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This page sets out the method for algebraic fractions in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Algebraic fractions topic page.
When you are asked to simplify a single algebraic fraction, factorise everything you can first.
When adding or subtracting two algebraic fractions, build a common denominator from both denominators.
When an equation contains algebraic fractions, multiply every term by each denominator to remove them.
A fraction simplifies when the same bracket divides the top and the bottom.
(x²-9)/(x-3) is x+3, because x²-9 = (x-3)(x+3).
(2x+6)/(x+3) = 2, because 2x+6 = 2(x+3).
1/(x+2) + 1/(x+3) has denominator (x+2)(x+3) and numerator 2x+5.
1/(x+2) - 1/(x+4) has numerator 2, the gap between the constants.
Cancel only a common factor. Do not cancel a single letter from a sum.
(x^2-9)/(x^2+5x+6) is not simplified by crossing out x^2. The top is (x-3)(x+3) and the bottom is (x+2)(x+3). The shared factor is (x+3), so the fraction is (x-3)/(x+2), provided x≠ -3 and x≠ -2. At x=1 the original is (1-9)/(1+5+6)=(-8)/12=-2/3, and (1-3)/(1+2) is the same. If those two numbers disagree, you cancelled a term.
Single-term fractions are the exception that causes the confusion. (6x^2)/3x=2x is legal because 3x is a factor of the numerator. (x^2+6)/x is x+6/x, not x+6. You can divide each term of a sum by the denominator, and you must keep both pieces. Cancelling the x in only one term of a sum is the same mistake as cancelling x^2 above.
Method: 1. Factorise the numerator and the denominator. 2. Cancel a common factor, never a common term. 3. State any value that makes a cancelled factor zero. 4. Check by substituting a number that is still allowed.
2/(x-1)+3/(x+2) has denominator (x-1)(x+2). The first numerator becomes 2(x+2) and the second becomes 3(x-1). That is 2x+4+3x-3=5x+1. So the sum is (5x+1)/((x-1)(x+2)). At x=2 the left side is 2+3/4=11/4 and the right side is 11/(1× 4). Same value. Do not expand the denominator unless a question asks for it. Factored form shows the excluded values x=1 and x=-2.
Subtraction needs a bracket around the whole second numerator. 3/(x-2)-(2x+1)/(x+4) becomes (3(x+4)-(2x+1)(x-2))/((x-2)(x+4)). Students who write 3(x+4)-2x+1 have only subtracted 2x. The +1 must change sign as well, and if the second numerator has two factors, the whole product is subtracted. Expand only after the bracket is in place.
Method: 1. Factorise each denominator. 2. The common denominator is the product of the distinct factors. 3. Multiply each numerator so the fractions match. 4. Add the numerators and simplify once at the end.
To multiply, factor everything that factorises, cancel, then multiply what remains. (x^2-4)/(x+1)×(x+1)/(x-2)=((x-2)(x+2))/(x+1)×(x+1)/(x-2). Cancel (x+1) and (x-2) to leave x+2, with x≠ -1 and x≠ 2. Dividing by (x-2)/(x+1) is the same as multiplying by (x+1)/(x-2). Flip once. Flipping both fractions, or flipping the first, undoes the division.
Equations are different from expressions. For 3/(x-2)-2/(x+1)=1, multiply through by (x-2)(x+1). You get 3(x+1)-2(x-2)=(x-2)(x+1). The right-hand side is the denominator times 1, so it is not left behind. Then 3x+3-2x+4=x^2-x-2, so x+7=x^2-x-2, and x^2-2x-9=0. The discriminant is 4+36=40, so x=1±√(10). Neither root is 2 or -1.
Method: 1. Multiply numerators and multiply denominators. 2. Factorise and cancel before you expand everything. 3. To divide, multiply by the reciprocal. 4. An equation is cleared by multiplying by the common denominator, then excluding its zeros.
Substitute a simple value that is not an excluded one. For the sum in the second section, x=2 worked. For a solution of an equation, substitute into the original fractions, not only into the quadratic, in case a cancelled factor hid an extra root. x=0 in 3/(x+1)+2/(x-1) is 3+(-2)=1, so x=0 really solves that equation equal to 1. x=5 gives 3/6+2/4=1 as well.
Write the excluded values even when neither of them is a solution. It shows you looked. If a quadratic root is excluded, throw it out in a sentence: "not x=2, because the original denominator is zero". An expression question does not have solutions, so the excluded values are part of the simplified fraction, not a separate equation. Keep the two jobs apart.
Method: 1. Substitute a permitted number into the original and into the simplified form. 2. The two values must match. 3. A value that makes any denominator zero is excluded, even if it appears to solve the equation. 4. Write those excluded values beside the solution.
Simplifying an algebraic fraction is the same idea as simplifying 12/18: find a common factor and divide it out. The difference is that the factors are brackets. Factorise the top and the bottom completely, then cross out any bracket that appears on both. For (x^2+x-12)/(x^2-9) the top is (x+4)(x-3) and the bottom is (x-3)(x+3), so the common bracket (x-3) cancels and the answer is (x+4)/(x+3).
Look for three kinds of factorising: a common factor such as 3x^2+6x=3x(x+2), a difference of two squares such as x^2-25=(x-5)(x+5), and an ordinary quadratic. Only whole brackets or whole factors can cancel. In (x+4)/(x+3) you must not cancel the x or the 3 and 4. A numerical check protects you: put x=5 in the original and in your answer and confirm they match.
Method: 1. Factorise the numerator completely, starting with any common factor. 2. Factorise the denominator completely. 3. Cancel whole factors that appear on both the top and the bottom. 4. Check by substituting a small value of x into the original and the answer.
Tip: If nothing cancels, look again for a difference of two squares or a common factor you missed.
To multiply algebraic fractions, multiply across, but do not expand first. Factorise every top and bottom and cancel before you multiply. To divide, keep the first fraction, change the division to a multiplication, and flip the second fraction. For (x^2-9)/(x+2)÷(x+3)/(x^2-4) write ((x-3)(x+3))/(x+2)×((x-2)(x+2))/(x+3).
Now cancel the matching brackets, (x+3) with (x+3) and (x+2) with (x+2). What is left is (x-3)(x-2). You can leave it as two brackets or expand it to x^2-5x+6, unless the question asks for a particular form. Exam questions often print a target form such as 1/(2x(x-3)), so check that your final line has the same structure. A tidy top-and-bottom factorised form makes the cancelling visible to the marker.
Method: 1. Change a division into a multiplication by flipping the second fraction. 2. Factorise every numerator and denominator. 3. Cancel matching brackets between any top and any bottom. 4. Multiply what is left and compare with the form asked for.
To add or subtract, you need a common denominator. The simplest choice is the product of the different denominators, with any repeated factor used once. For 5/(x+1)+2/3x use 3x(x+1). Rewrite each fraction over this denominator: 15x/(3x(x+1))+(2(x+1))/(3x(x+1)). Then add the numerators: (17x+2)/(3x(x+1)). Keep the denominator in factorised form.
Three fractions work in the same way. For 2/(x+2)+3/(x-1)-5/x the common denominator is x(x+2)(x-1). After multiplying each numerator by the missing brackets and expanding, the x^2 terms cancel and the numerator becomes 10-x. When a subtraction is involved, put the whole numerator in brackets before expanding, so that the minus changes every term. A numerical check at x=2 confirms the result.
Method: 1. Find a common denominator that contains every different factor once. 2. Multiply each numerator by the bracket or number that its denominator is missing. 3. Put a bracket around a numerator that follows a minus sign, then expand. 4. Collect like terms and check with a number.
A show-that question might read "show that 4x/(x+2)-(3x-1)/(x-2)-1 can be written as (ax+b)/(x^2-4)". The denominator x^2-4 is the product (x+2)(x-2), so use it as the common denominator. The x^2 terms cancel in the numerator, leaving the linear expression that matches the form. When an equation has fractions on one side, multiply every term, including the whole-number side, by the common denominator.
For 6/(x+1)+1/(x-2)=2 multiply by (x+1)(x-2) to get 6(x-2)+(x+1)=2(x+1)(x-2). This simplifies to 2x^2-9x+7=0, which factorises as (2x-7)(x-1)=0. Both x=1 and x=7/2 are valid here. A solution that makes a denominator zero must be rejected, so always substitute back into the original to check.
Method: 1. Use the printed denominator or the full product as the common denominator. 2. In an equation, multiply every term, including numbers, by that denominator. 3. Rearrange to a quadratic equal to zero and factorise or use the formula. 4. Reject any answer that makes an original denominator zero.
Tip: Multiply the whole of the right-hand side as well, including a plain number such as 2.
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