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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

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

Example 1

Solve the linear system of equation using the

substitution method.

Step 1

Solve one equation for one of its variables.

Step 2

Substitute the expression from step 1 into the

other equation and solve.

Step 3

Substitute the value from step 2 into the revised

equation from step 1 and solve.

The solution is (-8, 5)

Which equation should you choose in step 1?

In general, you should solve for a variable whose

coefficient is 1 or -1

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.

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.

Example 2

Solve the linear system using the Linear

Combination (Elimination) Method.

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

Step 2

After step 1 we now have

-4x 8y -26 4x 5y 8

Add the revised equations.

Solve for y

Step 3

Substitute the value obtained in Step 2 into

either of the original equations and solve for

the other variable.

Check your solution

The solution checks

Example 3

Linear Combination Multiply both

Equations Solve the linear system using the

Linear Combination method.

Step 1) Multiply one or both equations by a

constant to obtain coefficients that differ only

in sign for one variable.

Step 2) Add the revised equations

Step 3) Substitute the value obtained in Step 2

into either original equation

Solution (2, 3)

Example 4 Linear Systems with many or no

solutions

Solve the linear system.

Use the substitution method Step 1)

Step 2)

6 7 Because 6 is not equal to 7, there are no

solutions.

Two lines that do not intersect are parallel.

Solve the Linear System

Solve using the linear combination method

Step 1)

Step 2) Add revised equations

Because the equation 0 0 is always true, there

are infinitely many solutions.

Linear Equations that have infinitely many

solutions are equivalent equations for the same

line.

Home work page 153 12 20 even, 24 34 even, 38

52 even.