How much does the cow weigh?

Transcription

1 1 GRADE 11 PRE-CALCULUS UNIT G SYSTEMS UNIT NOTES 1. Solving Systems of equation is an important subject. To date you have only learned to solve for one unknown in an equation for example: when does 2x + 1 = 7; or when does x 2 + 4x 10 = 5; 2. You might be surprised to learn that sometimes two unknowns govern a certain situation. So for example the volume of an enclosed gas depends on its temperature and its pressure. Another example: Windchill depends on wind speed and temperature. You are likely familiar with lots of other relationships that depend on more than one variable. 3. In this unit we will learn to solve Systems of Equations. Both Linear and non-linear. In Grade 10 you likely learned solving Systems of Linear Equations, much of that is repeated here in the first few pages. SOLVING SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES How much does the cow weigh? 6 7 y? y y x x x Diagram A 4. A familiar example: Diagram B x is the weight of the Cow. y is the weight of the Beaver. 1 Cow and 1 Beaver weigh 6, 1 Cow and 2 Beavers weigh 7. How much does 1 Cow weigh by itself? How much does a beaver weigh? Kara goes to the furniture store and buys four chairs and a table for $ Clifton goes to the same store and buys six chairs and a table for $ How much does a chair cost? How much does a table cost? There is one solution but the solution has two parts, the solution has a chair cost and a table cost. C GR11Pre_G_UnitNotes.doc Revised:

2 2 5. Make a table of some possibilities. Guess at prices until Clifton and Kara pay the same Price of 1 Price of 1 Kara Clifton Chair Table Price 4 Chairs & One Table Price 6 Chairs & One table $20 $150 $80 + $150 = $230 $120 + $150 = $270 Surely there has to be a better way to do it than guessing and checking. DEFINITIONS 6. Systems of Linear Equations Definition A set of two or more linear equations with the same variables. 7. Solution to a System of Linear Equations Definition The set of all ordered pairs that satisfy all equations (make them true) 8. Systems of Linear Equations are also sometimes called Simultaneous Linear Equations if you google the subject. SOLVING SYSTEMS OF TWO EQUATIONS BY SUBSTITUTION 9. The substitution method steps: Label the equations Express one variable in terms of the other (isolating a variable) Substitute that expression into the other equation Solve for the single variable Substitute that found variable back into either equation to find the remaining unknown variable Check the answer in both original equations

3 3 10. Example 1 Substitution: Solve the following system of equations (note that two unknowns requires two distinct and independent equations!) 1) y = 2x 2) 2x + y = 8 Eqn 1) says that: y = 2x, so substitute that into y in eqn 2) to make eqn 3): 3) 2x + (2x) = 8: Now we only have one unknown! Easy to solve! 4x = 8; therefore x = Now substitute that x =2 into either original equation 1) or 2). Workspace Using eqn 1) : y = 2x = 2 (2) = 4 So an x of 2 and a y of 4 makes both equations true! 12. Check this answer in the original equations 1) does 4 = 2(2)?? YEP! 2) does 2(2) + 4 = 8? YEP! Both equations are solved when x = 2 and y = 4. The ordered pair (2, 4) 13. Example 2 Substitution. Solve the following system of equations (both in general form) 1) 2x + 6y = 12 2) x 3y = 9 Follow the steps! 14. We need to isolate one of the variables by itself. We can isolate either of the two x s or either of the two y s; it doesn t matter. For simplicity we will isolate the x in eqn 2). 2) x 3y = 9 adding 3y to both sides: x = 9 + 3y 15. So plug the expression 9 + 3y into eqn 1) in place of x 2(9 + 3y) + 6y = y + 6y = y = y = 6 y = 1/2

