Consistent and Dependent

Transcription

1 Graphing a System of Equations System of Equations: Consists of two equations. The solution to the system is an ordered pair that satisfies both equations. There are three methods to solving a system; Graphing, Substitution and Elimination. How to read graphs of a system of equations: Intersecting Lines Same Line Parallel Lines One solution (at intersection) Infinitely Many solutions NO Solutions! Consistent and Independent Consistent and Dependent Inconsistent

2 YOU TRY: Solve the following system by graphing: y = x + 1 y = -x - 3 NOTE: It may be necessary to put an equation in Slope-Intercept Form first before graphing! One Solution = (-2, 1) Consistent and Independent

3 Solve the following system by graphing: y = 2x + 1-2x + y = 3 NOTE: It may be necessary to put an equation in Slope-Intercept Form first before graphing! No Solutions - Inconsistent

4 Solve the following system by graphing: 2x + y = - 6 y = -2x - 6 NOTE: It may be necessary to put an equation in Slope-Intercept Form first before graphing! Infinitely Many Solutions Consistent and Dependent

5 Substitution System of Equations Solutions # Solutions Graphing Substitution / Elimination One After solving, we find one coordinate pair as our solution (x,y) No Solution After solving, we end up with a false statement: Ie: 3 = 4 or 2 = -2 Infinitely Many After solving, we end up with a true statement: Ie: 3 = 3 or x = x SOLVE A SYSTEM BY USING SUBSTITUTION:

6 Step 1: Step 2: Step 3: Step 4: Rewrite one of the equations so that you isolate ONE of the variables. (Easiest when one variable has a coefficient of 1) Put one equation in x = or y = form Substitute the isolated variable into the second equation. Solve for the remaining variable Substitute the solution from step 3 back into an original equation to find the remaining variable. YOU TRY! x + 2y = 6 x=-2y+6 y = 2x - 4 3x 4y = 28-6x + 3y = -12 3( 2y+6) 4y=28 6y+18 4y =28 10y+18=28 10y=10 y = 1 6x+3(2x 4)= 12 6x+6x 4= 12 4= 12 False statement so NO SOLUTION Now find x: x + 2(1)=6 x=4 Solution: (4, -1)

7 Real World Problems: 1. Harvey has some $1 bills and some $5 bills. In all he has 6 bills worth $22. Let x be the number of $1 bills and y be the number of $5 bills. Write a system of equations to represent the information & use substitution to determine how many bills of each denomination Harvey has. Let: x = # of $1 bills y = # of $5 bills Quantity equation: x + y = 6 x = 6-y Money equation: 1x + 5y = 22 Substitute: (6-y)+5y=22 6+4y=22 4y=16 y=4 x+y=6 x+4=6 x=2 ANSWER: Harvey has two $1 bills and four $5 bills.

8 2. A store sold a total of 125 car stereo systems and speakers in one week. The stereo systems sold for $ and the speakers sold for $ Total sales were $ How many of each item were sold? Let: x = # of stereo systems sold y = # of speakers sold Quantity equation: x + y = 125 y=125-x Value equation: x y = x (125-x) = x x = x = x = 4558 x=53 53+y=125 y = 72 The store sold 53 stereo systems and 72 speakers.

9 Elimination Using Addition/Subtraction The third method to solving systems of equations is Elimination. Elimination gives us exact solutions - just like the Substitution Method. SOLVE A SYSTEM BY USING ELIMINATION: Step 1: Put both equations in Standard Form (so like terms are aligned). One of the variables should have the same coefficient with opposite signs (additive inverse). (if the signs are not opposite, multiply one of the equations by -1). Step 2: Step 3: Step 4: Add the equations and one of the variables will cancel out (get eliminated). Solve for the remaining variable. Substitute the solution to Step 3 in one of the original equations to find the remaining variable.

