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Quiz Date 1/22/19 Change For version B #5

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1 Quiz Date 1/22/19 Change For version B #5 π’š=πŸπ’™+πŸ” π’š=πŸ“π’™
Pick up your homework from the back table Work on your quiz No EQ and No Warm-up. Change For version B #5 π’š=πŸπ’™+πŸ” π’š=πŸ“π’™ Essential Question: None Yes you can used your notes Warm Up: None When done, turn in your quiz and any work to the BACK TABLE Then, work on this week’s homework.

2 Tuesday 01/22/19 Homework solution
Solve the system of equation by any method 1. 𝑦=βˆ’ 1 2 π‘₯+2 𝑦=π‘₯+8 Solve by graphing because the equations are in slope intercept form of 𝑦=π‘šπ‘₯+𝑏 𝑦=βˆ’ 1 2 π‘₯+2 First equation Slope = βˆ’ y-intercept is 2 𝑦=π‘₯+8 Second equation Slope = y-intercept is 8 π‘Ίπ’π’π’–π’•π’Šπ’π’ ( βˆ’4, 4)

3 Tuesday 01/22/19 Homework solution
Solve the system of equation by any method 2. π‘¦βˆ’2π‘₯=3 2π‘¦βˆ’12=π‘₯ Solve by Substation because the x is already solve for Step 1: Determine the equation that is solve for π‘₯=2π‘¦βˆ’12 Step 2: Substitute the solve equation into the other equation π‘¦βˆ’2 2π‘¦βˆ’12 =3 Step 3: Simplify and solve for the variable π‘¦βˆ’4𝑦+24=3 Distribute βˆ’3𝑦+24=3 Combine the variable βˆ’3𝑦=βˆ’21 Subtract 24 to both sides 𝑦=7 Divide by -3 to both sides

4 WRONG An order pair is ( π‘₯, 𝑦) Correct order pair ( 2,7)
Step 4: Substitute the solved variable back into the original to get the other variable π‘¦βˆ’2π‘₯=3 First equation (7)βˆ’2π‘₯=3 Substitute 𝑦=7 7βˆ’2π‘₯=3 βˆ’2π‘₯=βˆ’4 Subtract 7 to both sides π‘₯=2 Divide by -2 to both sides The solution of a system is always in the format of an order pair (7,2) WRONG An order pair is ( π‘₯, 𝑦) Correct order pair ( 2,7)

5 Tuesday 01/22/19 Homework solution
Solve the system of equation by any method 3. 𝑦=2π‘₯+3 𝑦= 1 2 π‘₯+6 Solve by graphing because the equations are in slope intercept form of 𝑦=π‘šπ‘₯+𝑏 𝑦=2π‘₯+3 First equation Slope = y-intercept is 3 𝑦= 1 2 π‘₯+6 First equation Slope = y-intercept is 6 π‘Ίπ’π’π’–π’•π’Šπ’π’ ( 2, 7)

6 5.3 Solve System by Elimination Date 1/23/19
Copy down Essential Question. Work on Warm Up. Essential Question How would you describe the process of solving for a system using Elimination? Warm Up: Explain the different between the two things. π‘₯+3𝑦=βˆ’2 π‘₯=3𝑦+16 π‘₯+3𝑦=βˆ’2 π‘₯βˆ’3𝑦=16 Solve by substitution DON’T Solve by substitution

7 Understand when to solve by Elimination
π‘₯+3𝑦=βˆ’2 π‘₯=3𝑦+16 π‘₯+3𝑦=βˆ’2 π‘₯βˆ’3𝑦=16 Solve by substitution DON’T Solve by substitution 𝒙 is already solve for notice π‘₯=3𝑦+16 nothing is solve for For this we use Elimination

8 How to solve by Elimination
π‘₯+3𝑦=βˆ’2 π‘₯βˆ’3𝑦=16 Step 1: check that the coefficient of the one of the variable are opposites. π‘₯+3𝑦=βˆ’2 π‘₯βˆ’3𝑦=16 The coefficient of the y are opposites Step 2: Add the two equations(one variable should disappear) . One equation with one unknown

9 Step 3: Solve for the variable you have left Step 4: Substitute the solved variable back into one of the original equation to solve for other variable. The solution needs to be in a order pair.

