-1)(x+1).$$

Expand and simplify:

$$2x+2+x-1=x^2-1,$$ $$x^2-3x-2=0.$$

Using the quadratic formula,

$$x=\frac{3\pm\sqrt{(-3)^2-4(1)(-2)}}2 =\frac{3\pm\sqrt{17}}2.$$

Neither value is excluded, so

$$\boxed{x=\frac{3+\sqrt{17}}2\quad\text{or}\quad x=\frac{3-\sqrt{17}}2}.$$

Example 12: an excluded apparent solution

Solve

$$\frac{x+1}{x-2}=\frac3{x-2}.$$

The restriction is $x\ne2$. Multiplying by $x-2$ gives

$$x+1=3,$$

so the algebra suggests $x=2$. But $x=2$ is excluded from the original equation. Therefore the equation has

$$\boxed{\text{no solution}}.$$

Clearing denominators can produce a value at which the multiplication was not reversible. Domain checking is essential.

Identities involving algebraic fractions

An equation asks for particular values of $x$. An identity is true for every $x$ in its domain. To prove an identity, transform one side into the other or simplify both sides to the same expression.

Example 13: prove an identity

Show that

$$\frac1{x-1}-\frac1{x+1}\equiv\frac2{x^2-1}.$$

Starting from the left,

$$\frac1{x-1}-\frac1{x+1} =\frac{x+1-(x-1)}{(x-1)(x+1)} =\frac2{x^2-1}.$$

Hence the identity is true for $x\ne-1,1$.

Checking one numerical value may expose an error, but it cannot prove an identity for all permitted values.

Common mistakes

• Cancelling terms instead of factors: factorise first, then cancel common multiplicative factors.
• Adding denominators: $\frac1a+\frac1b$ is not $\frac2{a+b}$. Use a common denominator.
• Losing brackets in subtraction: write $A-B$ before expanding the second numerator.
• Inverting the wrong fraction: in $A\div B$, replace only $B$ by its reciprocal.
• Expanding too early: factorised form reveals cancellation and common denominators.
• Forgetting restrictions: obtain them from the original expression, not only the simplified answer.
• Clearing only some denominators: multiply every term on both sides of an equation.

Self-check

Try these without looking at the answers.

1. State the restrictions and simplify $\frac{x^2-7x+12}{x^2-16}.$
2. Simplify $\frac{2x}{x^2-9}\times\frac{x+3}{4x}.$
3. Simplify $\frac{x^2-1}{x^2+3x+2}\div\frac{x-1}{x+2}.$
4. Write as a single fraction $\frac3x-\frac2{x-1}.$
5. Write as a single simplified fraction $\frac2{x^2-x}+\frac3x.$
6. Simplify $\frac{\dfrac1x+\dfrac1y}{\dfrac1x-\dfrac1y}.$
7. Solve $\frac4x-1=\frac2x.$
8. Solve $\frac1{x-2}+\frac1{x+2}=\frac13.$
9. Explain why $\frac{x^2-1}{x-1}=x+1$ is not valid at every real value of $x$.

Answers

1. Factorising gives $$\frac{(x-3)(x-4)}{(x-4)(x+4)} =\boxed{\frac{x-3}{x+4}}, \qquad x\ne-4,4.$$
2. Cancel $2x$, $x+3$ and the matching factors: $$\boxed{\frac1{2(x-3)}}, \qquad x\ne-3,0,3.$$
3. Factorising first gives $$\frac{(x-1)(x+1)}{(x+1)(x+2)} \times\frac{x+2}{x-1}=\boxed1,$$ with $x\ne-2,-1,1$. The restriction $x\ne1$ is required because the divisor would be zero.
4. Using $x(x-1)$ as the common denominator, $$\boxed{\frac{x-3}{x(x-1)}}, \qquad x\ne0,1.$$
5. Since $x^2-x=x(x-1)$, $$\frac2{x(x-1)}+\frac{3(x-1)}{x(x-1)} =\boxed{\frac{3x-1}{x(x-1)}}, \qquad x\ne0,1.$$
6. Multiply top and bottom by $xy$: $$\boxed{\frac{x+y}{y-x}},$$ where $x\ne0$, $y\ne0$ and $x\ne y$.
7. Multiply by $x$, where $x\ne0$: $4-x=2,$ so $\boxed{x=2}$.
8. Here $x\ne2,-2$. Multiplying by $3(x-2)(x+2)$ gives $$3(x+2)+3(x-2)=x^2-4,$$ so $x^2-6x-4=0$. Therefore $$\boxed{x=3\pm\sqrt{13}}.$$ Neither solution is excluded.
9. The original denominator is zero at $x=1$. Cancellation shows equality only when $x\ne1$.

What to learn next

These techniques support much of A-level algebra. Next, use them when:

• solving quadratic and rational inequalities
• simplifying expressions before applying the product, quotient and chain rules
• reversing fraction addition in partial fractions
• analysing reciprocal and rational graphs

The central habit is simple: factorise, record restrictions, then operate.