Transcript 3.2 Lesson - FHS PAP Algebra 2

SOLVING SYSTEMS
USING ELIMINATION
September 28/29, 2011
Elimination Method
eliminate
The goal is to _____________
one of the variables by
adding your equations.
_________
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
4(25) – 2y = 80
100 – 2y = 80
-2y = -20
y = 10
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
-2( 3x + 3y = 18 )
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
So let’s multiply the first equation by 3 and
the bottom by -2.
3 (2x + 6y = 28 )
-2 (3x + 4y = 27 )
Elimination Method –Example 3
3 ( 2x + 6y = 28 )
-2 ( 3x + 4y = 27 )
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
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
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.
-479(x +
y=
9)
479x + 339y = 3611
-479x – 479y = -4311
+ 479x + 339y = 3611
0x - 140y = -700
y = 5….and x = 4
The solution is
4 electric and 5
acoustical
guitars!