4 4 16. So now we know the y = 1/2, plug that into either original equation. Workspace Plugging into eqn 1): 2x + 6(-1/2) = 12 2x 3 = 12; 2x = 15; x = 15 / 2 ( or 7.5 to earthlings) Solution: ( 1/2, 15/2) 17. Check the answer by evaluating with your hopefully correct solution: 1) 2x + 6y = 12 so 2(15/2) + 6( 1/2) = 12??? So 15 3 = 12?? Yep!! 2) x 3y = 9 so 15/2 3(-1/2) = 9?? 7.5 ( 1.5) = 9?? Yep!! (subtracting a negative is adding) 18. So the one solution that satisfies both eqn 1) and eqn 2) is an x = 15/2 and y = 1/2. Solution: ( 1/2, 15/2) 20. Here are two more examples for you to do; solve for unknowns x and for y: a. x + y = 7; and x 2y = 2 b. 3x = y + 5; and y = 4 Ans: x = 4; y = 3; so (4, 3) lol: x = 3; y = 4 so : (3,4)

5 5 SOLVING A SYSTEM OF EQUATIONS IN TWO VARIABLES BY ELIMINATION 21. The elimination method eliminates one of the variables by adding or subtracting the two equations or their multiples. The method is also sometimes called Solving by Addition or Subtraction. 22. STEPS FOR THE ELIMINATION METHOD Put equations into some standard order:(ax + by = c for example) Label the equations Multiply either or sometimes both equations by some number so that one variable can be eliminated when the equations are added or subtracted Add or subtract the equations to eliminate one variable and solve the other Find the eliminated variable now by substituting the solved variable back into either equation of the system Check your answer (x, y) in both original equations 23. Example 1 of Elimination Method. 1) 2x + y = 8 2) x y = 4 Simply add the two equations together! eqn1) + eqn 2) 2x + x + y y = x = 12 x = 4 As before plug in this 4 into either original equation for x to find the y. (4) y = 4. y = 0 So x = 4, y = 0 is the one (and only) solution to both equations. The solution is written as the ordered pair (4, 0)

6 6 Example 2: 24. Solve for x and y: 4x + y = 8 and y = x Label and put in some standard order: (A) 4x + y = 8 (B) 2x + y = 10 Add or subtract to eliminate a variable (A) (B) 6x = 18 Solve: x = 3 Plug in to either eqn. 4( 3) + y = 8; y = 4 So the solution for the ordered pair of (x, y) is ( 3, 4) Check! 4( 3) + 4 = 8 and 2( 3) + 4 = You try a couple here; solve the system of linear equations two variables x and y for: y = 6 x; and 2x + y = 11 4x + 3y = 19; and 3x + 5y = 17 Ans: x = 5 ;y = 1 (5, 1) Ans: x = 4 ;y = 1 ; (4, 1) Trick ; make one of the variables in each eqn have a Common Multiple coefficient. Graphing Systems of Linear Equations 26. It is pretty easy to graph both lines of a System of Linear Equalities in two unknowns and find out where they share the same x and y. Where they intersect is where they share the same x and y. The TI-83 Graphing Calculator has an intersect feature that finds where any two graphed functions intersect. Any Smart Phone App or graphing website does the same thing. 27. Solving a system of Linear Equations with a graphing tool will be demonstrated in class but it pretty much looks like the picture below:

7 7 4 3 y = - 4x y = 2x Solution: (1, -2) -3-4 An EXCEL scatter plot TI-83 Graphing Calculator SOLVING SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES (We will seldom have time to learn solving for 3 variables in class, it is here for completeness and interest. You will find if you go to university that there are other ways to solve such as the Gaussian Elimination method) 30. How would you solve a system of equations in three unknowns? Example: Solve the following system of three equations with three unknowns. (1) 2x y + z = 3 (2) x + 3y 2z = 11 (3) 3x 2y + 4z = 1 The solution is (3, 2, -1) for the x, y, z ordered triplet and you can check to see that it is true. 31. A system of three unknowns in linear equations is solved by taking two pairs of equations, and eliminating the same variable in each pair. Once that is done there remains two equations with two unknowns; so follow the normal steps for a system of two equations. 32. Solve the following system of three equations with three unknowns. (1) 2x y + z = 3 (2) x + 3y 2z = 11 (3) 3x 2y + 4z = 1

