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Published byGabriella Miles Modified over 5 years ago
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On your whiteboards write an equation that represents this bar model
π₯ π₯ π₯ π₯ 7 π¦ π¦ On your whiteboards write an equation that represents this bar model And anotherβ¦ And anotherβ¦
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Draw a pair of bar models to represent the following pair of simultaneous equations
3π₯βπ¦=9 2π₯βπ¦=5
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3π₯βπ¦=9 2π₯βπ¦=5
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Discuss: Why isnβt this helpful?
In an attempt to work out the values of π₯ and π¦ a student combines her bar model like this: π₯ π₯ π₯ π₯ π₯ 5 π¦ 9 π¦ Discuss: Why isnβt this helpful?
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3π₯βπ¦=9 2π₯βπ¦=5 ? What is the value of π₯? How do you know?
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3π₯βπ¦=9 2π₯βπ¦=5 β 4 π₯ = 4
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π₯=4 now work out the value of π¦
Using the first equation: Using the first bar model: 3π₯βπ¦=9 4 4 4 3Γ4βπ¦=9 9 π¦ 12βπ¦=9 π¦=12 β9 π₯=4 now work out the value of π¦
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On your whiteboards, solve:
5π₯β2π¦=45 3π₯β2π¦=25 Allow students to use bar model/elimination Share and compare whiteboards
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π₯ π₯ π₯ π₯ π₯ 45 π¦ π¦ π₯ π₯ π₯ π₯ π₯ 20 25 π¦ π¦ 2π₯=20 π₯=10
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10 10 10 10 10 1 45 π¦ π¦ 50β2π¦=45 2π¦=5 π¦=2.5
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10 10 10 2 25 π¦ π¦ 30β2π¦=25 2π¦=5 π¦=2.5
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Subtracting the equations
1 β 2 5π₯β2π¦=45 β 3π₯β2π¦=25 Allow students to use bar model/elimination Share and compare whiteboards 2π₯ = 20 π₯=10
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1 5π₯β2π¦=45 5Γ10β2π¦=45 50β2π¦=45 2π¦=5 π¦=2.5
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On your whiteboards, solve:
2π₯β3π¦=13 2π₯ β5π¦=9 Allow students to use bar model/elimination Share and compare whiteboards
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π₯ π₯ π₯ π₯ 13 π¦ π¦ π¦ 9 π¦ π¦ π¦ π¦ π¦ 2π₯β3π¦=13 2π₯ β5π¦=9
Allow students to use bar model/elimination Share and compare whiteboards
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13 13=9+2π¦ 4=2π¦ 2=π¦ 9 π¦ π¦ Allow students to use bar model/elimination Share and compare whiteboards π¦=2
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π₯ π₯ 13 2 2 2 π₯ π₯ 9 2 2 2 2 2 Allow students to use bar model/elimination Share and compare whiteboards
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π₯ π₯ 1 13 2 2 2 2π₯=13+6 2π₯=19 Allow students to use bar model/elimination Share and compare whiteboards π₯=9.5
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2π₯β3π¦=13 β 2π₯ β5π¦=9 2π¦=4 π¦=2 Allow students to use bar model/elimination Share and compare whiteboards
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2π₯β3π¦=13 1 2π₯ β3Γ2=13 2π₯ β6=13 2π₯ =19 π₯ =9.5
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On your whiteboards, solve:
2π₯=1+π¦ 7π₯β2π¦=9.5 Allow students to use bar model/elimination Share and compare whiteboards
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3π₯ β3π¦=36 β2π¦+3π₯=5 π₯ β3π¦=6 2π₯ β4π¦=2 3π₯ βπ¦=9 1 2 π₯βπ¦=2 3π₯ β2π¦=6
Solve the following using a bar model or elimination. Practise on whiteboards first, then in your books. 3π₯ β3π¦=36 β2π¦+3π₯=5 1. 4. π₯ β3π¦=6 2π₯ β4π¦=2 3π₯ βπ¦=9 1 2 π₯βπ¦=2 2. 5. 3π₯ β2π¦=6 3π₯β2π¦=24 2π₯ βπ¦=3 3. 4π₯=8.5+π¦
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Mark your work π₯=15, π¦=3 π₯=4, π¦=3 π₯=2.5, π¦=1.5 π₯=2, π¦=0.5 π₯=10, π¦=3
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Challenge Acknowledgements: UKMT
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