3-2 Solving Linear Systems Algebraically - PowerPoint PPT Presentation

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3-2 Solving Linear Systems Algebraically

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3-2 Solving Linear Systems Algebraically Objective: CA 2.0: Students solve system of linear equations in two variables algebraically. Substitution Method Example 1 ... – PowerPoint PPT presentation

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Title: 3-2 Solving Linear Systems Algebraically


1
3-2 Solving Linear Systems Algebraically
Objective CA 2.0 Students solve system of
linear equations in two variables algebraically.
2
Substitution Method
1. Solve one equation for one of its variables
2. Substitute the expression from step 1 into
the other equation and solve for the other
variable
3. Substitute the value from step 2 into the
revised equation from step 1 and solve
3
Example 1
Solve the linear system of equation using the
substitution method.
4
Step 1
Solve one equation for one of its variables.
5
Step 2
Substitute the expression from step 1 into the
other equation and solve.
6
Step 3
Substitute the value from step 2 into the revised
equation from step 1 and solve.
The solution is (-8, 5)
7
Which equation should you choose in step 1?
In general, you should solve for a variable whose
coefficient is 1 or -1
8
The Linear Combination Method
Step 1 Multiply one or both of the equations by
a constant to obtain coefficients that differ
only in sign for one variable.
9
Step 2 Add the revised equations from step 1.
Combining like terms will eliminate one variable.
Solve for the remaining variable.
Step 3 Substitute the value obtained in Step 2
into either of the original equations and solve
for the other variable.
10
Example 2
Solve the linear system using the Linear
Combination (Elimination) Method.
11
Step 1
Multiply one or both of the equations by a
constant to obtain coefficients that differ only
in sign for one variable
Multiply everything by -2
Leave alone
12
Step 2
After step 1 we now have
-4x 8y -26 4x 5y 8
Add the revised equations.
Solve for y
13
Step 3
Substitute the value obtained in Step 2 into
either of the original equations and solve for
the other variable.
14
Check your solution
The solution checks
15
Example 3
Linear Combination Multiply both
Equations Solve the linear system using the
Linear Combination method.
16
Step 1) Multiply one or both equations by a
constant to obtain coefficients that differ only
in sign for one variable.
17
Step 2) Add the revised equations
18
Step 3) Substitute the value obtained in Step 2
into either original equation
Solution (2, 3)
19
Example 4 Linear Systems with many or no
solutions
Solve the linear system.
20
Use the substitution method Step 1)
Step 2)
6 7 Because 6 is not equal to 7, there are no
solutions.
21
Two lines that do not intersect are parallel.
22
Solve the Linear System
Solve using the linear combination method
Step 1)
23
Step 2) Add revised equations

Because the equation 0 0 is always true, there
are infinitely many solutions.
24
Linear Equations that have infinitely many
solutions are equivalent equations for the same
line.
25
Home work page 153 12 20 even, 24 34 even, 38
52 even.
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