Equations and Inequalities Cheat Sheet PDF - Pure Year 1

Equations and Inequalities Cheat Sheet Pure Year 1...

Equations and Inequalities Cheat Sheet Pure Year 1 Quadratic Inequalities This chapter covers both linear and quadratic simultaneous equations and how Equation 2 can be rewritten as: Similar to the method we used to solve quadratic equations, we can also solve to solve them algebraically. You should also be able to interpret solutions of a 𝑦 2) 𝑥 = −1 − quadratic inequalities. given equation graphically. It also cover both linear and quadratic inequalities. 2 Let us look at the inequality 2𝑥 ! − 3𝑥 − 2 > 0. To solve this, we need to first find the We can substitute this rewritten equation into equation 1). critical (similar to roots of a function) values of this inequality by solving the quadratic Linear Simultaneous Equations 𝑦 1) 𝑦 ! + 2.−1 − / = 10 equation on the left-hand side. Linear simultaneous equations have two same unknowns in their respective 2 𝑦 ! − 2 − 𝑦 = 10 2𝑥 ! − 3𝑥 − 2 > 0 equation and has one set of values between them which makes both the (2𝑥 + 1)(𝑥 − 2) > 0 equations valid. 𝑦 ! − 𝑦 − 12 = 0 $ (𝑦 + 3)(𝑦 − 4) = 0 Hence one critical point is at 𝑥 = − and the other one at 𝑥 = 2. Now we can plot the ! In the two equations 1 and 2, 𝑥 and 𝑦 are 𝑦 = −3 and 𝑦 = 4 graph of 2𝑥 ! − 3𝑥 − 2. two unknowns. For the equations to be Substitute these two 𝑦 values into equation 2) to obtain the 𝑥 values. valid, the 𝑥 value in equation 1 has to be the "# $ % The set of values which corresponds to the inequality 2𝑥 ! − 𝑥 = −1 − ! = ! and 𝑥 = −1 − ! = −3 same 𝑥 value in equation 2; the same would 3𝑥 − 2 > 0, are the 𝑥 values of the plot which are above the 𝑥 apply for the unknown 𝑦. Simultaneous Equations on graphs axis. Hence, the solution to this inequality would be 𝑥 > 2 and 1 2 × 6 + (−6) = 6 The only set of values for which this would The solutions of a set of simultaneous equation can be represented on a graph. Simultaneous $ equations share the same set of values for the unknowns, hence if two given simultaneous 𝑥 < − !. be true is if the unknowns have the values 2 6 × 6 + 2 × −6 = 24 of 𝑥 = 6 & 𝑦 = −6 equations were illustrated on a graph then at some point on their respective plot, they would share the same coordinate and hence intersect. Hence, the intersection point on a graph of However, if the inequality were to be 2𝑥 ! − 3𝑥 − 2 < 0, then two lines would be the solution or at least one of the solutions for the curves’ or lines’ the set of values which would correspond to it, would be the 𝑥 In order to solve these simultaneous equations and find the set of values which values that are below the 𝑥 axis. Hence, the solution to this simultaneous equation. make the given equations valid, we can use either the method of elimination or $ inequality would be − ! < 𝑥 < 2 substitution. The graph on the left shows the plot of 𝑦 = 𝑥 ! and 𝑦 − 𝑥 − 2 = 0. Example 1: Solve the following simultaneous equation. If we take the point (2,4), we can check if this Inequalities on graphs and Regions set of values are valid for both the equations. You may come across question where you are asked to find the solutions to the 1) 2𝑥 + 𝑦 = 6 𝑦 = 𝑥 ! (4) = (2)! " # inequality by interpreting the functions graphically. 2) 6𝑥 + 2𝑦 = 24 𝑦−𝑥−2=0 (4) − (2) − 2 = 0 "# Example 4: The graph shows the plot of 𝑦 = 𝑥 ! and 𝑦 = 2𝑥 + 3. Determine the We can also check if the point (−1,1) agrees solutions to the inequality 2𝑥 + 3 > 𝑥 !. We can use the method of elimination to solve this. To eliminate one unknown, with the two equations. we need the coefficient of a single unknown to be the same for both the equations. First you will need to find the points of intersection, which can 𝑦 = 𝑥 ! (1) = (−1)! " # be done by equating the two equations and solving it. In order to achieve this, we can multiply the first equation by 2. This would give 𝑦−𝑥−2=0 (1) − (−1) − 2 = 0 " # use a third equation which is equivalent to equation 1: Hence from this we can see that we can use 𝑥 ! = 2𝑥 + 3 ! 𝑥 − 2𝑥 − 3 = 0 graphs to solve simultaneous equations by (𝑥 − 3)(𝑥 + 1) = 0 3) 4𝑥 + 2𝑦 = 12 plotting and observing any intersection points. 𝑥 = 3 𝑎𝑛𝑑 𝑥 = −1 Now we can eliminate the unknown 𝑦 by taking away equation 3) from 2). Note that when solving simultaneous equations that come in the form of a quadratic Hence the two points of intersection are (3,9) and (−1,1). The equation 𝑎𝑥 ! + 𝑏𝑥 + 𝑐 = 0, the discriminant of the equation after substituting can set of values that validates to the inequality 2𝑥 + 3 > 𝑥 ! is the 2 6𝑥 + 2𝑦 = 24 be used to determine the number of solutions that the simultaneous equations have. line 𝑦 = 2𝑥 + 3 is above the curve 𝑦 = 𝑥 !. Hence, the set of 3 4𝑥 + 2𝑦 = 12 Hence, on a graph it can also indicate the number of intersection points. values lie between the intersection points −1 < 𝑥 < 3 2𝑥 + 0 = 12 Linear Inequalities Regions on graphs can be shaded to identify the areas that satisfies given linear or 2𝑥 = 12 Similar to the methods we have learnt to solve linear equations, we can also solve linear quadratic inequalities. 𝑥=6 inequality problems using the same approach. When you solve an inequality, you find the set Example 5: Shade the regions which satisfy the inequalities. 1) 2 × 6 + 𝑦 = 6 of all real numbers that make the inequality valid. 𝑥 ! − 8𝑥 + 15 ≤ 𝑦 𝑦 = −6 Example 3: Find the set of values for which 𝑦−𝑥 𝑓(𝑥), then this would represent the region substitution. As a quadratic equation is involved there can be up to two sets of 𝑥 > −13 𝑥 ≤ −7 above the curve or line. values/solutions for the simultaneous equation. If 𝑦 < 𝑓(𝑥), then this would represent the region below the curve or line. Example 2: Solve the simultaneous equation. Therefore, for the inequality 𝑦 − 𝑥 < 3, the region satisfied is We can plot this inequality on a number line to find the set of values which agree with both represented by the area below the green dotted line and for 1) 𝑦 ! + 2𝑥 = 10 the inequalities. The area which overlaps on the number line is between -13 and -7. Hence the inequality 𝑥 ! − 8𝑥 + 15 ≤ 𝑦, the area above the red curve. 2) 2𝑥 + 𝑦 + 2 = 0 the solution for this inequality would be −13 < 𝑥 ≤ −7. Hence the region satisfied for both the inequalities is illustrated by the grey shaded area. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc