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Quiz Date 1/22/19 Change For version B #5 π=ππ+π π=ππ
Pick up your homework from the back table Work on your quiz No EQ and No Warm-up. Change For version B #5 π=ππ+π π=ππ 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 weekβs homework.
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Tuesday 01/22/19 Homework solution
Solve the system of equation by any method 1. π¦=β 1 2 π₯+2 π¦=π₯+8 Solve by graphing because the equations are in slope intercept form of π¦=ππ₯+π π¦=β 1 2 π₯+2 First equation Slope = β y-intercept is 2 π¦=π₯+8 Second equation Slope = y-intercept is 8 πΊπππππππ ( β4, 4)
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Tuesday 01/22/19 Homework solution
Solve the system of equation by any method 2. π¦β2π₯=3 2π¦β12=π₯ Solve by Substation because the x is already solve for Step 1: Determine the equation that is solve for π₯=2π¦β12 Step 2: Substitute the solve equation into the other equation π¦β2 2π¦β12 =3 Step 3: Simplify and solve for the variable π¦β4π¦+24=3 Distribute β3π¦+24=3 Combine the variable β3π¦=β21 Subtract 24 to both sides π¦=7 Divide by -3 to both sides
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WRONG An order pair is ( π₯, π¦) Correct order pair ( 2,7)
Step 4: Substitute the solved variable back into the original to get the other variable π¦β2π₯=3 First equation (7)β2π₯=3 Substitute π¦=7 7β2π₯=3 β2π₯=β4 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₯, π¦) Correct order pair ( 2,7)
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Tuesday 01/22/19 Homework solution
Solve the system of equation by any method 3. π¦=2π₯+3 π¦= 1 2 π₯+6 Solve by graphing because the equations are in slope intercept form of π¦=ππ₯+π π¦=2π₯+3 First equation Slope = y-intercept is 3 π¦= 1 2 π₯+6 First equation Slope = y-intercept is 6 πΊπππππππ ( 2, 7)
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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₯+3π¦=β2 π₯=3π¦+16 π₯+3π¦=β2 π₯β3π¦=16 Solve by substitution DONβT Solve by substitution
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Understand when to solve by Elimination
π₯+3π¦=β2 π₯=3π¦+16 π₯+3π¦=β2 π₯β3π¦=16 Solve by substitution DONβT Solve by substitution π is already solve for notice π₯=3π¦+16 nothing is solve for For this we use Elimination
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How to solve by Elimination
π₯+3π¦=β2 π₯β3π¦=16 Step 1: check that the coefficient of the one of the variable are opposites. π₯+3π¦=β2 π₯β3π¦=16 The coefficient of the y are opposites Step 2: Add the two equations(one variable should disappear) . One equation with one unknown
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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.
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Checking your answer The solution is (7, β3)
Equation 1 π₯+3π¦=β2 (7)+3(β3)=β2 β2=β2 Equation 2 π₯β3π¦=16 7 β3(β3)=16 16=16
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Wednesday 01/23/19 Homework solution
Solve the system of equation by elimination 1. 3π¦+2π₯=6 5π¦β2π₯=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. 3π¦+2π₯=6 5π¦β2π₯=10 Step 2: Add the two equations together (one variable should disappear) 3π¦+2π₯=6 5π¦β2π₯=10 8π¦ =16 Step 3: Simplify and solve for the variable 8π¦=16 π¦=2 Divide by 8 to both sides
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WRONG An order pair is ( π₯, π¦) Correct order pair ( 0, 2)
Step 4: Substitute the solved variable back into the original to get the other variable 3π¦+2π₯=6 First equation 3(2)β2π₯=6 Substitute π¦=2 6β2π₯=6 β2π₯=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₯, π¦) Correct order pair ( 0, 2)
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Wednesday 01/23/19 Homework solution
Solve the system of equation by elimination 2. 5π¦+4π₯=22 β12π¦β4π₯=β36 Solve by elimination because the coefficient of x are opposite Step 1: check that the coefficient of the one of the variable are opposites. 5π¦+4π₯=22 β12π¦β4π₯=β36 Step 2: Add the two equations together (one variable should disappear) 5π¦ π₯ = 22 β12π¦β4π₯ =β36 β7π¦ =β14 Step 3: Simplify and solve for the variable β7π¦=β14 π¦=2 Divide by -7 to both sides
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WRONG An order pair is ( π₯, π¦) Correct order pair ( 3, 2)
Step 4: Substitute the solved variable back into the original to get the other variable 5π¦+4π₯=22 First equation π₯=22 Substitute π¦=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₯, π¦) Correct order pair ( 3, 2)
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Wednesday 01/23/19 Homework solution
Solve the system of equation by elimination 3. 3π₯βπ¦=5 π₯+π¦=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₯βπ¦=5 π₯+π¦=3 Step 2: Add the two equations together (one variable should disappear) 3π₯ β π¦ = 5 π₯ + π¦ = 3 4π₯ =8 Step 3: Simplify and solve for the variable 4π₯=8 π₯=2 Divide by 4 to both sides
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Correct order pair ( 2, 1) An order pair is ( π₯, π¦)
Step 4: Substitute the solved variable back into the original to get the other variable π₯+π¦=3 second equation (2)+π¦=3 Substitute x=2 2+π¦=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₯, π¦)
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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₯β2π¦ = β19 5π₯+2π¦ = π₯+4π¦ =18 β2π₯+4π¦ = 8 6π₯ =β18 π₯+8π¦=26
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Describe the process of solving a system by Elimination in your own words
3π₯+2π¦=4 3π₯β2π¦=β4 Step 1: Step 2:
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Step 3: Step 4: The solution needs to be in a order pair.
