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P1 Chapter 3 :: Equations and Inequalities [email protected] www.drfrostmaths.com @DrFrostMaths Last modified: 26th August 2017 Use of DrFrostMaths for practice Register for free at: www.drfrostmaths.com/homework Practise questions by chapter, including past paper Edexcel questions and extension questions (e.g. MAT). Teachers: you can create student accounts (or students can register themselves). Chapter Overview There is little new content in this chapter since GCSE. 1:: Simultaneous Equations 2:: Simultaneous Equations using Graphs Solve: Find the points of intersection of π¦ = 3π₯ 2 β 2π₯ + 4 and 7π₯ + π¦ + 3 = 0 π₯ + π¦ = 11 π₯π¦ = 30 NEW! (since GCSE) You may have to use the discriminant to show that the two graphs have no points of intersection. 3:: Solving Inequalities 4:: Sketching Inequalities Find the set of values of π₯ for which: π₯ 2 β 11π₯ + 24 < 0 Sketch the region that satisfies the inequalities: 2π¦ + π₯ < 14 π¦ β₯ π₯ 2 β 3π₯ β 4 NEW! (since GCSE, and new to A Level 2017+) Use set notation to represent solutions to inequalities. Solutions sets The solution(s) to an equation may be: A single value: 2π₯ + 1 = 5 Multiple values: π₯ 2 + 3π₯ + 2 = 0 An infinitely large set of values: π₯>3 No (real) values! π₯ 2 = β1 Every value! π₯2 + π₯ = π₯ π₯ β 1 The point is that you shouldnβt think of the solution to an equation/inequality as an βanswerβ, but a set of values, which might just be a set of 1 value (known as a singleton set), a set of no values (i.e. the empty set β ), or an infinite set (in the last example above, this was β) ! The solutions to an equation are known as the solution set. Solutions sets For simultaneous equations, the same is true, except each βsolutionβ in the solution set is an assignment to multiple variables. All equations have to be satisfied at the same time, i.e. βsimultaneouslyβ. Scenario A single solution: Two solutions: Example π₯+π¦ =9 π₯βπ¦ =1 π₯2 π¦2 + = 10 π₯+π¦ =4 No solutions: π₯+π¦ =1 π₯+π¦ =3 Infinitely large set of solutions: π₯+π¦ =1 2π₯ + 2π¦ = 2 Solution Set Solution 1: π = π, π = π To be precise here, the solution set is of size 1, but this solution is an assignment to multiple variables, i.e. a pair of values. Solution 1: π = π, π = π Solution 2: π = π, π = π This time we have two solutions, each an π₯, π¦ pair. The solution set is empty, i.e. β , as both equation canβt be satisfied at the same time. Solution 1: π = π, π = π Solution 2: π = π, π = π Solution 3: π = π, π = βπ Solution 4: π = π. π, π = π. π β¦ Infinite possibilities! Textbook Error Pg39: βLinear simultaneous equations in two unknowns have one set of values that will make a pair of equations true at the same time.β There are two separate errors in this statement β Iβll let you work out what! 1 :: Simultaneous Equations Recap! Solve the simultaneous equations: Solve the simultaneous equations: 3π₯ + π¦ = 8 2π₯ β 3π¦ = 9 π₯ + 2π¦ = 3 π₯ 2 + 3π₯π¦ = 10 We can either use substitution (i.e. making π₯ or π¦ the subject of one equation, and substituting it into the other) or elimination, but the latter is easier for linear equations. ππ + ππ = ππ ππ β ππ = π Adding the two equations to βeliminateβ π: πππ = ππ β π = π Substituting into first equation: ππ + ππ = ππ β π = βπ We canβt use elimination this time as nothing would cancel. We instead: (1) Rearrange linear equation to make π₯ or π¦ the subject. (2) Substitute into quadratic equation and solve. π = π β ππ Substitute into other equation: π β ππ π + ππ π β ππ = ππ β¦ πππ + ππ + π = π ππ + π π + π = π π π=β β π=π π π = βπ β π = π Test Your Understanding Solve the simultaneous equations: 3π₯ 2 + π¦ 2 = 21 π¦ =π₯+1 3π₯ 2 + π₯ + 1 2 = 21 3π₯ 2 + π₯ 2 + 2π₯ + 1 = 21 4π₯ 2 + 2π₯ β 20 = 0 2π₯ 2 + π₯ β 10 = 0 2π₯ + 5 π₯ β 2 = 0 5 π₯ = β ππ π₯ = 2 2 3 π¦ = β ππ π¦ = 3 2 Exercise 3A/B Pearson Pure Mathematics Year 1/AS Pages 40, 41 Extension 2 [STEP 2010 Q1] Given that 5π₯ 2 + 2π¦ 2 β 6π₯π¦ + 4π₯ β 4π¦ 1 [MAT 2012 1G] There are β‘ π π₯ β π¦ + 2 2 + π ππ₯ + π¦ 2 + π positive real numbers π₯ and a) Find the values of π, π, π, π. π¦ which solve the equations b) Solve the simultaneous equations: 2π₯ + ππ¦ = 4, 5π₯ 2 + 2π¦ 2 β 6π₯π¦ + 4π₯ β 4π¦ = 9, π₯+π¦ =π 6π₯ 2 + 3π¦ 2 β 8π₯π¦ + 8π₯ β 8π¦ = 14 for: (Hint: Can we use the same method in (a) to rewrite the second equation?) A) All values of π; a) Expanding RHS: B) No values of π; π + πππ ππ + π + π ππ + βππ + πππ ππ + πππ β πππ C) π = 2 only; + (ππ + π ) D) Only π > β2 Comparing coefficients: π = π, π = π, π = βπ, π = βπ If π = π then ππ + ππ = π and π + π = π which are equivalent. This would give an infinite solution set, thus the answer is C. b) π β π + π π + βππ + π π β π = π Using method in (a): π π β π + π π + βππ + π π β π = ππ Subtracting yields π β ππ = ±π and π β π + π = ±π We have to consider each of 4 possibilities. Final solution set: π = βπ, π = βπ ππ π = π, π = π or π = π, π = π ππ π = π, π = ππ Simultaneous Equations and Graphs Recall that a line with a given equation is the set of all points that satisfy the equation. i.e. It is a graphical representation of the solution set where each (π₯, π¦) point represents each of the solutions π₯ and π¦ to the equation. e.g. π₯ = 3, π¦ = 7 Now suppose we introduced a second simultaneous equation: π¦ = 2π₯ + 1 π₯+π¦ =5 The second line again consists of all points (π₯, π¦) which satisfy the equation. So what point must satisfy both equations simultaneously? The point of intersection! Example a) On the same axes, draw the graphs of: 2π₯ + π¦ = 3 π¦ = π₯ 2 β 3π₯ + 1 b) Use your graph to write down the solutions to the simultaneous equations. π = βπ, π = π ππ π = π, π = βπ (We could always substitute into the original equations to check they work) c) What algebraic method (perhaps thinking about the previous chapter), could we have used to show the graphs would have intersected twice? Substituting linear equation into quadratic: π = π β ππ β΄ π β ππ = ππ β ππ + π ππ β π β π = π Since there were two points of intersection, the equation must have two distinct solutions. Thus ππ β πππ > π π = π, π = βπ, π = βπ π+π=π>π Thus the quadratic has two distinct solutions, i.e. we have two points of intersection. Another Example a) On the same axes, draw the graphs of: π¦ = 2π₯ β 2 π¦ = π₯ 2 + 4π₯ + 1 b) Prove algebraically that the lines never meet. When we try to solve simultaneously by substitution, the equation must have no solutions. ππ + ππ + π = ππ β π ππ + ππ + π = π π = π, π = π, π = π ππ β πππ = π β ππ = βπ βπ < π therefore no solutions, and therefore no points of intersection. Final Example The line with equation π¦ = 2π₯ + 1 meets the curve with equation ππ₯ 2 + 2π¦ + π β 2 = 0 at exactly one point. Given that π is a positive constant: a) Find the value of π. b) For this value of π, find the coordinates of this point of intersection. a Substituting: ππ₯ 2 + 2 2π₯ + 1 + π β 2 = 0 ππ₯ 2 + 4π₯ + 2 + π β 2 = 0 ππ₯ 2 + 4π₯ + π = 0 Since one point of intersection, equation has one solution, so π2 β 4ππ = 0. π = π, π = 4, π = π 16 β 4π 2 = 0 π = ±2 But π is positive so π = 2. b When π = 2, 2π₯ 2 We can breathe a sigh of relief as we were expecting one solution only. + 4π₯ + 2 = 0 π₯ 2 + 2π₯ + 1 = 0 π₯ + 1 2 = 0 β π₯ = β1 π¦ = 2 β1 + 1 = β1 β (β1, β1) Exercise 3C Pearson Pure Mathematics Year 1/AS Page 45 Set Builder Notation Recall that a set