Solve Systems with Eliminationwhsalgebra1b.weebly.com/.../7/4/1/8/7418584/concept_6.3_eliminati… · Solving Systems of Equations So far, we have solved systems using graphing and

Objective

You will be able to:

Solve systems of equations using

elimination with addition or (subtraction)

and multiplication.

UNIT 6

Concept 3

Solving Systems of Equations

So far, we have solved systems using

graphing and substitution. These notes

show how to solve the system

algebraically using ELIMINATION with

addition and subtraction.

Elimination is easiest when the

equations are in standard form.

The Linear Combination Method

aka The Addition Method, aka The

Elimination Method.

Add (or subtract) a multiple of

one equation to (or from) the

other equation, in such a way

that either the x-terms or the y-

terms cancel out.

Then solve for x (or y,

whichever's left) and substitute

back to get the other coordinate.

Solve this system of equations

using the addition or

subtraction method.

When solving systems of linear

equations using elimination, you

will sometimes need to multiply

one or both equations by a factor

in order to get the same

coefficients for a variable.

Best time to use this method is

when you have two coefficients

that the same

Steps in Elimination

1. Line up the two equations

using standard form

(Ax + By = C).

2. GOAL: The coefficients of

the same variable in both

equations should have the

same value but opposite

signs.

3. If this doesn’t exist, multiply

one or both of the equations

by a number that will make

the same variable

coefficients opposite

values.

4. Add the two equations

(like terms).

5. The variable with

opposite coefficients

should be eliminated.

6. Solve for the

remaining variable.

7. Substitute that

solution into either of

the two equations to

solve for the other

variable.

Color Coded Steps in solving a system of

equations by elimination.

Step 1: Put the equations in Standard Form.

Step 2: Determine which

variable to eliminate.

Step 3: Add or subtract the

equations.

Step 4: Plug back in to find

the other variable.

Step 5: Check your

solution.

Standard Form: Ax + By = C

Look for variables that have the

same coefficient.

Solve for the variable.

Substitute the value of the variable

into the equation.

Substitute your ordered pair into

BOTH equations.

1) Solve the system using elimination.

x + y = 5

3x – y = 7

Step 1: Put the equations in

Standard Form.



They already are!

The y’s have the same

coefficient.


equations.

Add to eliminate y.

x + y = 5

(+) 3x – y = 7

4x = 12

x = 3

Add into your

spirals



the other variable.

x + y = 5

(3) + y = 5

y = 2

Step 5: Check your

solution.

(3, 2)

(3) + (2) = 5

3(3) - (2) = 7

The solution is (3, 2). What do you think the answer

would be if you solved using substitution?

x + y = 5

3x – y = 7


4x + y = 7

4x – 2y = -2


Standard Form. They already are!



The x’s have the same

coefficient.


equations.

Subtract to eliminate x.

4x + y = 7

(-) 4x – 2y = -2

3y = 9

y = 3

Remember to

“keep-change-

change”

Add into your

spirals



the other variable.

4x + y = 7

4x + (3) = 7

4x = 4

x = 1

Step 5: Check your

solution.

(1, 3)

4(1) + (3) = 7

4(1) - 2(3) = -2

4x + y = 7

4x – 2y = -2


y = 7 – 2x

4x + y = 5


Standard Form.

2x + y = 7

4x + y = 5



The y’s have the same

coefficient.


equations.

Subtract to eliminate y.

2x + y = 7

(-) 4x + y = 5

-2x = 2

x = -1

Add into your

spirals



the other variable.

y = 7 – 2x

y = 7 – 2(-1)

y = 9

Step 5: Check your

solution.

(-1, 9)

(9) = 7 – 2(-1)

4(-1) + (9) = 5

y = 7 – 2x

4x + y = 5

Example 1 (In note packet pg 10)


Standard Form.




equations.

x - 2y = 14

x + 3y = 9

Both equations are already in

Standard Form.

We will eliminate the variable “x”

Since they both have same coefficient.

Subtraction is easiest, just remember this

Is like multiplying the entire equation by

-1.

x - 2y = 14

-( x + 3y = 9)

-5y = 5

y = -1

Example 1 continued…


the other variable.

