Download 3.3 Solving Systems with Elimination

3.3 Solving Systems with Elimination
Sometimes it is easier to eliminate a variable entirely from a system of equations rather than use the
substitution method. We do this by adding opposite coefficients together to get zero of one variable.
Subtracting to Eliminate
We first need to decide which variable is easiest to eliminate. Consider the following system of equations.
3 + 1
3 + 2 = 4
Notice that in this case the coefficients for are the same. This means that they will be easily eliminated.
If we subtract 3 from both sides of the first equation, we will eliminate the variable from the left side but will still
have are the right side. However, if we instead subtracted 3 + 2 from the left side we could subtract 4 from
the right side because we know that 3 + 2 is exactly equal to 4 thanks to the second equation. (Remember that
if we’re going to solve a system of equations, we’ll have to use both equations somehow, which is what we just
did.)
This is sort of like repossession in a way. If you don’t have the money to pay the bank, they can repossess
your property up to an equivalent value of what you owe. In the same way, if we don’t want to take away 3 +
2 from the right side, we can take away something equivalent which is 4 in this case. Let’s take a look.
3 + =
1
−3 + 2 − 4
0 − 1 = −3
Notice that it almost looks like we subtracted the second equation from the first. What we actually did was
subtract expressions that are equal from both sides to keep the first equation balanced. Now we can solve since
we have zero ’s left.
−1 = −3
−1 −3
=
−1
−1
=3
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Now that we know what equals, we can substitute that back into either equation to find the value of
the solution point.
3 + = 1
3 + 3 = 1
3 + 3 − 3 = 1 − 3
3 = −2
3 −2
=
3
3
=−
2
3
So we get the solution A− C , 3B which can verify is in the second equation by substituting both values in to
make sure it is a true mathematical statement.
Adding to Eliminate
Adding to eliminate a variable will work the same way. In this case we should find one variable with the
opposite coefficient of the same variable in the other equation. For example, consider this system of equations:
3 + 2 4
− 2 4
Notice that the first equation has 2 as the coefficient for and the second equation has a -2 as the
coefficient. That means we should be able to add ( − 2) to both sides of the first equation. However, remember
that we don’t want to end up with more of the variable on the right side, so we will add something equivalent to
it. In this case that will be 4.
3 + 2 +( − 2)
4
+4
4 = 8
4 8
4
4
2
We get 0 which means that the
variable is eliminated
We’ll leave it as an exercise to show that from here we can get −1 which means that our solution to
this system of equations is the point 2, −1.
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When the Coefficients Don’t Match
The elimination method works fine when the coefficients match or are opposites, but what about when it
is just a messy system of equations like this?
2 + 3 −1
4 + 5 = −1
Solving this system by the substitution method would mean dealing with fractions and the coefficients don’t
match so it looks like the elimination method won’t work either. However, is there a way we can get the coefficients
to match?
Notice that in the first equation we have a 2 and in the second we have a 4. Wouldn’t it be nice if the
first equation had a 4 instead of the 2? Is there any way we can make that happen? If we multiply both sides of
the first equation by 2, we will maintain equality and have 4 to match the second equation. Let’s do so.
22 + 3 = 2−1
4 + 6 = −2
Now that we have matching coefficients we can use the elimination method to continue to solve by
subtracting 4 + 6 from the left side of our new equation and subtracting -2 from the right side since that is
equal to 4 + 6.
4 + 6 −2
−(4 + 5) − (−1
0 + 1 = −1
= −1
From here we can again substitute = −1 into either original equation to find that = 1 which gives us
the solution of 1, −1.
Infinite and No Solutions
It is still possible to get infinite solutions or no solution for a system of equations. If both variables get
eliminated and we get down to a number equals a number statement that is always true, there are infinite solutions.
If we get down to a number equals a number statement that is false, there are no solutions.
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Lesson 3.3
Solve the following systems using the elimination method.
1.
3.
5.
7.
9.
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+ =1
− =5
3 + = 6
3 − 2 = 9
+ = −3
− =1
D
+ 2 = −10
2 + 2 = −10
−4 = 4
4 − 3 = −10
2.
4.
6.
8.
10.
2 + 3 = 7
−2 + = 5
+ 3 = 1
3 + 3 = 6
4 + = −9
4 + 2 = −10
−2 + = 10
4 + = −8
=1
6 − 5 = 11
11.
13.
15.
17.
19.
− 2 = 5
3 − 2 = 9
=5
2 − 3 = 16
4 − 3 = 12
+ 2 = 12
C
3 = 6
4 − = −2
+ =4
2 + 2 = 8
12.
14.
16.
18.
20.
3 + = 5
2 + = 10
C
3 + = 6
3 − 2 = −1
−5 + 3 = 6
− =4
3 + = 2
6 + 3 = 5
+ =2
2 + 2 = 8
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21.
23.
25.
27.
29.
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+ 3 = 12
2 − 3 = 12
5 + 4 = −3
10 − 2 = −3
4 − 7 = 10
3 + 2 = −7
3 − 4 = −10
5 + 8 = −2
C
4 + = 17
6 + 5 = 20
22.
24.
26.
28.
30.
2 + 3 = 10
5 + 7 = 24
5 − 4 = −8
3 + 8 = 3
− 3 = −4
4 = 8
4 + 3 = 19
3 − 4 = 8
3 + 4 = −25
2 = −6
Write and solve a system of equations using any method (graphing, elimination, or substitution) for each of the
following situations.
31. The sum of two numbers is 82 and their difference is 26. Find each of the numbers.
32. Kathryn buys 8 cups of coffee and 2 bagels one day and pays $31. Harry buys 3 cups of coffee and 3 bagels
the same day and pays $17.25. How much is each cup of coffee and each bagel?
33. Farmer Deanna looks out her window and counts a total of 64 legs on a total of 20 animals. If she has only
sheep and chickens, how many of each does she have? (Hint: Sheep have 4 legs each and chickens 2 legs each.)
34. Tyler and Pearl went on a 20-kilometer bike ride that lasted 3 hours. Because there were so many steep hills
on the bike ride, they had to walk for most of the trip. Their walking speed was 4 kilometers per hour. Their riding
speed was 12 kilometers per hour. How much time did they spend walking?
35. A used book store also started selling used CDs and videos. In the first week, the store sold 40 used CDs and
videos at $4.00 per CD and $6.00 per video. The sales for both CDs and videos totaled $180.00. How many CDs
and videos did the store sell in the first week?
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36. A metal alloy is 25% copper. Another metal alloy is 50% copper. How much of each alloy should be used to
make 1000 grams of a metal alloy that is 45% copper?
37. Dried apricots worth $3.25 a pound were mixed with dried prunes worth $4.75 a pound to produce a mixture
of dried fruit worth $3.79 a pound. How much of each kind of fruit was used to produce 25 pounds of mixture?
38. One number added to twice another number is 23. Four times the first number added to twice the other
number is 38. What are the numbers?
39. The owners of the River View Restaurant have hired enough servers to handle 17 tables of customers, and
the fire marshal has approved the restaurant for a limit of 56 customers. How many two-seat and how many
four-seat tables should the owners purchase?
40. The Rodriguez family and the Wong family went to a brunch buffet. The restaurant charges one price for
adults and another price for children. The Rodriguez family has two adults and three children, and their bill was
$40.50. The Wong family has three adults and one child, and their bill was $38.00. Determine the price of the
buffet for an adult and the price for a child.
160