SOLVING SYSTEMS USING ELIMINATION September 28/29, 2011

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

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

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

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

Elimination Method –Example 3 3. 2x + 6y = 28 3x + 4y = 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 2x + 6y = 28 3x + 4y = 27 So let’s multiply the first equation by 3 and the bottom by -2. 3 ( ) -2 ( )

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

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 479x + 339y = 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. x + y = 9 479x + 339y = 3611 -479( ) -479x – 479y = -4311 479x + 339y = 3611 + The solution is 4 electric and 5 acoustical guitars! 0x - 140y = -700 y = 5….and x = 4