Step 5: Check your

solution.

Since y = -1, we will substitute in

x - 2y = 14

x- 2 (-1) = 14 Therefore

x+2=14 my solution

x=12 is ( 12, -1)

Substitute ( 12, -1) into:

x - 2y = 14

12 – 2(-1) = 14

12+2 = 14

14 = 14

x - 2y = 14

x + 3y = 9

Example 2 (In note packet pg 11)


Standard Form.




equations.

4x + 3y = -1 5x + 4y = 1

Both equations are already in

Standard Form.

So for this one we need to multiply

each equation to create same

Coefficients. So we will multiply

EQ1 by 4 and EQ2 by -3,

then add to eliminate y.

4(4x + 3y = -1) -3(5x + 4y = 1)

16x + 12y = -4) -15x - 12y = -3) x = -7



the other variable.

Step 5: Check your

solution.

4x + 3y = -1

4(-7) + 3y = -1 Therefore

-28 +3y = -1 my solution is

3y = 27

y = 9 (-7, 9)

4x + 3y = -1 5x + 4y = 1

4(-7) +3(9) = -1 (-7) + 4(9) = 1

-28 + 27 = -1 -35 + 36 = 1

-1 = -1 (check!) 1 = 1

4x + 3y = -1 5x + 4y = 1

The third example we will do together in class.

Example 1:

Equation ‘a’: 2x - 4y = 13

Equation ‘b’: 4x - 5y = 8

Multiply equation ‘a’ by –2 to

eliminate the x’s:

Equation ‘a’: -2(2x - 4y = 13)

Equation ‘b’: 4x - 5y = 8

EXTRA EXAMPLES

EXTRA EXAMPLES

Extra Example 1, continued:

Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26

Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8

Add the equations (the x’s are

eliminated): -4x + 8y = -26

4x - 5y = 8

3y = -18

y = -6

EXTRA EXAMPLES

-22

4-11

2

Example 1, continued:

Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26

Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8

Substitute y = -6 into either equation:

-11

2

4x - 5(-6) = 8

4x + 30 = 8

4x = -22

x =

x =

Solution: ( , -6)

EXTRA EXAMPLES

Extra Example 2: Equation ‘a’: -9x + 6y = 0

Equation ‘b’: -12x + 8y = 0

Equation ‘a’: - 4(-9x + 6y = 0)

Equation ‘b’: 3(-12x + 8y = 0)

Multiply equation ‘a’ by –4 and equation ‘b’

by 3 to eliminate the x’s:

36x - 24y = 0

-36x + 24y = 0

0 = 0

Equation ‘a’: - 4(-9x + 6y = 0)

Equation ‘b’: 3(-12x + 8y = 0)

What does this answer mean?

Is it true?


EXTRA EXAMPLES

36x - 24y = 0

-36x + 24y = 0

0 = 0

Example 2, continued:

When both variables are eliminated,

if the statement is TRUE (like 0 = 0), then they are the same lines and there are infinite solutions.

if the statement is FALSE (like 0 = 1), then they are parallel lines and there is no solution.

36x - 24y = 0

-36x + 24y = 0

0 = 0

Since 0 = 0 is TRUE, there are infinite solutions.

EXTRA EXAMPLES

Summary

When solving systems of linear equations using elimination, you

will sometimes need to multiply one or both equations by a factor

in order to get the same coefficients for a variable. This process is

very similar to getting a common denominator for fractions.

So to summarize, this is the most complex method of solving systems of

of equations. Although you just need to be sure to follow the steps to

ensure success.

Remember:

ASSIGNMENT

• Do your summary from the guided questions online.

• Do practice problems in note packet on page 12

• Do practice quiz on page 13.

• Section 6.3 pages 347-34

Do Problems 1-10; 33 & 34, and 43-45 9

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Solve Systems with Eliminationwhsalgebra1b.weebly.com/.../7/4/1/8/7418584/concept_6.3_eliminati… · Solving Systems of Equations So far, we have solved systems using graphing and