TODAY IN ALGEBRA…



Warm Up: Solving a system by Elimination



Learning Goal: 7.4 You will solve systems of linear equations by Elimination with multiplication first



Independent Practice



MID CH.7 TEST-MONDAY/TUESDAY!

WARM UP: Solve the systems using ELIMINATION.

1. 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 4𝑥 + 3𝑦 = 2 5𝑥 + 3𝑦 = −2 2. 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 − 2𝑥 + 𝑦 = −8 2𝑥 − 3𝑦 = 8

WARM UP: Solve the systems using ELIMINATION.

1. Solve the linear system STEP 1: Move all variables to one side.

All variables on one side already!

− Same signs: SUBTRACT 4𝑥 − 3𝑦 = 15 2𝑥 − 3𝑦 = 9 2𝑥 = 6 STEP 2: Solve for x: 2𝑥 = 6 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 4 4𝑥 − 3𝑦 = 15 2𝑥 − 3𝑦 = 9 STEP 3: Substitute into 𝑬𝒒𝒏 𝟏 𝑥 = 3 𝑬𝒒𝒏 𝟏 4𝑥 − 3𝑦 = 15 3 − 3𝑦 = 15 12 − 3𝑦 = 15 −12 − 12 −3𝑦 = 3 2 𝒙 = 𝟑 2 −3 −3 𝒚 = −𝟏

SOLUTION:

(𝟑, −𝟏)

WARM UP: Solve the systems using ELIMINATION.

2. Solve the linear system STEP 1: Move all variables to one side.

All variables on one side already!

Different signs: ADD 𝑬𝒒𝒏 𝟏 − 2𝑥 + 𝑦 = −8 𝑬𝒒𝒏 𝟐 2𝑥 − 2𝑦 = 8 STEP 3: Substitute into 𝑬𝒒𝒏 𝟏 𝑦 = 0 𝑬𝒒𝒏 𝟏 − 2𝑥 + 𝑦 = −8 + −2𝑥 + 𝑦 = −8 2𝑥 − 2𝑦 = 8 −𝑦 = 0 STEP 2: Solve for x: −𝑦 = 0 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 −2𝑥 + 0 = −8 −2𝑥 = −8 −2 −2 𝒙 = 𝟒 −1 𝒚 = 𝟎 −1

SOLUTION:

(𝟑, −𝟏)

7.3 SOLVING LINEAR SYSTEMS BY ELIMINATION

STEPS FOR SOLVING A LINEAR SYSTEM BY ELIMINATION WITH ELIMINATION FIRST: STEP 1: Align the variables with x and y on the same side of each equation. Choose a variable to eliminate and multiply each equation by the coefficients. Then ADD or SUBTRACT the equations.

STEP 2: SOLVE the equation for the remaining variable.

STEP 3: Substitute the solution from STEP 2 into either equation to solve for given variable.

THIS IS MUCH EASIER WITH AN EXAMPLE!

7.4 ELIMINATION WITH MULTIPLICATION FIRST

EXAMPLE 1: Solve the linear system 𝑬𝒒𝒏 𝟏 6𝑥 + 5𝑦 = 19 𝑬𝒒𝒏 𝟐 2𝑥 + 3𝑦 = 5 STEP 1: Move all variables to one side.

All variables on one side already!

NOTICE: there are no matching coefficients. Choose one of variable to eliminate Multiply opposite coefficients on each equation.

𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 I’ll eliminate the x’s first 𝟐( 6𝑥 + 5𝑦 = 19 𝟔( 2𝑥 + 3𝑦 = 5 ) ) 12𝑥 + 10𝑦 = 38 12𝑥 + 18𝑦 = 30 Now I have matching coefficients!

Solve this system using elimination.