10 Now find x: 2x + 3(6) = 20 2x + 18 = 20 2x = 2 X = 1 Solution: (1, 6) YOU TRY: 1. 4x + 6y = x 5y = 11-3x - 5y = 11 3x 6y = 3 3x + 7y = 1 3x + 7y = -1 7x = 35 2y = 10 x = 5 y = 5 Find y: Find x: 3(5) 6y = 3 3x 5(5) = y = 3 3x 25 = 11-6y = -12 3x = 36 y = 2 x = 12 Solution: (5, 2) Solution (12, 5)

11 Elimination Using Multiplication Sometimes, to get a variable to eliminate when adding the two equations together, we need to multiply one or both equations by a number to allow that variable to cancel. Step 1: Step 2: Step 3: Step 4: Step 5: Make sure both equations are in Standard Form. Create a pair of opposite terms by multiplying at least one equation by a constant. Add the equations to eliminate a variable. Solve for the remaining variable. Substitute solution to step 4 in one of the original equations to solve for the remaining variables. See the next page for an example:

12 Example: y = 8 3x x + 5y = -2 Original System Step 1: Rewrite equations in Standard Form. y = 8 3x x + 5y = -2 3x + y = 8 x + 5y = -2 Step 2: Multiply the second equation by -3 NOTE: You could have multiplied the first equation by -5 and get the same results. Step 3: Add the equations together. Step 4: Solve 3x + y = 8 3x + y = 8 x + 5y = -2-3x 15y = 6 3x + y = 8-3x 15y = 6-14y = 14 14y 14 = Y = -1 Step 5: Substitute that value back in to an original equation to solve for remaining variable Solution: (3, -1) x + 5(-1) = -2 x 5 = -2 x = 3

13 YOU TRY: -5y + 4x = 49 (7y + 2x = -23) (-2) 5x + 12y = 10 (2x 3y = -35) (4) -5y + 4x = 49-14y -4x = 46-19y = 95 y = -5 5x + 12y = 10 8x 12y = x = -130 x = -10 7(-5) + 2x = x = -23 2x = 12 x = 6 Solution: (6, -5) 9x + 13y = 2 (-3x 5y = 2) (3) 5(-10) + 12y = y = 10 12y = 60 Y = 5 Solution: (-10, 5) (3x + 4y = 10) (2) (2x + 3y = 7) (-3) 9x + 13y = 2-9x 15y = 6-2y = 8 y = -4 6x + 8y = 20-6x-9y = -21 -y = -1 y = 1-3x 5(-4) = 2-3x + 20 = 2-3x = -18 x = -6 Solution: (-6, -4) 3x + 4(1) = 10 3x + 4 = 10 3x = 6 x = 2 Solution: (2, 1)

14 Applying Systems of Linear Equations Three types of solutions: One Solution: Intersecting lines. Two lines that meet at one point. That one coordinate is the solution to the system. No solution: Parallel lines. When solving by substitution or elimination, we end up with a false statement. Infinite Many Solutions: The same line. When we solve by substitution or elimination, we end up with a true statement. Best Method to use when solving systems: Graphing: Use graphing when you don t need an exact answer, but rather when you can estimate. Substitution: Use substitution when one of the coefficients in either equation equals 1. This allows us to get a variable alone easily. Elimination: Use elimination when we can eliminate a variable. We may need to multiply one or both equations by a constant to facilitate this.

15 Determine the best method to solve each system, then solve. 2x + 3y = -11 Multiply by 4-8x 5y = 9 Use elimination. 8x + 12y = -44-8x 5y = 9 7y = -35 y = -5 2x + 3(-5) = -11 2x 15 = -11 2x = 4 x = 2 3x + 4y = 11 2x + y = -1 y = -2x - 1 Use substitution. 3x + 4(-2x 1) = 11 3x 8x 4 = 11-5x 4 = 11-5x = 15 x = -3 2(-3) + y = y = -1 y = 5 Solution: (2, -5) 3x 4y = -5-3x + 2y = 3 Use elimination. -2y = -2 y = 1 Solution: (-3, 5) 3x + 7y = 4 5x 7y = 12 Use elimination. 8x = 16 x = 2 3x 4(1) = -5 3x 4 = -5 Solution: 3x = -1 x = 1 3 ( 1 3, 1) 3(2) + 7y = y = 4 Solution: 7y = -2 y = 2 7 (2, 2 7 )

16 Systems of Inequalities Solve a system of inequalities by graphing Step 1: Graph first inequality and shade appropriate area. Step 2: Graph the second inequality and shade appropriate area. Step 3: The region where the shading overlaps is the solution to the inequality system: EXAMPLE: y > 3x + 5 y x 2

17 YOU TRY: Graph the system of inequalities: y > -x 1 y x + 3