10 Checking your answer The solution is (7, βˆ’3)
Equation 1 π‘₯+3𝑦=βˆ’2 (7)+3(βˆ’3)=βˆ’2 βˆ’2=βˆ’2 Equation 2 π‘₯βˆ’3𝑦=16 7 βˆ’3(βˆ’3)=16 16=16

11 Wednesday 01/23/19 Homework solution
Solve the system of equation by elimination 1. 3𝑦+2π‘₯=6 5π‘¦βˆ’2π‘₯=10 Solve by elimination because the coefficient of x are opposite Step 1: check that the coefficient of the one of the variable are opposites. 3𝑦+2π‘₯=6 5π‘¦βˆ’2π‘₯=10 Step 2: Add the two equations together (one variable should disappear) 3𝑦+2π‘₯=6 5π‘¦βˆ’2π‘₯=10 8𝑦 =16 Step 3: Simplify and solve for the variable 8𝑦=16 𝑦=2 Divide by 8 to both sides

12 WRONG An order pair is ( π‘₯, 𝑦) Correct order pair ( 0, 2)
Step 4: Substitute the solved variable back into the original to get the other variable 3𝑦+2π‘₯=6 First equation 3(2)βˆ’2π‘₯=6 Substitute 𝑦=2 6βˆ’2π‘₯=6 βˆ’2π‘₯=0 Subtract 6 to both sides π‘₯=0 Divide by -2 to both sides The solution of a system is always in the format of an order pair (2,0) WRONG An order pair is ( π‘₯, 𝑦) Correct order pair ( 0, 2)

13 Wednesday 01/23/19 Homework solution
Solve the system of equation by elimination 2. 5𝑦+4π‘₯=22 βˆ’12π‘¦βˆ’4π‘₯=βˆ’36 Solve by elimination because the coefficient of x are opposite Step 1: check that the coefficient of the one of the variable are opposites. 5𝑦+4π‘₯=22 βˆ’12π‘¦βˆ’4π‘₯=βˆ’36 Step 2: Add the two equations together (one variable should disappear) 5𝑦 π‘₯ = 22 βˆ’12π‘¦βˆ’4π‘₯ =βˆ’36 βˆ’7𝑦 =βˆ’14 Step 3: Simplify and solve for the variable βˆ’7𝑦=βˆ’14 𝑦=2 Divide by -7 to both sides

14 WRONG An order pair is ( π‘₯, 𝑦) Correct order pair ( 3, 2)
Step 4: Substitute the solved variable back into the original to get the other variable 5𝑦+4π‘₯=22 First equation π‘₯=22 Substitute 𝑦=2 10+4π‘₯=22 4π‘₯=12 Subtract 10 to both sides π‘₯=3 Divide by 4 to both sides The solution of a system is always in the format of an order pair (2,3) WRONG An order pair is ( π‘₯, 𝑦) Correct order pair ( 3, 2)

15 Wednesday 01/23/19 Homework solution
Solve the system of equation by elimination 3. 3π‘₯βˆ’π‘¦=5 π‘₯+𝑦=3 Solve by elimination because the coefficient of y are opposite Step 1: check that the coefficient of the one of the variable are opposites. 3π‘₯βˆ’π‘¦=5 π‘₯+𝑦=3 Step 2: Add the two equations together (one variable should disappear) 3π‘₯ βˆ’ 𝑦 = 5 π‘₯ + 𝑦 = 3 4π‘₯ =8 Step 3: Simplify and solve for the variable 4π‘₯=8 π‘₯=2 Divide by 4 to both sides

16 Correct order pair ( 2, 1) An order pair is ( π‘₯, 𝑦)
Step 4: Substitute the solved variable back into the original to get the other variable π‘₯+𝑦=3 second equation (2)+𝑦=3 Substitute x=2 2+𝑦=3 y=1 Subtract 2 to both sides The solution of a system is always in the format of an order pair Correct order pair ( 2, 1) An order pair is ( π‘₯, 𝑦)

17 5.3 Solve System by Elimination Day 2 Date 1/24/19
Copy down Essential Question. Work on Warm Up. Fix error on homework Essential Question How is solving a system using elimination different that solving using substitution? Warm Up: Add the two equation in each problem. π‘₯βˆ’2𝑦 = βˆ’19 5π‘₯+2𝑦 = π‘₯+4𝑦 =18 βˆ’2π‘₯+4𝑦 = 8 6π‘₯ =βˆ’18 π‘₯+8𝑦=26

18 Describe the process of solving a system by Elimination in your own words
3π‘₯+2𝑦=4 3π‘₯βˆ’2𝑦=βˆ’4 Step 1: Step 2:

19 Step 3: Step 4: The solution needs to be in a order pair.

20 Practice on solving by Elimination
2π‘₯βˆ’π‘¦=9 4π‘₯+𝑦=21 Step 1: check that the coefficient of the one of the variable are opposites. 2π‘₯βˆ’1𝑦=9 4π‘₯+1𝑦=21 The coefficient of the y are opposites Step 2: Add the two equations(one variable should disappear) .