8 8 Solution: (steps, there are several!) 33. Eliminate the y between (1) and (2) 3*(1) 6x 3y + 3z = 9 add (2) x + 3y 2z = 11 (4) 7x + z = Eliminate the y again between a different pair of equations 2*(2) 2x + 6y 4z = 22 add 3*(3) 9x 6y + 12z = 3 (5) 11x + 8z = Solve equations (4) and (5) by any method (4a) z = 20 7x substitute (4a) into (5) (5) 11x + 8(20 7x) = 25 11x x = 25 45x = 135 x = +3 Substitute x = 3 into either eqn (4) or (5) to find second unknown. 7(3) + z = z = 20 z = Substitute x = +3 and z = 1 into either of equations (1) to (3). (Eqn1) 2(3) y + ( 1) = 3 6 y 1 = 3 5 y = 3 y = 2

9 9 STEPS TO SOLVE LINEAR EQUATIONS IN THREE UNKNOWNS 37. The aim is to get two equations in two unknowns 38. Eliminate one of the variables in two pairs of equations Select one pair of equations, eliminate one variable Select a different pair of equations, eliminate the same variable as in the first pair. Use the two new equations in two unknowns and solve using any method from the method of Solving Systems of Linear Equations in 2 Unknowns SOLVING NON-LINEAR SYSTEMS 39. Nothing says that we had to apply these techniques of systems of equations to just Linear functions. The idea of substitution can readily be applied to a system of non-linear equations such as the following: 1) y = x 2 2) 5x + y = Of course one could always graph the functions and find the solution of the two points they share in common Or we could use the substitution method to solve the linear - quadratic system of equations using algebra ) y = x 2 (a quadratic equation) 2) 5x + y = 4 (a line) so: 5x + x 2 = 4; so x 2 + 5x + 4 = 0. So factoring provides that x = 4 which means y = 16 or x= 1 which means y = 1. So the set of ordered pairs { ( 4, 16), ( 1, 1) } are the solutions Or you could of course complete the square or use the quadratic formula.

10 10 Interesting example 42. Solve the system y = x 2 and y = x By either substitution or elimination you get the combined formula: x 2 = x which of course, dividing by x leaves x = 1 and consequently y = 1 as the solution. But doesn t x = zero and y = zero work as a solution too!?? isn t 0 = 0 2 and 0 = 0? 43. Zero and zero is clearly a solution but our combined formula did not reveal that. Just another reason why doing a quick sketch of the problem might twig you that there are two solutions. The reason you did not get 0 as an answer was because it was an answer, however, when you divided by it, it gave you no answer! Dividing by zero is forbidden, it is an undefined operation, so don t expect it to give you a useful or complete solution. Of course had you solved the combined formula of x 2 x = 0 by subtracting x from both sides, you would have found both solutions. This is another good reason why you should always equate quadratics to zero to solve them! Practice 44. Solve the following linear-quadratic or quadratic-quadratic systems of equations. A sketch is always advisable too! The answers will be some number of (x, y) ordered pairs a. (A) y = 2x; and (B) y = x 2 3 b. (A) x y = 5; and (B) y = 2x 2 + 9x + 3

11 11 c. (A) y = x 2 + 4; and (B) x + y = 2 d. (A) y = x 2 + 4; and (B) x + y = 4 e. (A) y = x 2 + 6x + 5; and (B) y 6x = x f. (A) y = x 2 ; and (B) y = x 2 + 4x + 2 Answers: a. { ( 1, 2), (3, 6) } b. ( 2, 7) c. No solution d. (0, 4) e. same curve; dependent system; infinite number of ordered pair solutions f. ( ½, ¼) Notice we put each x, y solution in round parentheses (x, y), and the entire set of solutions in braces { }; ie: { (x 1, y 1 ) ; (x 2, y 2 ),. (x n, y n ) } CONCLUDING MATERIEL Now you have learned to solve for functions that have two unknowns. They require two independent equations of course. Just like in life if you have two unknown things to determine you need two different facts.