solving systems of equations algebraically chapter 3.2
TRANSCRIPT
Solving Systems of Equations Algebraically
Chapter 3.2
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
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.
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)
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)
Elimination
Elimination:– Eliminate one of the variables by adding or
subtracting the two equations together
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)
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)
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.
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)
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)
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
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!!
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!!
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