21 Step 3: Solve for the variable you have left Step 4: Substitute the solved variable back into one of the original equation to solve for other variable. The solution needs to be in a order pair. The solution is (5, 1)

22 How to solve by Elimination when there is the same number but they are not opposite
3π‘₯+4𝑦=18 βˆ’2π‘₯+4𝑦=8

23 How to solve by Elimination when there is the same number but they are not opposite
Step 1: check that the coefficient of the one of the variable are opposites. 3π‘₯+4𝑦=18 βˆ’2π‘₯+4𝑦=8 Step 1b: 3π‘₯+4𝑦=18 βˆ’1(βˆ’2π‘₯+4𝑦=8) Result 3π‘₯+4𝑦=18 2π‘₯βˆ’4𝑦=βˆ’8

24 3π‘₯+ 4𝑦 = 18 2π‘₯ βˆ’4𝑦 =βˆ’8 5π‘₯ =10 (2, 3) Solution An order pair is ( π‘₯, 𝑦)
Step 2: Add the two equations together (one variable should disappear) 3π‘₯+ 4𝑦 = π‘₯ βˆ’4𝑦 =βˆ’8 5π‘₯ =10 Step 3: Simplify and solve for the variable 5π‘₯=10 π‘₯=2 Divide by 5 to both sides Step 4: Substitute the solved variable back into the original to get the other variable 3π‘₯+4𝑦=18 First equation 3(2)+4𝑦=18 Substitute x=2 6+4𝑦=18 4𝑦=12 Subtract 6 to both sides 𝑦=3 Divide by 4 to both sides The solution of a system is always in the format of an order pair (2, 3) Solution An order pair is ( π‘₯, 𝑦)

25 Practice on solving a system by Elimination
3π‘₯+3𝑦=15 βˆ’2π‘₯+3𝑦=βˆ’5

26 Practice on solving a system by Elimination
Step 1: check that the coefficient of the one of the variable are opposites. 3π‘₯+3𝑦=15 βˆ’2π‘₯+3𝑦=βˆ’5 Step 1b: 3π‘₯+3𝑦=15 βˆ’1(βˆ’2π‘₯+3𝑦=βˆ’5) Result 3π‘₯+3𝑦=15 2π‘₯βˆ’3𝑦=5

27 3π‘₯+3𝑦 = 15 2π‘₯βˆ’3𝑦 = 5 5π‘₯ =20 (4, 1) Solution An order pair is ( π‘₯, 𝑦)
Step 2: Add the two equations together (one variable should disappear) 3π‘₯+3𝑦 = π‘₯βˆ’3𝑦 = 5 5π‘₯ =20 Step 3: Simplify and solve for the variable 5π‘₯=20 π‘₯=4 Divide by 5 to both sides Step 4: Substitute the solved variable back into the original to get the other variable 3π‘₯+3𝑦=15 First equation 3(4)+3𝑦=15 Substitute x=4 12+3𝑦=15 3𝑦=3 Subtract 12 to both sides 𝑦=1 Divide by 4 to both sides The solution of a system is always in the format of an order pair (4, 1) Solution An order pair is ( π‘₯, 𝑦)

28 Thursday 01/24/19 Homework solution
Solve the system of equation by elimination 1. 7π‘₯+4𝑦=2 9π‘₯βˆ’4𝑦=30 Solve by elimination because the coefficient of y are opposite Step 1: check that the coefficient of the one of the variable are opposites. 7π‘₯+4𝑦=2 9π‘₯βˆ’4𝑦=30 Step 2: Add the two equations together (one variable should disappear) 7π‘₯+ 4𝑦 = 2 9π‘₯ βˆ’4𝑦 = π‘₯ =32 Step 3: Simplify and solve for the variable 16π‘₯=32 π‘₯=2 Divide by 16 to both sides

29 Correct order pair (2, βˆ’3) An order pair is ( π‘₯, 𝑦)
Step 4: Substitute the solved variable back into the original to get the other variable 7π‘₯+4𝑦=2 First equation 7(2)+4𝑦=2 Substitute x=2 14+4𝑦=2 4𝑦=βˆ’12 Subtract 14 to both sides 𝑦=βˆ’3 Divide by 4 to both sides The solution of a system is always in the format of an order pair Correct order pair (2, βˆ’3) An order pair is ( π‘₯, 𝑦)