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Practice on solving by Elimination
2π₯βπ¦=9 4π₯+π¦=21 Step 1: check that the coefficient of the one of the variable are opposites. 2π₯β1π¦=9 4π₯+1π¦=21 The coefficient of the y are opposites Step 2: Add the two equations(one variable should disappear) .
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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)
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How to solve by Elimination when there is the same number but they are not opposite
3π₯+4π¦=18 β2π₯+4π¦=8
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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₯+4π¦=18 β2π₯+4π¦=8 Step 1b: 3π₯+4π¦=18 β1(β2π₯+4π¦=8) Result 3π₯+4π¦=18 2π₯β4π¦=β8
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3π₯+ 4π¦ = 18 2π₯ β4π¦ =β8 5π₯ =10 (2, 3) Solution An order pair is ( π₯, π¦)
Step 2: Add the two equations together (one variable should disappear) 3π₯+ 4π¦ = π₯ β4π¦ =β8 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₯+4π¦=18 First equation 3(2)+4π¦=18 Substitute x=2 6+4π¦=18 4π¦=12 Subtract 6 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) Solution An order pair is ( π₯, π¦)
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Practice on solving a system by Elimination
3π₯+3π¦=15 β2π₯+3π¦=β5
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Practice on solving a system by Elimination
Step 1: check that the coefficient of the one of the variable are opposites. 3π₯+3π¦=15 β2π₯+3π¦=β5 Step 1b: 3π₯+3π¦=15 β1(β2π₯+3π¦=β5) Result 3π₯+3π¦=15 2π₯β3π¦=5
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3π₯+3π¦ = 15 2π₯β3π¦ = 5 5π₯ =20 (4, 1) Solution An order pair is ( π₯, π¦)
Step 2: Add the two equations together (one variable should disappear) 3π₯+3π¦ = π₯β3π¦ = 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₯+3π¦=15 First equation 3(4)+3π¦=15 Substitute x=4 12+3π¦=15 3π¦=3 Subtract 12 to both sides π¦=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₯, π¦)
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Thursday 01/24/19 Homework solution
Solve the system of equation by elimination 1. 7π₯+4π¦=2 9π₯β4π¦=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₯+4π¦=2 9π₯β4π¦=30 Step 2: Add the two equations together (one variable should disappear) 7π₯+ 4π¦ = 2 9π₯ β4π¦ = π₯ =32 Step 3: Simplify and solve for the variable 16π₯=32 π₯=2 Divide by 16 to both sides
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Correct order pair (2, β3) An order pair is ( π₯, π¦)
Step 4: Substitute the solved variable back into the original to get the other variable 7π₯+4π¦=2 First equation 7(2)+4π¦=2 Substitute x=2 14+4π¦=2 4π¦=β12 Subtract 14 to both sides π¦=β3 Divide by 4 to both sides The solution of a system is always in the format of an order pair Correct order pair (2, β3) An order pair is ( π₯, π¦)
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Thursday 01/24/19 Homework solution
Solve the system of equation by elimination 2. 3π₯β4π¦=β5 5π₯β2π¦=β6 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₯β4π¦=β5 βπβ(5π₯β2π¦=β6) Multiple the second equation by -2 Step 1 again: check that the coefficient of the one of the variables are opposites. 3π₯β4π¦=β5 β10π₯+4π¦=12) We are good now Step 2: Add the two equations together (one variable should disappear) 3π₯ + β4π¦ = β5 β10π₯ +4π¦ =12 β7π₯ =7 Step 3: Simplify and solve for the variable β7π₯=7 π₯=β1 Divide by -7 to both sides
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Correct order pair (β1, 1 2 ) An order pair is ( π₯, π¦)
Step 4: Substitute the solved variable back into the original to get the other variable 3π₯β4π¦=β5 First equation 3(β1)β4π¦=β5 Substitute x=β1 β3β4π¦=β5 β4π¦=β2 Add 3 to both sides π¦= 2 4 ππ 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 (β1, 1 2 ) An order pair is ( π₯, π¦)
