SOLVING SYSTEMS USING ELIMINATION September 2829 2011 Elimination

SOLVING SYSTEMS USING ELIMINATION September 28/29, 2011

Elimination Method The goal is to eliminate _______ one of the adding by _____ your equations. variables

Elimination Method - Steps Step 1: MULTIPLY one or both of the equations by a constant to obtain coefficients that differ only in sign for one of the variables. Step 2: ADD the revised equations from Step 1. Combining like terms will eliminate one of the variables. Solve for the remaining variables. Step 3: SUBSTITUTE the value obtained in Step 2 into either of the original equations and solve for the other variable.

Elimination Method –Example 1 1. 4 x - 2 y = 80 x + 2 y = 45 First, look to see if any terms cancel out. In this case, they can! Which one, x or y? y! 4 x - 2 y = 80 + x + 2 y = 45 5 x + 0 = 125 5 x = 125 x = 25 Now that you know x = 25, plug it back in to either equation to solve for y: 4 x - 2 y = 80 4(25) – 2 y = 80 100 – 2 y = 80 -2 y = -20 y = 10 The solution is (25, 10)

Elimination Method –Example 2 2. 6 x – 2 y = 20 3 x + 3 y = 18 In this case, neither variable can be cancelled out. Which term is going to be easier to work x! with? 6 x – 2 y = 20 -2( 3 x + 3 y = 18 ) To cancel out the x, we will need to multiply the second equation by a “-2”. You should get…. 6 x – 2 y = 20 -6 x - 6 y = -36 Now, you can add them like on the previous example!

Elimination Method –Example 2 6 x – 2 y = 20 + -6 x - 6 y = -36 0 - 8 y = -16 -8 y = -16 y=2 Now that you know y = 2 plug it back in to either equation to solve for x: 3 x + 3 y = 18 3 x + 3 (2) = 18 3 x + 6 = 18 3 x = 12 x =4 The solution to the systems of equations is (4, 2)

Elimination Method –Example 3 3. 2 x + 6 y = 28 3 x + 4 y = 27 By looking at the variables, neither the x or the y is going to be easy to cancel. In this case, we will have to multiply both equations by something! Let’s cancel out the x… 2 and 3 are both factors of … 6 So let’s multiply the first equation by 3 and the bottom by -2. 3 (2 x + 6 y = 28 ) -2 (3 x + 4 y = 27 )

Elimination Method –Example 3 3 ( 2 x + 6 y = 28 ) -2 ( 3 x + 4 y = 27 ) 6 x + 18 y = 84 + -6 x – 8 y = -54 0 + 10 y = 30 y=3 Now that you know y = 3 plug it back in to either equation to solve for x: 2 x + 6 y = 28 2 x + 6 (3) = 28 x=5 The solution is (5, 3)

Elimination Method –Word Problem In one week, a music store sold 9 guitars for a total of $3611. Electric guitars sold for $479 each and acoustic guitars sold for $339 each. How many of each type of guitar were sold? Step 1: Identify your variables x = number of electric guitars y = number of acoustic guitars Step 2: Write a systems of equations x+y=9 479 x + 339 y = 3611

Elimination Method –Word Problem In one week, a music store sold 9 guitars for a total of $3611. Electric guitars sold for $479 each and acoustic guitars sold for $339 each. How many of each type of guitar were sold? Step 3: Solve the systems using elimination. -479(x + y= 9) 479 x + 339 y = 3611 -479 x – 479 y = -4311 + 479 x + 339 y = 3611 0 x - 140 y = -700 y = 5…. and x = 4 The solution is 4 electric and 5 acoustical guitars!