30 Thursday 01/24/19 Homework solution
Solve the system of equation by elimination 2. 3π‘₯βˆ’4𝑦=βˆ’5 5π‘₯βˆ’2𝑦=βˆ’6 Solve by elimination because the coefficient of y are opposite Step 1: check that the coefficient of one of the variables are opposites. They are not, so we need to make one of them something else 3π‘₯βˆ’4𝑦=βˆ’5 βˆ’πŸβˆ™(5π‘₯βˆ’2𝑦=βˆ’6) Multiple the second equation by -2 Step 1 again: check that the coefficient of the one of the variables are opposites. 3π‘₯βˆ’4𝑦=βˆ’5 βˆ’10π‘₯+4𝑦=12) We are good now Step 2: Add the two equations together (one variable should disappear) 3π‘₯ + βˆ’4𝑦 = βˆ’5 βˆ’10π‘₯ +4𝑦 =12 βˆ’7π‘₯ =7 Step 3: Simplify and solve for the variable βˆ’7π‘₯=7 π‘₯=βˆ’1 Divide by -7 to both sides

31 Correct order pair (βˆ’1, 1 2 ) An order pair is ( π‘₯, 𝑦)
Step 4: Substitute the solved variable back into the original to get the other variable 3π‘₯βˆ’4𝑦=βˆ’5 First equation 3(βˆ’1)βˆ’4𝑦=βˆ’5 Substitute x=βˆ’1 βˆ’3βˆ’4𝑦=βˆ’5 βˆ’4𝑦=βˆ’2 Add 3 to both sides 𝑦= 2 4 π‘œπ‘Ÿ 1 2 Divide by -4 to both sides The solution of a system is always in the format of an order pair Correct order pair (βˆ’1, 1 2 ) An order pair is ( π‘₯, 𝑦)

32 Thursday 01/24/19 Homework solution
Solve the system of equation by elimination 3. 2π‘₯+3𝑦=8 3π‘₯+2𝑦=7 Solve by elimination because the coefficient of y are opposite Step 1: check that the coefficient of one of the variables are opposites. They are not, so we need to make both of them something else πŸ‘βˆ™ 2π‘₯+3𝑦=8 Multiple the first equation by 3 βˆ’πŸβˆ™ 3π‘₯+2𝑦=7 Multiple the second equation by βˆ’2 Step 1 again: check that the coefficient of the one of the variables are opposites. 6π‘₯+9𝑦=24 βˆ’6π‘₯βˆ’4𝑦=βˆ’14 We are good now Step 2: Add the two equations together (one variable should disappear) 6π‘₯+9𝑦=24 βˆ’6π‘₯βˆ’4𝑦=βˆ’14 5𝑦 =10 Step 3: Simplify and solve for the variable 5𝑦=10 𝑦=2 Divide by 5 to both sides

33 Correct order pair (1, 2) An order pair is ( π‘₯, 𝑦)
Step 4: Substitute the solved variable back into the original to get the other variable 2π‘₯+3𝑦=8 First equation 2π‘₯+3(2)=8 Substitute y=2 2π‘₯+6=8 2π‘₯=2 Subtract 6 to both sides π‘₯=1 Divide by 2 to both sides The solution of a system is always in the format of an order pair Correct order pair (1, 2) An order pair is ( π‘₯, 𝑦)

34 5.3 Solve System by Elimination Day 3 Date 1/25/19
Turn in your homework to the back table. Copy down Essential Question. Work on Warm Up. Essential Question Why do the coefficient in a system of equation need to be opposite values of each other? Warm Up: Add the two equation in each problem. 1. βˆ’π‘₯ +𝑦=5 π‘₯βˆ’5𝑦=βˆ’9 2. π‘₯βˆ’10𝑦=60 π‘₯+14𝑦=12 3. 2π‘₯+3𝑦=12 5π‘₯βˆ’π‘¦=13 βˆ’4𝑦=βˆ’4 2π‘₯+4𝑦=72 7π‘₯+2𝑦=25