is a collection of values such that: a) The order of values does not matter. b) There are no duplicates. Recap from GCSE: Froflection: Sets seem sensible for listing solutions to an equation, as the order doesnβt matter. β’ We use curly braces to list the values in a set, e.g. π΄ = 1,4,6,7 β’ If π΄ and π΅ are sets then π΄ β© π΅ is the intersection of π΄ and π΅, giving a set which has the elements in π΄ and π΅. β’ π΄ βͺ π΅ is the union of π΄ and π΅, giving a set which has the elements in π΄ or in π΅. β’ β is the empty set, i.e. the set with nothing in it. β’ Sets can also be infinitely large. β is the set of natural numbers (all positive integers), β€ is the set of all integers (including negative numbers and 0) and β is the set of all real numbers (including all possible decimals). β’ We write π₯ β π΄ to mean βπ₯ is a member of the set Aβ. So π₯ β β would mean βπ₯ is a real numberβ. 1,2,3 β© 3,4,5 = π 1,2,3 βͺ 3,4,5 = π, π, π, π, π 1,2 β© 3,4 =β Set Builder Notation It is possible to construct sets without having to explicitly list its values. We use: The | or : means βsuch thatβ. or ππ₯ππ ππππππ‘πππ } {ππ₯ππ βΆ ππππππ‘πππ } Can you guess what sets the following give? (In words βAll numbers 2π₯ such that π₯ is an integer) 2π₯ βΆ π₯ β β€ = {0,2, β2,4, β4,6, β6, β¦ } i.e. The set of all even numbers! 2π₯ βΆ π₯ β β = {2,4,8,16,32, β¦ } π₯π¦: π₯, π¦ πππ πππππ = {4,6,10,14,15, β¦ } i.e. All possible products of two primes. We previously talked about βsolutions setsβ, so set builder notation is very useful for specifying the set of solutions! Set Builder Notation Can you use set builder notation to specify the following sets? All odd numbers. {2π₯ + 1 βΆ π₯ β β€} All (real) numbers greater than 5. {π₯: π₯ > 5} All (real) numbers less than 5 or greater than 7. All (real) numbers between 5 and 7 inclusive. Technically it should be {π₯: π₯ > 5, π₯ β β} but the π₯ > 5 by default implies real numbers greater than 5. π₯: π₯ < 5 βͺ {π₯: π₯ > 7} We combine the two sets together. π₯: 5 β€ π₯ β€ 7 While we could technically write π₯: π₯ β₯ 5 β© {π₯: π₯ β€ 7}, we tend to write multiple required conditions within the same set. Recap of linear inequalities Inequality Solution Set 2π₯ + 1 > 5 {π₯ βΆ π₯ > 2} 3 π₯ β 5 β₯ 5 β 2(π₯ β 8) {π₯ βΆ π₯ β₯ 7.2} {π₯ βΆ π₯ β€ β2} βπ₯ β₯ 2 Fro Note: Multiplying or both sides of an inequality by a negative number reverses the direction. Combining Inequalities: If π₯ < 3 and 2 β€ π₯ < 4, what is the combined solution set? 2 3 4 2β€π₯<3 If both inequalities have to be satisfied, we have to be on both lines. Place your finger vertically and scan across. RECAP :: Solving Quadratic Inequalities Solve π₯ 2 + 2π₯ β 15 > 0 Step 1: Get 0 on one side (already done!) π₯+5 π₯β3 >0 Step 2: Factorise Step 3: Sketch and reason π¦ π¦ = (π₯ + 5)(π₯ β 3) β5 3 π₯ Since we sketched π¦ = (π₯ + 5)(π₯ β 3) weβre interested where π¦ > 0, i.e. the parts of the line where the π¦ value is positive. Click to Fro-Bolden > What can you say about the π₯ values of points in this region? π < βπ What can you say about the π₯ values of points in this region? π>π π₯: π₯ < β5 βͺ {π₯: π₯ > 3} Fro Note: If the π¦ value is βstrictlyβ greater than 0, i.e. > 0, then the π₯ value is strictly less than -5. So the < vs β€ must match the original question. Solving Quadratic Inequalities Solve π₯ 2 + 2π₯ β 15 β€ 0 Step 1: Get 0 on one side (already done!) π₯+5 π₯β3 β€0 Step 2: Factorise Step 3: Sketch and reason π¦ π¦ = (π₯ + 5)(π₯ β 3) β5 3 Again, what can we say about the π₯ value of any point in this region? π₯ {π₯ βΆ β5 β€ π₯ β€ 3} Bro Note: As discussed previously, we need β€ rather than < to be consistent with the original inequality. Further Examples Solve ππ + ππ β₯ βπ π₯ 2 + 5π₯ + 4 β₯ 0 π₯+4 π₯+1 β₯0 π¦ = (π₯ + 4)(π₯ + 1) β4 Solve ππ < π π₯2 β 9 < 0 π₯+3 π₯β3 <0 π¦ β1 π¦ π₯ π₯ β€ β4 or π₯ β₯ β1 Fro Note: The most common error Iβve seen students make with quadratic inequalities is to skip the βsketch stepβ. Sodβs Law states that even though you have a 50% chance of getting it right without a sketch (presuming youβve factorised correctly), you will get it wrong. β3 π¦ = (π₯ + 3)(π₯ β 3) 3 π₯ β3 < π₯ < 3 βUse of Technologyβ Monkey says: When Iβm not busy flinging poo at other monkeys, I use the quadratic inequality solver on my ClassWiz. Just go to Menu β Inequalities, then choose βorder 2β (i.e. quadratic) Test Your Understanding Edexcel C1 May 2010 Q3 Edexcel C1 June 2008 Q8 Fro Note: What often confuses students is that the original equation has no solutions, but the inequality π 2 + 8π < 0 did have solutions. But think carefully what weβve done: Weβve found the solutions for π that result in the original equation not having any solutions for π₯. These are different variables, so have different solutions sets, even if the solution set of π influences the solution set of π₯. Deal with inequalities with a division by π₯ Spec Note: This is an example in the textbook, although it is ambiguous whether this type of question is in the new specification. Dealing with an π₯ in the denominator within an inequality is a skill previously in the old Further Pure 2 module. But you never really know! 6 π₯ Find the set of values for which > 2, π₯ β 0 Why canβt we just multiply both sides by π? We earlier saw that multiplying by a negative number would flip the inequality, but multiplying by a positive number would not. Since we donβt know π₯, we donβt know whether the inequality would flip or not! 6 Once solution is to sketch π¦ = and π¦ = 2, find the points of π₯ intersection and reason about the graph (see next section, βInequalities on Graphsβ), but an easier way is to multiply both sides by ππ , because it is guaranteed to be positive: π¦ ππ > πππ πππ β ππ < π ππ β ππ < π π πβπ <π π<π<π π¦ = π₯(π₯ β 3) 0 3 π₯ Exercise 3D/3E Pearson Pure Mathematics Year 1/AS Page 47-48, 50-51 Inequalities on Graphs (New to the 2017 spec) π¦ π¦ = (π₯ + 5)(π₯ β 3) β5 π₯ 3 When we solved quadratic inequalities, e.g. π₯ + 5 π₯ β 3 > 0 We plotted π¦ = (π₯ + 5)(π₯ β 3) and observed the values of π₯ for which π¦ > 0. Can we use a similar method when we donβt have 0 on one side? Example: πΏ1 has equation π¦ = 12 + 4π₯. πΏ2 has equation π¦ = π₯ 2 . The diagram shows a sketch of πΏ1 and πΏ2 on the same axes. a) Find the coordinates of π1 and π2 , the points of intersection. b) Hence write down the solution to the inequality 12 + 4π₯ > π₯ 2 . π¦ a Solve simultaneously to find points of intersection: π2 π₯ 2 = 12 + 4π₯ π₯ 2 β 4π₯ β 12 = 0 π₯ = 6, π₯ = β2 β π¦ = 36, π¦ = 4 π1 6,36 , π2 β2,4 π1 β2 6 π₯ b When 12 + 4π₯ > π₯ 2 the πΏ1 graph is above the πΏ2 graph. This happens when β2 < π₯ < 6. Inequality Regions On graph paper, shade the region that satisfies the inequalities: 2π¦ + π₯ < 14 π¦ β₯ π₯ 2 β 3π₯ β 4 You did this at GCSE, the only difference here being that the graphs involved might not be straight lines. Step 2: An inequality involving π₯ and π¦ represents a 2D region in space. Identify the correct side of each line each inequality represents. π¦ β1 4 Fro Tip: To quickly sketch 2π¦ + π₯ = 14, consider what happens when π₯ is 0 and when π¦ is 0. π₯ Click to Frosketch > Fro Tip: Make sure π¦ is on the side where it is positive. If π¦ is on the smaller side, youβre below the line. If π¦ is on the greater side, youβre above the line. Exercise 3F/3G Pearson Pure Mathematics Year 1/AS Page 53, 55