3-2 Solving Linear Systems Algebraically

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

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