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Solving Systems of Equations Algebraically
Chapter 3.2
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Alternatives to Graphing
Sometimes, graphing systems is not the best way to go– Lines don’t intersect at a discernable point– Don’t have enough room to graph– Etc.
Alternate methods of solving systems– Substitution– Elimination
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Substitution
Substitution: – One equation is solved for one variable in terms
of the other. This expression can be substituted for the variable in the other equation.
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Example 1
Solve either equation for one variable
Substitute this expression into the other equation
– Now, there should only be one variable
After solving for one variable, plug it into either equation to solve for other variable
x + 4y = 26
x - 5y = -10
x = 5y – 10
5y – 10 + 4y = 26
9y – 10 = 26
9y = 36
y = 4
x + 4y = 26x + 4(4) = 26x + 16 = 26x = 10
Solution: (10, 4)
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Example 2
Solve either equation for one variable
Substitute this expression into the other equation
– Now, there should only be one variable
After solving for one variable, plug it into either equation to solve for other variable
2x + y = 43x + 2y = 1
y = -2x + 4
3x + 2(-2x + 4) = 13x - 4x + 8 = 1-x + 8 = 1-x = -7x = 7
2x + y = 42(7) + y = 414 + y = 4y = -10
Solution: (7, -10)
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Elimination
Elimination:– Eliminate one of the variables by adding or
subtracting the two equations together
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Example 3
Instead of using Substitution, you can subtract one equation from the other
– Subtract the 2nd from the 1st
After solving for one variable, plug it into either equation to find the other
x + 2y = 10
x + y = 6
0 + y = 4
y = 4
x + y = 6
x + 4 = 6
x = 2
Solution:
(2, 4)
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Example 4
Instead of using Substitution, you can add one equation to the other
After solving for one variable, plug it into either equation to find the other
2x + y = 5
3x - y = 20
5x + 0 = 25
x = 5
2x + y = 5
2(5) + y = 5
10 + y = 5
y = -5
Solution:
(5, -5)
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Elimination with Multiplication
When one variable cannot be easily eliminated using simple addition or subtraction, multiply one or both equations by constants so that a variable CAN be eliminated.
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Example 5
Simple addition or subtraction isn’t going to help here.
Decide which variable to eliminate
– Let’s do x
Now subtract 2nd from 1st
Use one variable to solve for the other
2x + 3y = 12
5x - 2y = 11
10x + 15y = 60
10x - 4y = 22
0 + 19y = 38
y = 2
x5 x2
2x + 3y = 12
2x + 3(2) = 12
2x + 6 = 12
2x = 6
x = 3
Solution:
(3, 2)
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Example 6
Simple addition or subtraction isn’t going to help here.
Decide which variable to eliminate
– Let’s do h
Now add together
Use one variable to solve for the other
2g + h = 6
3g - 2h = 16
4g + 2h = 12
3g - 2h = 16
7g + 0 = 28
g = 4
x2
2g + h = 6
2(4) + h = 6
8 + h = 6
h = -2
Solution:
(4, -2)
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Inconsistent and Dependent Systems
If you add or subtract two equations in a system and the result is an equation that is never true, then the system is inconsistent and it has no solution.
– Examples 1 = 2 -1 = 1
If the result is an equation that is always true, then the system is dependent and has infinitely many solutions.
– Examples 1 = 1 9 = 9
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Example 7 (substitution)
Solve either equation for one variable
Substitute this expression into the other equation– Now, there
should only be one variable
x + 3y = 8
1/3 x + y = 9
x = -3y + 8
1/3(-3y + 8) + y = 9
-y + 8/3 + y = 9
8/3 = 9
Not True No Solutions!!
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Example 8 (substitution)
Solve either equation for one variable
Substitute this expression into the other equation– Now, there
should only be one variable
2a - 4b = 6
-a + 2b = -3
a = 2b + 3
2(2b + 3) - 4b = 6
4b + 6 - 4b = 6
6 = 6
Always True
Infinitely Many Solutions!!
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Example 9 (elimination)
Simple addition or subtraction isn’t going to help here.
Decide which variable to eliminate
– Let’s do x
Now add together
4x - 2y = 5
-2x + y = 1
-4x + 2y = 2
4x - 2y = 5
0 + 0 = 7
Not True
x2
No Solutions