35 Exploration activity: Solving by Elimination
2. 1π‘₯βˆ’10𝑦=60 1π‘₯+14𝑦=12 2x + 4y = 72 2. π‘₯βˆ’10𝑦=60 π‘₯+14𝑦=12 2x + 4y = 72 3. 2π‘₯+3𝑦=12 5π‘₯βˆ’1𝑦=13 7x + 2y = 25 3. 2π‘₯+3𝑦=12 5π‘₯βˆ’1𝑦=13 7x + 2y = 25 Step 1: check that the coefficient of the one of the variable are opposites. Step 1: check that the coefficient of the one of the variable are opposites. Step 1b: multiple by a -1 to get the opposite value. Step 1b: multiple by a -1 to get the opposite value. 2π‘₯+3𝑦=12 βˆ’1βˆ™(5π‘₯βˆ’1𝑦=13) 1π‘₯βˆ’10𝑦=60 βˆ’1βˆ™(1π‘₯+14𝑦=12) 2π‘₯+3𝑦=12 βˆ’5π‘₯+1𝑦=βˆ’13 2π‘₯+3𝑦=12 βˆ’5π‘₯+1𝑦=βˆ’13 1π‘₯βˆ’10𝑦=60 βˆ’1π‘₯βˆ’14𝑦=βˆ’12 1π‘₯βˆ’10𝑦=60 βˆ’1π‘₯βˆ’14𝑦=βˆ’12 Step 1: check that the coefficient of the one of the variable are opposites. Step 1: check that the coefficient of the one of the variable are opposites.

36 How to Solve by Elimination by multiplying first
2π‘₯+3𝑦=12 5π‘₯βˆ’1𝑦=13 Step 1: check that the coefficient of the one of the variable are opposites. 2π‘₯+3𝑦=12 5π‘₯βˆ’1𝑦=13 Step 1b: multiple by (something) to get the opposite value. 2π‘₯+3𝑦=12 3βˆ™(5π‘₯βˆ’π‘¦=13) Multiple by 3 to the second equation 2π‘₯+3𝑦=12 15π‘₯βˆ’3𝑦=39 Step 1: check that the coefficient of the one of the variable are opposites. 2π‘₯+3𝑦=12 15π‘₯βˆ’3𝑦=39

37 2π‘₯ + 3𝑦 = 12 15π‘₯ βˆ’ 3𝑦 = 39 17π‘₯ = 51 (3, 2) Solution
Step 2: Add the two equations together (one variable should disappear) 2π‘₯ + 3𝑦 = 12 15π‘₯ βˆ’ 3𝑦 = π‘₯ = 51 Step 3: Simplify and solve for the variable 17π‘₯=51 π‘₯=3 Divide by 17 to both sides Step 4: Substitute the solved variable back into the original to get the other variable 2π‘₯+3𝑦=12 First equation 2(3)+3𝑦=12 Substitute x=3 6+3𝑦=12 3𝑦=6 Subtract 6 to both sides 𝑦=2 Divide by 3 to both sides The solution of a system is always in the format of an order pair (3, 2) Solution An order pair is ( π‘₯, 𝑦)

38 or Practice on solving by Elimination by multiplying first
2π‘₯+𝑦=3 π‘₯βˆ’3𝑦=12 Step 1: check that the coefficient of the one of the variable are opposites. 2π‘₯+𝑦=3 1π‘₯βˆ’3𝑦=12 or 2π‘₯+1𝑦=3 1π‘₯βˆ’3𝑦=12 Step 1b: multiple by (something) to get the opposite value. 2π‘₯+𝑦=3 βˆ’2βˆ™(1π‘₯βˆ’3𝑦=12) Multiple by -2 to the second equation 2π‘₯+𝑦=3 βˆ’2π‘₯+6𝑦=βˆ’24 Step 1: check that the coefficient of the one of the variable are opposites. 2π‘₯+𝑦=3 βˆ’2π‘₯+6𝑦=βˆ’24

39 2π‘₯ + 𝑦 =3 βˆ’2π‘₯ + 6𝑦 =βˆ’24 7𝑦=βˆ’21 (3, βˆ’3) Solution
Step 2: Add the two equations together (one variable should disappear) 2π‘₯ + 𝑦 =3 βˆ’2π‘₯ + 6𝑦 =βˆ’24 7𝑦=βˆ’21 Step 3: Simplify and solve for the variable 7𝑦=βˆ’21 𝑦=βˆ’3 Divide by 7 to both sides Step 4: Substitute the solved variable back into the original to get the other variable 2π‘₯+𝑦=3 First equation 2π‘₯+ βˆ’3 =3 Substitute y=βˆ’3 2xβˆ’3 =3 2π‘₯=6 Add 3 to both sides π‘₯=3 Divide by 2 to both sides The solution of a system is always in the format of an order pair (3, βˆ’3) Solution An order pair is ( π‘₯, 𝑦)

40 Math Talk Which is incorrect? Explain the error


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