3.2 Solving Linear Systems Algebraically

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# 3.2 Solving Linear Systems Algebraically - PowerPoint PPT Presentation

3.2 Solving Linear Systems Algebraically. 2 Methods for Solving Algebraically. Substitution Method (used mostly when one of the equations has a variable with a coefficient of 1 or -1) Elimination Method. Substitution Method.

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### 3.2 Solving Linear Systems Algebraically

2 Methods for Solving Algebraically
• Substitution Method

(used mostly when one of the equations has a variable with a coefficient of 1 or -1)

• Elimination Method
Substitution Method
• Solve one of the given equations for one of the variables. (whichever is the easiest to solve for)
• Substitute the expression from step 1 into the other equation and solve for the remaining variable.
• Substitute the value from step 2 into the revised equation from step 1 and solve for the 2nd variable.
• Write the solution as an ordered pair (x,y).
3x-y=13

2x+2y= -10

Solve the 1st eqn for y.

3x-y=13

-y= -3x+13

y=3x-13

Now substitute 3x-13 in for the y in the 2nd equation.

2x+2(3x-13)= -10

Now, solve for x.

2x+6x-26= -10

8x=16

x=2

Now substitute the 2 in for x in for the equation from step 1.

y=3(2)-13

y=6-13

y=-7

Solution: (2,-7)

Plug in to check soln.

Ex: Solve using substitution method
Elimination Method
• Multiply one or both equations by a real number so that when the equations are added together one variable will cancel out.
• Add the 2 equations together. Solve for the remaining variable.
• Substitute the value form step 2 into one of the original equations and solve for the other variable.
• Write the solution as an ordered pair (x,y).
2x-6y=19

-3x+2y=10

Multiply the entire 2nd eqn. by 3 so that the y’s will cancel.

2x-6y=19

-9x+6y=30

-7x=49

and solve for the variable.

x=-7

Substitute the -7 in for x in one of the original equations.

2(-7)-6y=19

-14-6y=19

-6y=33

y= -11/2

Now write as an ordered pair.

(-7, -11/2)

Plug into both equations to check.

Ex: Solve using elimination method.
9x-3y=15

-3x+y= -5

Which method?

Substitution!

Solve 2nd eqn for y.

y=3x-5

9x-3(3x-5)=15

9x-9x+15=15

15=15

OK, so?

What does this mean?

Both equations are for the same line!

¸ many solutions

Means any point on the line is a solution.

Ex: Solve using either method.
6x-4y=14

-3x+2y=7

Which method?

Linear combo!

Multiply 2nd eqn by 2.

6x-4y=14

-6x+4y=14

0=28

Huh?

What does this mean?

It means the 2 lines are parallel.

No solution

Since the lines do not intersect, they have no points in common.

Ex: Solve using either method.