GCSE: Algebraic Fractions Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com Last modified: 23rd April 2017

GCSE Specification Simplify algebraic fractions. Add and subtract algebraic fractions. Solve equations involving algebraic fractions which lead to quadratic equations. And from the AQA IGCSE Further Maths specification…

Starter 40 3 =2𝑥+4 40−𝑥 3 =𝑥+4   2 4−2𝑥 =3𝑥−2 2 4 =5𝑥−2   (Click your answer) Are these algebraic steps correct? 40 3 =2𝑥+4 40−𝑥 3 =𝑥+4 Fail  Win!  2 4−2𝑥 =3𝑥−2 2 4 =5𝑥−2 Ask students why they think the step might have been made (in addition to pointing out why it’s incorrect!). Emphasise that when a term is contained within either a bracket, fraction or root, applying the reverse of the term to the whole expression doesn’t generally have the desired effect.  Fail Win!  2−𝑥 =2𝑥+3 2 =3𝑥+3  Fail  Win!

𝑎 2 𝑏 𝑎+𝑏 𝑎𝑏 𝑏 Starter   Are these algebraic steps correct? Fail (Click your answer) Are these algebraic steps correct? 𝑎 2 𝑏 𝑎+𝑏 𝑎𝑏 𝑏 Fail  Win! 

𝑥 2 +2 =𝑦+2 Starter 𝑦 2 +𝑥 2+𝑥 𝑠(4+𝑧) 𝑠 (2𝑥+1)(𝑥−2) 𝑥−2 𝑝𝑞 𝑟+2 +1 𝑝𝑞 To cancel or not to cancel, that is the question? (Click your answer) 𝑦 2 +𝑥 2+𝑥 𝑠(4+𝑧) 𝑠 𝑥 2 +2 =𝑦+2 Fail  Win!  Fail  Win!  Fail  Win!  (2𝑥+1)(𝑥−2) 𝑥−2 𝑝𝑞 𝑟+2 +1 𝑝𝑞 1+𝑟 2 −1 Fail  Win!  Fail  Win!  Fail  Win! 

What did we learn? 𝑎+𝑏 𝑐+𝑏 → 𝑎 𝑐  Bro Tip #1: You can’t add or subtract a term which is ‘trapped’ inside a bracket, fraction or root. 2 𝑎−𝑥 =2𝑥+1 → 2 𝑎 =3𝑥+1  Bro Tip #2: In a fraction, we can only divide top and bottom by something, not add/subtract. (e.g. 5 7 is not the same as 6 8 !) 𝑎+𝑏 𝑐+𝑏 → 𝑎 𝑐 

Simplifying Algebraic Fractions Bro Tip: Just factorise top and bottom, then cancel! 𝑥 2 +𝑥 𝑥+1 = 𝑥 𝑥+1 𝑥+1 =𝑥 ? Bro Note: Do NOT leave your answer as 𝑥 1 ! ? 2 𝑥 2 +4𝑥 𝑥 2 −4 = 2𝑥 𝑥+2 𝑥+2 𝑥−2 = 2𝑥 𝑥−2 3𝑥+3 𝑥 2 +3𝑥+2 = 3 𝑥+2 ? 2 𝑥 2 −5𝑥−3 6 𝑥 3 −2 𝑥 4 =− 2𝑥+1 2 𝑥 3 ?

𝑥(𝑥+1) 𝑥 2 −1 → 𝑥 𝑥−1 2 𝑥 2 +5𝑥−3 𝑥 2 −9 → 2𝑥−1 𝑥−3 Test Your Understanding Bro Tip: Sometimes they’ve done part of the factorising for you! 𝑥(𝑥+1) 𝑥 2 −1 → 𝑥 𝑥−1 2 𝑥 2 +5𝑥−3 𝑥 2 −9 → 2𝑥−1 𝑥−3 𝑥 2 +2𝑥𝑦+ 𝑦 2 2𝑥+2𝑦 → 𝑥+𝑦 2 ? ? ?

− 4−𝑦 =𝑦−4 − 2𝑥−9 =9−2𝑥 1−𝑥 𝑥−1 =−1 3−2𝑥 2−𝑥 2𝑥−3 𝑥+1 = 𝑥−2 𝑥+1 Negating a difference − 4−𝑦 =𝑦−4 − 2𝑥−9 =9−2𝑥 1−𝑥 𝑥−1 =−1 3−2𝑥 2−𝑥 2𝑥−3 𝑥+1 = 𝑥−2 𝑥+1 ? ? ? ?