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Thursday 01/24/19 Homework solution
Solve the system of equation by elimination 3. 2π₯+3π¦=8 3π₯+2π¦=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 πβ 2π₯+3π¦=8 Multiple the first equation by 3 βπβ 3π₯+2π¦=7 Multiple the second equation by β2 Step 1 again: check that the coefficient of the one of the variables are opposites. 6π₯+9π¦=24 β6π₯β4π¦=β14 We are good now Step 2: Add the two equations together (one variable should disappear) 6π₯+9π¦=24 β6π₯β4π¦=β14 5π¦ =10 Step 3: Simplify and solve for the variable 5π¦=10 π¦=2 Divide by 5 to both sides
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Correct order pair (1, 2) An order pair is ( π₯, π¦)
Step 4: Substitute the solved variable back into the original to get the other variable 2π₯+3π¦=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₯, π¦)
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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. βπ₯ +π¦=5 π₯β5π¦=β9 2. π₯β10π¦=60 π₯+14π¦=12 3. 2π₯+3π¦=12 5π₯βπ¦=13 β4π¦=β4 2π₯+4π¦=72 7π₯+2π¦=25
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Exploration activity: Solving by Elimination
2. 1π₯β10π¦=60 1π₯+14π¦=12 2x + 4y = 72 2. π₯β10π¦=60 π₯+14π¦=12 2x + 4y = 72 3. 2π₯+3π¦=12 5π₯β1π¦=13 7x + 2y = 25 3. 2π₯+3π¦=12 5π₯β1π¦=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₯+3π¦=12 β1β(5π₯β1π¦=13) 1π₯β10π¦=60 β1β(1π₯+14π¦=12) 2π₯+3π¦=12 β5π₯+1π¦=β13 2π₯+3π¦=12 β5π₯+1π¦=β13 1π₯β10π¦=60 β1π₯β14π¦=β12 1π₯β10π¦=60 β1π₯β14π¦=β12 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.
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How to Solve by Elimination by multiplying first
2π₯+3π¦=12 5π₯β1π¦=13 Step 1: check that the coefficient of the one of the variable are opposites. 2π₯+3π¦=12 5π₯β1π¦=13 Step 1b: multiple by (something) to get the opposite value. 2π₯+3π¦=12 3β(5π₯βπ¦=13) Multiple by 3 to the second equation 2π₯+3π¦=12 15π₯β3π¦=39 Step 1: check that the coefficient of the one of the variable are opposites. 2π₯+3π¦=12 15π₯β3π¦=39
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2π₯ + 3π¦ = 12 15π₯ β 3π¦ = 39 17π₯ = 51 (3, 2) Solution
Step 2: Add the two equations together (one variable should disappear) 2π₯ + 3π¦ = 12 15π₯ β 3π¦ = 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₯+3π¦=12 First equation 2(3)+3π¦=12 Substitute x=3 6+3π¦=12 3π¦=6 Subtract 6 to both sides π¦=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₯, π¦)
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or Practice on solving by Elimination by multiplying first
2π₯+π¦=3 π₯β3π¦=12 Step 1: check that the coefficient of the one of the variable are opposites. 2π₯+π¦=3 1π₯β3π¦=12 or 2π₯+1π¦=3 1π₯β3π¦=12 Step 1b: multiple by (something) to get the opposite value. 2π₯+π¦=3 β2β(1π₯β3π¦=12) Multiple by -2 to the second equation 2π₯+π¦=3 β2π₯+6π¦=β24 Step 1: check that the coefficient of the one of the variable are opposites. 2π₯+π¦=3 β2π₯+6π¦=β24
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2π₯ + π¦ =3 β2π₯ + 6π¦ =β24 7π¦=β21 (3, β3) Solution
Step 2: Add the two equations together (one variable should disappear) 2π₯ + π¦ =3 β2π₯ + 6π¦ =β24 7π¦=β21 Step 3: Simplify and solve for the variable 7π¦=β21 π¦=β3 Divide by 7 to both sides Step 4: Substitute the solved variable back into the original to get the other variable 2π₯+π¦=3 First equation 2π₯+ β3 =3 Substitute y=β3 2xβ3 =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, β3) Solution An order pair is ( π₯, π¦)
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Math Talk Which is incorrect? Explain the error
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