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# Chapter 4

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1. Chapter 4 Linear and nonlinear relationships

2. Previously… • We had been looking at single linear graphs: • Sketching a linear graph using various methods • Finding the equation of a linear graph • Finding the distance between two points • Finding the midpoint between two points • Now we are looking at what happens when there are two linear graphs – known as simultaneous equations because they occur together at the same time • What we are ultimately trying to find is the point at which these lines cross or intersect – this is known as the simultaneous solution • We can find this solution graphically (by sketching the two graphs and reading the point of intersection), or algebraically by using substitution. • Note: if the gradient of the two lines is the same (same ‘m’ value, or the number in front of the x in y=mx+c), then the lines are parallel and have no solution as they never intersect In this case, (-2, -4) is the solution of these simultaneous equations

3. Ex 4A – graphical solutions of simultaneous equations • To solve simultaneous equations graphically, it means reading the point of intersection off the graph • For this, it is important that the graph is accurate (use graph paper and plot the points carefully) – a rough sketch won’t do • Sometimes the question will give you the graph and you just have to read the point of intersection. Other times you will need to draw the graph yourself to read the point of intersection.

4. Question 1(a) • Use thegraph to writethepoint of intersection (i.e. thesolution to thesimultaneousequations) • Theequationsforthesetwolines are: • x + y = 3 • x − y = 1 • Looking at thegraph, we can seethepoint of intersectionishere… • Thiscoordinateis (2, 1), whichisouranswer! • You can CHECK thisbysubstituting x=2 and y=1 intotheequationsgiven at thestart • x + y = 3 2 + 1 = 3  thisgivesusthecorrectanswer so weknowitisright! • x – y = 1 2 – 1 = 1  thisisalsocorrect, so we can be confidentwehavereadthegraphscorrectly

5. Question 3(a) • Solve each of the following simultaneous equations using a graphical method • This means you have to sketch the two graphs first (using whatever method you like from exercise 3A – I recommend the x and y intercept method) • You should use graph paper for this as your graphs need to be super accurate in order to read the point of intersection correctly • For equation: x + y = 5… • To find x-intercept, let y=0 x + 0 = 5 x = 5 Plot point (5, 0) • To find y-intercept, let x=0 0 + y = 5 y = 5 Plot point (0, 5) and draw the line connecting the points • For equation: 2x + y = 8… • x-int: let y=0 2x + 0 = 8 2x = 8 x = 4 Plot point (4, 0) • y-int: let x=0 2(0) + y = 8 y = 8 Plot point (0, 8) and draw the line connecting the points • Now that both lines are sketched, write the point at which they intersect… • Which is (3, 2)

6. Question 4 • Question 4: Solve each of the following pairs of simultaneous equations • This question uses the same steps as question 3, it’s just that the equations are a bit more complicated • Use whatever method you like to sketch the two graphs, then find the point of intersection • Use previous question as a guide • However, there is another way…

7. Alternative method for question 4 • You can use substitution rather than drawing the graphs for question 4 • Let’s look at 4(a) • y = 8 − xy = x + 2 • Both of theseequations are equal to y, thereforewe can saythat: y = y 8 − x = x + 2 Youcannot use thismethodifthegraphsdon’tequalthesamething • Then we can use algebra to find what x is… 8 − x = x + 2 (we need the x’s to be on the same side of the = sign) 8 = x + x +2 (nowsimplify) 8 = 2x + 2 (nowwewantthe 2 to go to theotherside) 8 – 2 = 2x (simplify) 6 = 2x (nowget x onitsown) 6 ÷ 2 = x 3 = x • Then, put x = 3 back intoone of theequationsfromthestart to getthe y value y = x + 2 y = 3 + 2 y = 5 • Thereforeoursolutionis(3, 5)

8. Question 7 (reasoning) • Jet skiquestion • Here are somehints to helpyou: • Whenitsays ‘writethe rules relating to thecost and length of rental’, itmeansyouneed to writetwoequations in theformy = mx + c • Isthe ‘y’ the time, orthecost? (whatvaluegoes up the y-axis?) • Isthe ‘x’ the time, orthecost? (whatvaluegoesalongthe x-axis?) • Whatisthe ‘c’ part of theequationbasedontheinformationgiven? (whatistheinitialpriceyoupayevenifyouhaven’thadany time onthe jet skiyet?) • Sketch thetwographs and use theevidencefromthesegraphs to answerquestions c and d