Exercise 1 𝑥+10 5 𝑥+10 4 =𝑥+10 [AQA] Simplify fully 9 𝑥 3 −16𝑥 6𝑥+8 = 𝑥 3𝑥−4 2 ? 2𝑥+6 2𝑥 = 𝑥+3 𝑥 4𝑥+8 3𝑥+6 = 4 3 𝑥 2 +𝑥−6 𝑥 2 −7𝑥+10 = 𝑥+3 𝑥−5 2𝑥+10 𝑥 2 −25 = 2 𝑥−5 𝑝 2 −9 2𝑝+6 = 𝑝−3 2 𝑥 2 +𝑥−2 𝑥 2 −4 = 𝑥−1 𝑥−2 6 𝑥 2 +3𝑥 4 𝑥 2 −1 = 3𝑥 2𝑥−1 ? 15 1 8 ? 𝑥 2 +2𝑥+1 𝑥 2 +3𝑥+2 = 𝑥+1 𝑥+2 𝑥 2 −8𝑥+15 2 𝑥 2 −7𝑥−15 = 𝑥−3 2𝑥+3 𝑥 2 −9 2 𝑥 2 −7𝑥+3 = 𝑥+3 2𝑥−1 6 𝑥 2 −𝑥−1 4 𝑥 2 −1 = 3𝑥+1 2𝑥+1 2 𝑦 2 +4𝑦 3 𝑦 2 +7𝑦+2 × 9 𝑦 2 −1 3 𝑦 2 −𝑦 =2 ? ? 9 2 16 ? 2 𝑥 2 −𝑥𝑦− 𝑦 2 𝑥 2 𝑦−𝑥 𝑦 2 = 𝟐𝒙+𝒚 𝒙𝒚 ? ? 10 3 ? N Simplify 𝑥 3 −𝑥 𝑥 2 +𝑥𝑦+𝑥+𝑦 = 𝒙 𝒙−𝟏 𝒙+𝒚 ? 11 4 ? ? ? 12 5 ? 13 ? 6 14 [AQA Set 4 Paper 2 Q18] Simplify fully 24𝑚−9 𝑚 2 64−9 𝑚 2 = 𝟑𝒎 𝟖−𝟑𝒎 𝟖+𝟑𝒎 𝟖−𝟑𝒎 = 𝟑𝒎 𝟖+𝟑𝒎 ? 7 ?

2 3 + 1 2 = 4+3 6 = 7 6 Adding/Subtracting Fractions ? ? ? ? What’s our usual approach for adding fractions? Since denominator of first fraction got multiplied by 2, so must numerator. We can say we are “cross multiplying the numerators”. 2 3 + 1 2 = 4+3 6 = 7 6 ? We can get a common denominator by just multiplying the denominators. Bro Tip: There’s a good reason why I’m writing as a single fraction 4+3 6 immediately rather than initially as two separate fractions 4 6 + 3 6 . It’ll avoid sign problems when we come to subtracting algebraic fractions. Quickfire Questions: ? ? ? 2 5 + 5 6 = 𝟏𝟐+𝟐𝟓 𝟑𝟎 = 𝟐𝟕 𝟑𝟎 6 7 − 2 3 = 𝟏𝟖−𝟏𝟒 𝟐𝟏 = 𝟒 𝟐𝟏 5 8 + 3 5 = 𝟐𝟓−𝟐𝟒 𝟒𝟎 = 𝟏 𝟒𝟎

1 3 + 1 9 = 𝟑 𝟗 + 𝟏 𝟗 = 𝟒 𝟗 Adding/Subtracting Fractions ? ? Sometimes we don’t need to multiply the denominators. We can find the Lowest Common Multiple of the denominators. ? ? 1 3 + 1 9 = 𝟑 𝟗 + 𝟏 𝟗 = 𝟒 𝟗 The technique (of NOT just blindly multiplying the denominators) will become important for harder algebraic fraction questions.

! Adding/Subtracting Algebraic Fractions The same principle can be applied to algebraic fractions. ! 𝑥 3 + 2𝑥+1 2 = 𝟐𝒙+𝟑 𝟐𝒙+𝟏 𝟔 = 𝟖𝒙+𝟑 𝟔 ? 𝑥+1 2 − 𝑥−1 3 = 𝟑 𝒙+𝟏 −𝟐(𝒙−𝟏) 𝟔 = 𝟖𝒙+𝟑 𝟔 ? Bro Note: Bracketing here is important to avoid sign errors. Note that −2×−1=+2, i.e. positive! ? Bro Tip: Notice that with this one, we didn’t need to times 𝑥 and 𝑥 2 together: 𝑥 2 is a multiple of both denominators.

