Transcript Solving Systems of Linear Equations By Elimination

Solving Systems of Linear
Equations By Elimination
What is Elimination?
• To eliminate means to get rid of or remove.
• You solve equations by eliminating one of the
variables (x or y) using addition or subtraction.
Example 1
Solve the following system of linear
equations by elimination.
Add equation (1) to
equation (2)

2x – 3y = 15
(1)
5x + 3y = 27
(2)
7x + 0y = 42
7x = 42
x=6
 By eliminating y, we can now
solve for x
Example 1
Substitute x= 6 into equation (1) to
solve for y
2x – 3y = 15
2(6) – 3y = 15
12 – 3y = 15
– 3y = 15 – 12
– 3y = 3
y = -1
Check your solution x = 6 and y = -1
in equation (2)
5x + 3y = 27
5(6) + 3(-1) = 27
30 – 3 = 27
27 = 27
LS = RS
Therefore, the solution set = (6,-1)
You Try
5x – 6y = -32
3x + 6y = 48
One More
7x + 2y = −19
−x + 2y = 21
If you have noticed in the last few examples that to
eliminate a variable, it’s coefficients must have a
sum or difference of zero.
Sometimes you may need to multiply one or both of the equations
by a nonzero number first so that you can then add or subtract
the equations to eliminate one of the variables.
We can add these two equations together to
eliminate the y variable.
2x + 5y = 17
6x – 5y = -19
We can add these two equations
together to eliminate the x variable.
7x + 2y = 10
-7x + y = -16
2x + 5y = -22
10x + 3y = 22
What are we going to do with these
equations, can’t eliminate a variable the way
they are written?
Multiplying One Equation
Solve by Elimination
2x + 5y = -22
10x + 3y = 22
2x + 5y = -22
10x + 3y = 22
5(2x + 5y = -22)
10x + 3y = 22
10x + 25y = -110
- (10x + 3y = 22)
0 + 22y = -132
y = -6
Step 2
y = -6
Solve for the eliminated variable using either of the original
equations.
2x + 5y = -22
2x + 5(-6) = -22
2x – 30 = -22
2x = 8
x=4
The solution is (4, -6).
Choose the first equation.
Substitute -6 for y.
Solve for x.
Solve by elimination.
-2x + 5y = -32
7x – 5y = 17
2x – 3y = 61
2x + y = -7
3x – 10y = -25
4x + 40y = 20
5x + 4y = -28
3x + 10y = -13
Multiplying Both
Equations
To eliminate a variable, you may need to multiply both equations
in a system by a nonzero number. Multiply each equation by
values such that when you write equivalent equations, you
can then add or subtract to eliminate a variable.
In these two equations you
cannot use graphing or
substitution very easily. However
ever if we multiply the first
equation by 3 and the second by
2, we can eliminate the y variable.
4x + 2y = 14
7x + 3y = -8
4 x 7 = 28
2x3=6
Find the least common
multiple LCM of the
coefficients of one variable,
since working with smaller
numbers tends to reduce
the likelihood of errors.
4x + 2y = 14
7x – 3y = - 8
3(4x + 2y = 14)
2(7x – 3y = -8)
12x + 6y = 42
14x – 6y = -16
26x + 0 = 26
26x = 26
x = 1
Solve for the eliminated variable y using either of the original
equations.
4x + 2y = 14
4(1) + 2y = 14
4 + 2y = 14
2y = 10
y = 5 The solution is (1, 5).
Solving Systems of Equations
The Best Time to Use Which Method
Graphing:
Substitution:
Elimination
Using
Addition:
Elimination
Using
Subtraction:
Used to estimate the solution, since graphing usually does
not give an exact solution.
Y = 2x - 3
y=x-1
If one of the variables in either equation has a
coefficient of 1 or –1
3y + 2x = 4
-6x + y = -7
If one of the variables has opposite coefficients in the
two equations 5x – 6y = -32
3x + 6y = 48
If one of the variables has the same coefficient in the
two equations
2x + 3y = 11
2x + 9y = 1
Elimination
Using
Multiplication: If none of the coefficients are 1 or –1 and neither of the
variables can be eliminated by simply adding and
subtracting the equations.
5(2x + 5y = -22)
10x + 3y = 22
3(4x + 2y = 14)
2(7x = 3y = -8)
Closure
Solve by your method of choice:
1) 2x + 5y = 17
6x + 5y = -9
2) 7x + 2y = 10
-7x + y = -16
3) 2x – 3y = 61
2x + y = -7
4) 24x + 2y = 52
6x – 3y = -36
5) y = 2x
y=x–1
6) 9x + 5y = 34
8x – 2y = -2
Solve Systems of Equation by
Elimination in Real World Context
• Two groups of people went to see Guardians
of the Galaxy in IMAX 3‐D. The first group
spent $73.50 on two adult and three children
tickets. The other group spent $109.50 on five
adult and two children tickets. What is the
cost for each type of ticket?
• For dinner, Pat had a double cheeseburger and
two medium fries totaling 1200 calories. Matt
has two double cheeseburgers and one
medium fries totaling 1260 calories. How
many calories are in one double
cheeseburger? One order of medium fries?
• For breakfast, Bill had a bacon, egg, and
cheese biscuit and two hotcakes totaling 660
calories. Phil had a bacon, egg, and cheese
biscuit and three hotcakes totaling 780
calories. How many calories are in one
hotcake? One bacon, egg, and cheese biscuit?
• On December 9th, Trevor Ariza, of the
Houston Rockets, scored 34 points. What is
odd about this is that he scored all 34 points
only on 2‐pointers and 3‐pointers. He made a
total of 15 shots. How many 3‐pointers did he
make?
• At McDonalds, a cheeseburger has 200 fewer
calories than a large fries. Two cheeseburgers
and a large fries have 1100 calories. How
many calories are in each item?
• At Billy’s preschool, they have a total of 25
bicycles and tricycles. Among them all, there
are 57 wheels. How many of each are there?