7.3 SOLVING LINEAR SYSTEMS BY ELIMINATION

EXAMPLE 1 Continued:

STEP 1:

Same signs: SUBTRACT − 12𝑥 + 10𝑦 = 38 12𝑥 + 18𝑦 = 30 −8𝑦 = 8 STEP 2: Solve for y: −8𝑦 = 8 −8 −8 𝒚 = −𝟏 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 STEP 3: Substitute 𝑦 = −1 into 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟏 12𝑥 + 10𝑦 = 38 12𝑥 + 10 −1 = 38 12𝑥 − 10 = 38 +10 + 10 12𝑥 = 48 12 12 𝒙 = 𝟒

SOLUTION:

(𝟒, −𝟏)

7.4 ELIMINATION WITH MULTIPLICATION FIRST

EXAMPLE 2: Solve the linear system 𝑬𝒒𝒏 𝟏 2𝑥 − 9𝑦 = 1 𝑬𝒒𝒏 𝟐 7𝑥 − 12𝑦 = 23 STEP 1: Move all variables to one side.

All variables on one side already!

NOTICE: there are no matching coefficients. Choose one of variable to eliminate Multiply opposite coefficients on each equation.

𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 I’ll eliminate the x’s first 𝟕( 2𝑥 − 9𝑦 = 1 ) 𝟐( 7𝑥 − 12𝑦 = 23 ) 14𝑥 − 63𝑦 = 7 14𝑥 − 24𝑦 = 46 Now I have matching coefficients!

Solve this system using elimination.

7.3 SOLVING LINEAR SYSTEMS BY ELIMINATION

EXAMPLE 2 Continued:

STEP 1:

Same signs: SUBTRACT 14𝑥 − 63𝑦 = 7 − 14𝑥 − 24𝑦 = 46 −39𝑦 = −39 STEP 2: Solve for y: −39𝑦 = −39 −39 𝒚 = 𝟏 −39 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 STEP 3: Substitute 𝑦 = 1 into 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟏 14𝑥 − 63𝑦 = 7 14𝑥 − 63 1 = 7 14𝑥 − 63 = 7 +63 + 63 14𝑥 = 70 14 14 𝒙 = 𝟓

SOLUTION:

(𝟓, 𝟏)

7.4 ELIMINATION WITH MULTIPLICATION FIRST

EXAMPLE 3: Solve the linear system 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 2𝑥 + 𝑦 = −9 4𝑥 + 11𝑦 = 9 STEP 1: Move all variables to one side.

All variables on one side already!

NOTICE: there are no matching coefficients. Choose one of variable to eliminate Multiply opposite coefficients on each equation.

I’ll eliminate the y’s first Sometimes you only need to multiply one equation!

𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 𝟏𝟏( 2𝑥 + 𝑦 = −9 ) 4𝑥 + 11𝑦 = 9 22𝑥 + 11𝑦 = −99 4𝑥 + 11𝑦 = 9 Now I have matching coefficients!

Solve this system using elimination.

7.3 SOLVING LINEAR SYSTEMS BY ELIMINATION

EXAMPLE 3 Continued:

STEP 1:

Same signs: SUBTRACT − 22𝑥 + 11𝑦 = −99 𝑬𝒒𝒏 𝟏 4𝑥 + 11𝑦 = 9 𝑬𝒒𝒏 𝟐 18𝑥 = −108 STEP 2: Solve for x: 18𝑦 = −108 18 18 𝒙 = −𝟔 STEP 3: Substitute 𝑥 = −6 into 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟏 22𝑥 + 11𝑦 = −99 22 −6 + 11𝑦 = −99 −132 + 11𝑦 = −99 +132 + 132 11𝑦 = 33 11 11 𝒚 = 𝟑

SOLUTION:

(−𝟔, 𝟑)

7.4 ELIMINATION WITH MULTIPLICATION FIRST

EXAMPLE 4: Solve the linear system 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 3𝑥 − 7𝑦 = 5 9𝑦 = 5𝑥 + 5 STEP 1: Move all variables to one side for 𝑬𝒒𝒏 𝟐 .