Further Example 3 𝑥−1 − 4 𝑥 = 3𝑥−4 𝑥−1 𝑥 𝑥−1 = 3𝑥−4𝑥+4 𝑥 𝑥−1 = −𝑥+4 𝑥 𝑥−1 ? ? ? Bro Tip: Be careful with your negatives! Bro Tip: The numerator needn’t be expanded out because it is factorised – you get the marks either way.

Test Your Understanding ? 2 𝑥+1 + 3 𝑥−1 = 5𝑥+1 𝑥+1 𝑥−1 ? ? ? ? 2 𝑥−2 − 𝑥 𝑥+1 = 2− 𝑥 2 +4𝑥 𝑥−2 𝑥+1 “To learn secret way of ninja, add or subtract fractions you must.”

Exercise 2 Write as a single fraction in its simplest form. 1 𝑥 + 1 𝑦 = 𝑦+𝑥 𝑥𝑦 1 𝑥+4 + 2 𝑥−4 = 3𝑥+4 𝑥+4 𝑥−4 𝑥+3 4 + 𝑥−5 3 = 7𝑥−11 12 2 𝑥+4 − 1 𝑥−4 = 𝑥−12 𝑥+4 𝑥−4 2 𝑥−1 − 1 𝑥+1 = 𝑥+3 𝑥−1 𝑥+1 4 2𝑥−1 − 3 2𝑥+1 = 2𝑥+7 2𝑥−1 2𝑥+1 2 𝑥+1 − 𝑥 𝑦 = 2𝑦− 𝑥 2 −1 𝑦 𝑥+1 ? 2 𝑦+3 − 1 𝑦−6 = 𝑦−15 𝑦+3 𝑦−6 2 𝑥 2 −9 + 1 𝑥+3 = 𝑥−1 𝑥+3 𝑥−3 2 2−𝑥 − 4 4−𝑥 = 2𝑥 2−𝑥 4−𝑥 3 𝑥+1 + 4 𝑥+1 2 = 3𝑥+7 𝑥+1 2 1 𝑥−3 − 2 3𝑥−1 = 𝑥+5 𝑥−3 3𝑥−1 1 𝑥 + 1 𝑥+1 + 1 𝑥+2 = 3 𝑥 2 +6𝑥+2 𝑥(𝑥+1)(𝑥+2) 1 𝑥+1 10 + 1 𝑥+1 9 = 𝑥+2 𝑥+1 10 ? 1 8 ? ? 9 2 ? ? 3 10 ? ? 11 4 ? ? 5 12 ? N1 ? 6 ? ? 7 N2

Harder Examples ? 2𝑥+1 𝑥 2 −1 + 1 𝑥−1 = 𝟐𝒙+𝟏 𝒙+𝟏 𝒙−𝟏 + 𝟏 𝒙−𝟏 = 𝟐𝒙+𝟏 𝒙+𝟏 𝒙−𝟏 + 𝒙+𝟏 𝒙+𝟏 𝒙−𝟏 = 𝟑𝒙+𝟐 𝒙+𝟏 𝒙−𝟏 1 𝑥 2 −𝑥 + 1 𝑥𝑦−𝑦 = 𝟏 𝒙 𝒙−𝟏 + 𝟏 𝒚 𝒙−𝟏 = 𝒚 𝒙𝒚 𝒙−𝟏 + 𝒙 𝒙𝒚 𝒙−𝟏 = 𝒙+𝒚 𝒙𝒚 𝒙−𝟏 Bro Hint: Factorise the denominators first where applicable. ?

Test Your Understanding 2 𝑥 2 −4 + 1 𝑥+2 = 𝟐 𝒙+𝟐 𝒙−𝟐 + 𝟏 𝒙+𝟐 = 𝟐 𝒙+𝟐 𝒙−𝟐 + 𝒙−𝟐 𝒙+𝟐 𝒙−𝟐 = 𝒙 𝒙+𝟐 𝒙−𝟐 ? 𝑥 𝑥 2 +5𝑥−6 − 6 𝑥 2 −𝑥 = 𝒙 𝒙+𝟔 𝒙−𝟏 − 𝟔 𝒙 𝒙−𝟏 = 𝒙 𝟐 𝒙 𝒙+𝟔 𝒙−𝟏 − 𝟔 𝒙+𝟔 𝒙 𝒙+𝟔 𝒙−𝟏 = 𝒙 𝟐 −𝟔𝒙−𝟑𝟔 𝒙 𝒙+𝟔 𝒙−𝟏 ?

