Transcript x - Benjamin N. Cardozo High School

Section 4.4
Review of
Methods for
Solving Systems
of Equations
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Choosing an Appropriate Method
Method
Substitution
Addition
Advantage
Works well if one or
more variable has a
coefficient of 1 or 1.
Disadvantage
Often becomes difficult
to use if no variable
has a coefficient of 1 or
1.
Works well if equations
have fractional or
decimal coefficients, or
if no variable has a
coefficient of 1 or 1.
None
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Example
Select an appropriate method for solving the system.
a.
5x – 3y = 13
0.9x + 0.4y = 30
b.
3x + 6y = 12
x + 2y = 7
c.
6x – 4y = 8
– 9x + 6y = –12
The addition method should be used
since none of the variables have a
coefficient of 1 or 1.
The substitution method should be used
since x has a coefficient of 1.
The addition method should be used
since none of the variables have a
coefficient of 1 or 1.
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Example
Solve the system algebraically.
5x – 3y = 10
0.9x + 0.4y = 30
(10)0.9x + (10)0.4y = (10)30
9x + 4y = 300
Multiply each term in equation (2)
by 10 to remove the decimal point.
This equation is equivalent to
equation (2).
(4)5x – (4)3y = (4)10
Multiply each term in (1) by 4.
(3)9x + (3)4y = (3)300
Multiply each term in (2) by 3.
Continued
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Example (cont)
5x – 3y = 10
0.9x + 0.4y = 30
20x – 12y = 40
27x + 12y = 900
47x = 940
These equations are now ready to add.
Add the equations.
x = 20
5(20) – 3y = 10
Substitute x = 20 into one of the equations.
100 – 3y = 10
–3y = –90
y = 30
The solution is (20, 30).
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Continued
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Example (cont)
5x – 3y = 10
0.9x + 0.4y = 30
Check:
5(20) – 3(30) = 10
100 – 90 = 10
10 = 10 
0.9(20) + 0.4(30) = 30
18 + 12 = 30
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30 = 30 
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Example
Solve the system algebraically.
3x + 6y = 12
x + 2y = 7
x = 2y + 7
3(2y + 7) + 6y = 12
6y + 21 + 6y = 12
21 = 12
Solve equation (2) for x.
Substitute into equation (1).
Simplify.
This results in a false statement.
There is no solution to this system of equations. If graphed,
these lines would be parallel.
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Example
Solve the system algebraically.
6x – 4y = 8
– 9x + 6y = –12
(6)6x – (6)4y = (6)8
4(–9x) + 4(6y) = 4(–12)
36x – 24y = 48
–36x + 24y = –48
0=0
Multiply each term in (1) by 6.
Multiply each term in (2) by 4.
This equation is equivalent to (1).
This equation is equivalent to (2).
Add the equations; this is always a
true statement.
There are an infinite number of solutions to this system of
equations. If graphed, these lines would be the same.
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Possible Solutions
Graph
Number of Solutions
Algebraic Interpretation
One unique solution
You obtain one value for x
and one value for y. For
example,
x = 3, y = 5.
No solution
You obtain an equation
that is inconsistent with
known facts. The system
is inconsistent.
Infinite number of
solutions
You obtain an equation
that is always true. These
equations are dependent.
(3, 5)
Two lines intersect at one point.
Parallel lines
Lines coincide
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