𝑬𝒒𝒏 𝟐 9𝑦 = 5𝑥 + 5 −5𝑥 − 5𝑥 −5𝑥 + 9𝑦 = 5 NOTICE: there are no matching coefficients. Choose one of variable to eliminate Multiply opposite coefficients on each equation.

I’ll eliminate the x’s first 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 −𝟓( 3𝑥 − 7 = 5 𝟑( − 5𝑥 + 9𝑦 = 5 ) ) −15𝑥 + 35𝑦 = −25 −15𝑥 + 27𝑦 = 15 Now I have matching coefficients!

Solve this system using elimination.

7.3 SOLVING LINEAR SYSTEMS BY ELIMINATION

EXAMPLE 4 Continued:

STEP 1:

Same signs: SUBTRACT −15𝑥 + 35𝑦 = −25 𝑬𝒒𝒏 𝟏 − −15𝑥 + 27𝑦 = 15 8𝑦 = −40 𝑬𝒒𝒏 𝟐 STEP 2: Solve for y: 8𝑦 = −40 8 8 𝒚 = −𝟓 STEP 3: Substitute 𝑦 = −5 into 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 − 15𝑥 + 35𝑦 = −25 −15𝑥 + 35 −5 = −25 −15𝑥 − 175 = −25 +175 + 175 −15𝑥 = 150 −15 −15 𝒙 = −𝟏𝟎

SOLUTION:

(−𝟏𝟎, −𝟓)

7.4 ELIMINATION WITH MULTIPLICATION FIRST

EXAMPLE 5: Write two linear equations that represent the situation, then solve the system.

There’s a sale on soccer balls. A coach purchases 10 soccer balls and 2 soccer bags for $155. Another soccer coach purchases 12 soccer balls and 3 soccer bags for $189.

Find the cost of a soccer ball and the cost of a soccer bag.

Assign the variables: x – cost of soccer balls y – cost of soccer bags 𝑬𝒒𝒏 𝟏 10𝑥 + 2𝑦 = 155 𝑬𝒒𝒏 𝟐 12𝑥 + 3𝑦 = 189

7.4 ELIMINATION WITH MULTIPLICATION FIRST

EXAMPLE 2: Solve the linear system 𝑬𝒒𝒏 𝟏 10𝑥 + 2𝑦 = 155 𝑬𝒒𝒏 𝟐 12𝑥 + 3𝑦 = 189 STEP 1: Move all variables to one side.

All variables on one side already!

NOTICE: there are no matching coefficients. Choose one of variable to eliminate Multiply opposite coefficients on each equation.

𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 I’ll eliminate the y’s first 𝟑( 10𝑥 + 2𝑦 = 155 ) 𝟐( 12𝑥 + 3𝑦 = 189 ) 30𝑥 + 6𝑦 = 465 24𝑥 + 6𝑦 = 378 Now I have matching coefficients!

Solve this system using elimination.

7.3 SOLVING LINEAR SYSTEMS BY ELIMINATION

EXAMPLE 4 Continued:

STEP 1:

Same signs: SUBTRACT 30𝑥 + 6𝑦 = 465 − 24𝑥 + 6𝑦 = 378 6𝑥 = 87 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟐 STEP 2: Solve for x: 6𝑥 = 87 6 6 𝒙 = 𝟏𝟒. 𝟓 STEP 3: Substitute 𝑥 = 14.5

into 𝑬𝒒𝒏 𝟏 𝑬𝒒𝒏 𝟏 30 30𝑥 + 6𝑦 = 465 14.5

+ 6𝑦 = 465 435 + 6𝑦 = 465 −435 − 435 6𝑦 = 30 6 6 𝒚 = 𝟓

x: soccer balls cost $14.50 each, y: bags cost $5.00 each

HOMEWORK #4: Pg. 454: 4-26 even, 38

If finished, work on other assignments: HW #1: Pg. 430: 3-10, 12-15, 22, 23, 32 HW #2: Pg. 439: 4-24 even, 31 HW #3: Pg. 447: 4-22 even