Multiplying and Dividing ? 1 𝑥 × 𝑥 2 2 = 𝒙 𝟐 𝟐𝒙 = 𝒙 𝟐 1 𝑥 ÷ 𝑥 2 2 = 𝟏 𝒙 × 𝟐 𝒙 𝟐 = 𝟐 𝒙 𝟑 𝑥 2 −𝑥 2𝑥𝑦 × 4 𝑥 2 𝑥−1 = 𝒙 𝒙−𝟏 𝟐𝒙𝒚 × 𝟒 𝒙 𝟐 𝒙−𝟏 = 𝒙−𝟏 𝟐𝒚 × 𝟒 𝒙 𝟐 𝒙−𝟏 = 𝟒 𝒙 𝟐 𝒙−𝟏 𝟐𝒚 𝒙−𝟏 = 𝟐 𝒙 𝟐 𝒚 ? ? AQA Set 1 Paper 1 Q10 Bro Tip: Just factorise first. = 𝟑𝒙−𝟕 𝒙+𝟐 𝟑𝒙+𝟐 𝟑𝒙−𝟐 ÷ 𝒙+𝟐 𝒙 𝟑𝒙+𝟐 = 𝟑𝒙−𝟕 𝒙+𝟐 𝟑𝒙+𝟐 𝟑𝒙−𝟐 × 𝒙 𝟑𝒙+𝟐 𝒙+𝟐 = 𝒙 𝟑𝒙−𝟕 𝟑𝒙−𝟐 ?

Test Your Understanding 𝑥 2 +4𝑥 𝑥 2 −1 ÷ 𝑥+4 𝑥+1 = 𝒙 𝒙+𝟒 𝒙+𝟏 𝒙−𝟏 × 𝒙+𝟏 𝒙+𝟒 = 𝒙 𝒙−𝟏 ? 1 ? 2 𝑥 2 −𝑥−6 16 𝑥 2 −25 ÷ 𝑥−2 4 𝑥 2 −5𝑥 = 𝟐𝒙+𝟑 𝒙−𝟐 𝟒𝒙+𝟓 𝟒𝒙−𝟓 ÷ 𝒙−𝟐 𝒙 𝟒𝒙−𝟓 = 𝟐𝒙+𝟑 𝒙−𝟐 𝟒𝒙+𝟓 𝟒𝒙−𝟓 × 𝒙 𝟒𝒙−𝟓 𝒙−𝟐 = 𝒙 𝟐𝒙+𝟑 𝟒𝒙+𝟓 2

Exercise 3 1 Fully simplify the following: 3 2𝑥+4 + 1 𝑥+2 = 𝟓 𝟐𝒙+𝟒 1 𝑥+1 + 1 𝑥 2 +𝑥 = 𝒙+𝟏 𝒙 𝒙+𝟏 = 𝟏 𝒙 3 𝑥 2 −1 + 2 𝑥+1 = 𝟐𝒙+𝟏 𝒙+𝟏 𝒙−𝟏 [AQA FM June 2013 Paper 2 Q6] Show that 𝑐 2 +5𝑐+4 3𝑐+3 simplifies to 𝑐+4 3 𝒄+𝟒 𝒄+𝟏 𝟑 𝒄+𝟏 = 𝒄+𝟒 𝟑 Hence or otherwise simplify fully 𝑐 2 +5𝑐+4 3𝑐+3 + 3−2𝑐 6 𝒄+𝟒 𝟑 + 𝟑−𝟐𝒄 𝟔 = 𝟐𝒄+𝟖+𝟑−𝟐𝒄 𝟔 = 𝟏𝟏 𝟔 [AQA Set 2 Paper 1 Q9] Simplify fully: 3𝑥 𝑥−3 𝑥+6 − 2 𝑥+6 = 𝟏 𝒙−𝟑 4 Simplify: 𝑥 𝑥+1 × 3 𝑥 2 = 𝟑 𝒙 𝒙+𝟏 𝑥+3 𝑥 ÷ 𝑥 2 −9 𝑥 3 = 𝒙 𝟐 𝒙−𝟑 𝑥 2 +2𝑥+1 𝑥+5 × 𝑥 2 −25 𝑥+1 =(𝒙+𝟏)(𝒙−𝟓) [Jan 2012 Paper 1 Q13] Simplify 𝑥 2 +4𝑥−12 𝑥 2 −25 ÷ 𝑥+6 𝑥 2 −5𝑥 = 𝒙 𝒙−𝟐 𝒙+𝟓 Simplify fully 3 𝑥 2 −5𝑥−2 𝑥 2 −4𝑥 ÷ 3𝑥+1 𝑥 2 −8𝑥+16 = 𝒙−𝟐 𝒙−𝟒 𝒙 Simplify fully: 𝑥 3 −𝑥 𝑥 4 −1 ÷ 𝑥+1 𝑥 3 +𝑥 = 𝒙 𝟐 𝒙+𝟏 ? ? a a ? b ? b ? c ? c 2 5 a ? ? b 6 ? ? 3 N ? ?

