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3.2 Solving Linear Systems Algebraically

3.2 Solving Linear Systems Algebraically. 2 Methods for Solving Algebraically 1.Substitution Method (used mostly when one of the equations has a variable

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Page 1: 3.2 Solving Linear Systems Algebraically. 2 Methods for Solving Algebraically 1.Substitution Method (used mostly when one of the equations has a variable

3.2 Solving Linear Systems Algebraically

Page 2: 3.2 Solving Linear Systems Algebraically. 2 Methods for Solving Algebraically 1.Substitution Method (used mostly when one of the equations has a variable

2 Methods for Solving Algebraically2 Methods for Solving Algebraically

1. Substitution Method

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

2. Elimination Method

Page 3: 3.2 Solving Linear Systems Algebraically. 2 Methods for Solving Algebraically 1.Substitution Method (used mostly when one of the equations has a variable

Substitution MethodSubstitution Method1. Solve one of the given equations for one of

the variables. (whichever is the easiest to solve for)

2. Substitute the expression from step 1 into the other equation and solve for the remaining variable.

3. Substitute the value from step 2 into the revised equation from step 1 and solve for the 2nd variable.

4. Write the solution as an ordered pair (x,y).

Page 4: 3.2 Solving Linear Systems Algebraically. 2 Methods for Solving Algebraically 1.Substitution Method (used mostly when one of the equations has a variable

ExEx: Solve using substitution method: Solve using substitution method3x-y=13

2x+2y= -10

1. Solve the 1st eqn for y.

3x-y=13

-y= -3x+13

y=3x-13

2. 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

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

y=3(2)-13

y=6-13

y=-7

4. Solution: (2,-7)

Plug in to check soln.

Page 5: 3.2 Solving Linear Systems Algebraically. 2 Methods for Solving Algebraically 1.Substitution Method (used mostly when one of the equations has a variable

Elimination MethodElimination Method1. Multiply one or both equations by a

real number so that when the equations are added together one variable will cancel out.

2. Add the 2 equations together. Solve for the remaining variable.

3. Substitute the value form step 2 into one of the original equations and solve for the other variable.

4. Write the solution as an ordered pair (x,y).

Page 6: 3.2 Solving Linear Systems Algebraically. 2 Methods for Solving Algebraically 1.Substitution Method (used mostly when one of the equations has a variable

ExEx: Solve using elimination method.: Solve using elimination method.2x-6y=19

-3x+2y=101. Multiply the entire 2nd

eqn. by 3 so that the y’s will cancel.

2x-6y=19 -9x+6y=30

2. Now add the 2 equations.

-7x=49and solve for the variable.

x=-7

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

2(-7)-6y=19 -14-6y=19 -6y=33

y= -11/24. Now write as an

ordered pair.(-7, -11/2)

Plug into both equations to check.

Page 7: 3.2 Solving Linear Systems Algebraically. 2 Methods for Solving Algebraically 1.Substitution Method (used mostly when one of the equations has a variable

ExEx: Solve using either method.: Solve using either 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.

Page 8: 3.2 Solving Linear Systems Algebraically. 2 Methods for Solving Algebraically 1.Substitution Method (used mostly when one of the equations has a variable

ExEx: Solve using either method.: 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

Add together.

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.