1. Simultaneous Equations Solve simultaneous equations using elimination In this section you will learn how to Solve simultaneous equations using substitution Solve simultaneous linear equations using the TI83 Graphics Calculator Select

2. If x + y = 7, what are x and y? xy 16 25 34 43 52 61 Etc….  If y = x – 1, what are x and y? xy 0 10 21 32 43 54 Key question…. Is there a pair of numbers that satisfies BOTH x + y = 7 AND y = x – 1 ?

3. If we drew these two lines on the ONE GRAPH, we would get…. x + y = 7 y = x – 1 WHAT DO YOU NOTICE ?????????  (4, 3)

4. The solution to the two simultaneous equations is the point where the lines intersect!!

5. Simultaneous equations This is where you have two equations containing x and y, and you have to find a pair of values (x, y) that fits BOTH Example Look at the two equations x – y = 1 x + y = 5 Can you guess a value of x and a value of y that fit BOTH of these at the same time (i.e. simultaneously)? x = 3, y = 2 or you can write (3, 2) We worked this out by guessing the answers. (fairly easy in this case!). But we need a reliable process that will work for all types – even much harder ones!

6. Let’s now draw graphs of these two lines: x + y = 5 x – y = 1 WHERE DO THEY MEET? (3, 2) The solution to the simultaneous equations is the point where the two lines intersect!

7. Coming up..... You’ll meet three methods of solving a pair of simultaneous equations: 1. Graphics calculator (TI-83) 2. Algebra – substitution 3. Algebra – elimination

8. Method 1 – Graphics calculator Example Use the GC to solve the simultaneous equations x + y = 5, x – y = 1 Step 1Make y the subject of each equation y = 5 – x y = x – 1 Step 2 Press “Y=“ and enter each of these formulae

9. Step 3Press “WINDOW” and enter these values. This sets the scope of your view window. Step 4 Press “GRAPH” Now to find this point Key point: If your graph doesn’t show the intersection point, try to estimate where it is and then change the window settings

10. Step 5Press 2 nd “TRACE” Step 6Press “5” [Intersect] Step 7Press “ENTER” 3 times x = 3, y = 2

11. Method 2 Algebra (Substitution)

12. Example 1 Use substitution to solve the equations y = 2x – 5 2x – 3y = – 1 Before starting, it’s a good idea to label each equation...........(1).........(2) Step 1 Find where a letter is (or can easily be made) the subject of one of the equations. We see that y is the subject of Equation (1)

13. Step 2 Substitute 2x – 5 into equation (2) in place of y. Equation (2) now changes from 2x – 3y = – 1to 2x – 3(2x – 5) = – 1 Step 3.........(3) Note this last equation (3) only has one letter (x). This means we can work it out!! 2x – 3(2x – 5) = – 1 2x – 6x + 15 = – 1 Expanding brackets – 4x = – 16 Tidying up Dividing through by – 4 x = 4 Step 4 We need to find y. Substitute x = 4 into EITHER equation (1) OR (2). Smarter to use Equation (1) is because y is the subject. y = 2x – 5 y = 3 Ans (4, 3)

14. Example 2 Use substitution to solve the equations 3x – y = 14 4x + 2y – 2 = 0....(1)....(2) Step 1 Find where a letter is the subject of one of the equations – easy to make y the subject of (1) 3x – y = 14y = 3x – 14 Step 2 Substitute 3x – 14 into equation (2) in place of y. 4x + 2(3x – 14) – 2 = 0 Step 3 Expand & work out 4x + 6x – 28 – 2 = 010x – 30 = 0 x = 3 Step 4 Find y by substituting into EITHER equation y = – 5 Ans (3,–5)

15. Method 3 Algebra (Elimination)

16. As the name implies, this method hinges on expressions being eliminated, or removed. Background Do you agree that 5 + 3 = 8 ?...............and that 7 – 3 = 4 ? Now what happens when we add both sides of the equals sign? We get 5 + 7 = 8 + 4 which is a true statement, and in the process both the 3’s have been eliminated. The good news is, we can make letters disappear like the 3’s just did!

17. The key to doing this successfully is getting the same letter in each equation with the same number in front. ONE MUST BE POSITIVE AND THE OTHER NEGATIVE. In other words, we aim to get the pairs of equations looking like this.......... 5x – 3y = 7 4x + 3y = 9 7x – y = 10 –7x + 3y = 9 or.... with the idea in mind that if we ADD the equations, the circled terms will cancel out and leave you with an equation with just one unknown.

18. Use elimination to solve the equations 3x – y = 11........... (1) x + y = 5............ (2) Example 3 Step 1 Identify a letter which has the same number in front of it in both equations (if possible). Each of the y terms has a 1 in front (one is negative and the other is positive) Step 2 Add the two equations together, noting the y terms cancel each other out as – y + y = 0 4x = 16 x = 4 Step 3 Substitute x = 4 into EITHER equation to find y. Pick the one where y is on its own, i.e. (2) x + y = 54 + y = 5 y = 1 Ans (4,1)

19. Use elimination to solve the equations 2x + 3y = –2........... (1) 5y + 2x = 2............ (2) Example 4 Step 1 Identify a letter which has the same number in front of it in both equations (if possible). Each of the x terms has a 2 in front (both positive) Step 2 Change all signs in one of the equations to create opposite signs in front of the letters to be eliminated. You could choose either one. Let’s choose (1) –2x – 3y = 2........... (3) 5y + 2x = 2............ (2) By having opposite signs in front of the x terms it means they will cancel when we add in the next step The x terms will cancel!

20. Step 3 Add the equations (3) and (2) together, noting the x terms cancel each other out as – 2x + 2x = 0 2y = 4 y = 2 Step 4 Substitute y = 2 into EITHER equation to find x. Let’s choose (2) 5y + 2x = 210 + 2x = 2 2x = – 8 x = – 4 Ans (-4,2)

21. Example 5 Use elimination to solve the equations 2x + 4y = 8........... (1) 7x + 2y = 34............ (2) Step 1 Identify a letter which has the same number in front of it in both equations (if possible). If not, can it easily be done by multiplying? The y terms have a 2 and 4 in front. If we multiply (2) through by – 2, then the y s will have a “+4” and a “– 4”. 2x + 4y = 8............... (1) – 14x – 4y = –68.......... (3) Step 2 We can now add them together as we know the y terms will cancel out: –12x = – 60 x = 5 subst into either equation 7(5)+ 2y = 34 y = – ½ Ans (5, –½)

22. Use elimination to solve the equations 3x – 2y = 21........... (1) 4x + 7y = – 1............ (2) Example 6 Step 1 Identify a letter which has the same number in front of it in both equations (if possible). We need to multiply BOTH equations by different numbers to make a pair of x’s OR a pair of y’s with the same number in front! Step 2Let’s try to make the y’s match up. What do we do?  Multiply (1) through by 7 21x – 14y = 147........(3) Multiply (2) through by 2 8x + 14y = – 2........(4) Step 3 What have we achieved? Add and solve 29x = 145 x = 5y = –3