Solving Equations with Algebraic Fractions Bro Tip: In general, whenever you have fractions in an equation, your intuition should be “multiply through by the denominator”. When asked to solve an equation with fractions: Combine fractions into single fraction then multiply through by denominator. But if multiplying everything by 𝑥 2 turns equation into a quadratic, this is simpler. 𝑥 2𝑥−3 + 4 𝑥+1 =1 𝒙 𝟐 +𝟗𝒙−𝟏𝟐 𝟐𝒙−𝟑 𝒙+𝟏 =𝟏 𝒙 𝟐 +𝟗𝒙−𝟏𝟐= 𝟐𝒙−𝟑 𝒙+𝟏 𝒙 𝟐 +𝟗𝒙−𝟏𝟐=𝟐 𝒙 𝟐 −𝒙−𝟑 𝒙 𝟐 −𝟏𝟎𝒙+𝟗=𝟎 𝒙−𝟏 𝒙−𝟗 =𝟎 𝒙=𝟏 𝒐𝒓 𝒙=𝟗 8 𝑥 2 + 4 𝑥 =4 8+4𝑥=4 𝑥 2 2+𝑥= 𝑥 2 𝑥 2 −𝑥−2=0 𝑥+1 𝑥−2 =0 𝒙=−𝟏 𝒐𝒓 𝒙=𝟐 ? ?

Test Your Understanding 𝑥 3 + 10 𝑥−1 =4 𝑥 2 −13𝑥+42=0 𝑥−6 𝑥−7 =0 𝑥=6 𝑜𝑟 𝑥=7 Solve, giving your answer to 3sf. 1 𝑥 +5= 2 𝑥 2 𝒙+𝟓 𝒙 𝟐 =𝟐 𝟓 𝒙 𝟐 +𝒙−𝟐 𝒙=−𝟎.𝟕𝟒𝟎 𝒐𝒓 𝒙=𝟎.𝟓𝟒𝟎 ? ? 4𝑥−1 5 + 𝑥+4 2 =3 𝒙= 𝟏𝟐 𝟏𝟑 ? 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎

Exercise 4 Give exact answers unless otherwise specified. Find exact solutions to: 𝑥+ 3 𝑥 =7 𝒙= 𝟕± 𝟑𝟕 𝟐 𝑥 3 − 4 𝑥−5 =2 𝒙=𝟐, 𝟗 𝑥−2 5 − 6 𝑥 =1 𝒙=−𝟑, 𝟏𝟎 1 𝑥 +3= 2 𝑥 2 𝒙=−𝟏, 𝟐 𝟑 ℎ+7 3 + 2ℎ−1 2 = 5 6 𝒉=−𝟎.𝟕𝟓 𝑥 2 − 2 𝑥+1 =1 𝒙=𝟑, −𝟐 Give your answer to 3sf: 2 𝑦 2 + 9 𝑦 −7=0 𝒚=−𝟎.𝟏𝟗𝟑 𝒐𝒓 𝟏.𝟒𝟖 5 2𝑥+1 2 4𝑥+5 =5𝑥−1 𝒙=𝟏𝟎 ? ? 1 7 ? ? 2 8 ? ? 9 3 10 [AQA FM Jan 2013 Paper 2 Q13] a) Show that 4 𝑥 + 2 𝑥−1 simplifies to 6𝑥−4 𝑥 𝑥−1 = 𝟒 𝒙−𝟏 +𝟐𝒙 𝒙 𝒙−𝟏 = 𝟔𝒙−𝟒 𝒙 𝒙−𝟏 b) Hence or otherwise, solve 4 𝑥 + 2 𝑥−1 =3, giving your solutions correct to 3sf. 𝟔𝒙−𝟒 𝒙 𝒙−𝟏 =𝟑 𝟔𝒙−𝟒=𝟑 𝒙 𝟐 −𝟑𝒙 𝟑 𝒙 𝟐 −𝟗𝒙+𝟒=𝟎 𝒙=𝟎.𝟓𝟒𝟑 𝒐𝒓 𝒙=𝟐.𝟒𝟔 ? 4 ? ? 5 ? [AQA FM June 2013 Paper 1 Q17] Solve 4 𝑥−2 + 1 𝑥+3 =5 𝒙=± 𝟖 6 ?