9. Ex 4E – linear inequations See www.mrssloman.weebly.com for video tutorial and worked examples from this exercise • An ‘inequation’ just means a normal equation (e.g. 2x + 3 = 10), but instead of an equals sign (=), it uses an inequality sign (either <, >, ≤, or ≥) • < is ‘less than’ • > is ‘greater than’ • ≤ is ‘less than or equal to’ • ≥ is ‘greater than or equal to’ • For the most part, you can treat these inequality signs just like a normal equals sign when you are working out what x is by undoing the things that have been done to it • The only exception is if you are multiplying or dividing by a negative number – in this case, you flip the direction of the inequality sign (< becomes >, and ≤ becomes ≥ and vice versa) Refer to your notes from exercise 2D (solving linear equations) for more examples

10. Ex 4F – sketching linear inequations • Linear inequations replace the equals sign in a typical equation (y=mx+c) with an inequality sign(>, ≥, < and ≤) • The graph of linear inequations is a half plane (meaning it shades in half of the Cartesian plane) • The line that divides the plane is the line of the inequation • Consider the linear inequationy ≥ x + 2. There are many points (x, y) that satisfy this inequation– i.e., anything in the shaded area • What if it was y < x + 2? The solutions would be everything below the line (as it uses the ‘less than’ sign ‘<’) • Use a solid line if it is ≥ or ≤ to show that the numbers on the line are included in the solutions • Use a dotted line if it is < or > to show that the numbers on the line aren’t included in the solutions

11. y ≥ x + 2 • y = x + 2 • Sketch this linear graph using the x and y intercept method • To find the x-intercept: let y=0 0 = x + 2 x = 0 – 2 x = -2 x –intercept is (-2, 0) • To find the y-intercept: let x=0 or read it off the equation (y = mx + c where c is the y-intercept) y-intercept is (0, 2) • As it uses a ‘greater than or equal to’ sign, connect the points with a solid line. • Shade the required region – in this case, what is above the line Steps • Write the question • Write the linear equation form (i.e. swap the inequality sign for an = sign) • Sketch this linear graph using the x and y intercept method • To find the x-intercept: let y=0 • To find the y-intercept: let x=0 OR read it off the equation (y = mx + c where c is the y-intercept) • Look at the original question – what sign does it use? If it is < or >, connect these points using a dotted line. If it is ≥ or ≤ connect these points using a solid line. • Shade the required region – either above the line (if > or ≥) or below the line (if < or ≤) Done!

12. Let’s do Ex 4F question 5 • Happy Yaps Dog Kennels charges \$35 per day for large dogs (dogs over 20 kg) and \$20 per day for small dogs (less than 20 kg). On any day, Happy Yaps Kennels can only accommodate a maximum of 30 dogs. • Iflrepresents the number of large dogs and s represents the number of small dogs. Write down an inequation, in terms of l and s, that represents the total number of dogs at Happy Yaps. Note: don’t be fooled! This question is just looking at the number of dogs, nothing about the cost! Total number of dogs (regardless of whether they are small or large) must be less than or equal to 30 at any time (so we know which sign to use) Answer: l + s ≤ 30 • Another inequation can be written as s ≥ 12. In the context of this problem, write down what this inequation represents. What was s? Small dogs So this inequation means that the number of small dogs must be 12 or more (or, at least 12)

13. The inequationl ≤ 15 represents the number of large dogs that Happy Yaps can accommodate on any day. This inequation is shown as a bold line on the graph, clearly shade in the area that is not within the region for this inequation. • Explore the maximum number of small and large dogs Happy Yaps Kennels can accommodate to receive the maximum amount in fees. Think back: what do we know? • The fee is \$35 a day for large dogs • The fee is \$20 per day for small dogs • On any day, they cannot have more than 30 dogs total • They cannot accept more than 15 large dogs • They must have at least 12 small dogs So, to get the highest fees, they want to maximise the number of large dogs as they cost more. So, 15 large dogs. That leaves room for 15 small dogs to be at capacity and to get the largest income.