TA: Geoff Williams
Tutorials: Thursday 1:30-2:20 in TSH-120
E-mail:williagg@math.mcmaster.ca
Office hours: Wednesday 1:30-3:30 at the Math Help Centre
INTRODUCTION
Example 1. You’re recycling a collection of bicycles, tandem bikes and unicycles. All are complete to begin with. You’ve stripped off the parts and melted down the frames, only to find out
you were supposed to have counted how many of each type of bike there was to begin with. You
do a quick count of some of the parts:
Count wheels: 340 wheels
Count seats: 220 seats
Is this enough to solve the problem? Why?
Solution:
Info:
• bicycle: 2 wheels, 1 seat, 2 pedals, 1 handlebar
• tandem bike: 2 wheels, 2 seats, 4 pedals, 2 handlebars
• unicycle: 1 wheel, 1 seat, 2 pedals, 0 handlebars
Let
b = number of bicycles
t = number of tandem bikes
u = number of unicycles
• Wheels: 2b + 2t + 1u = 340
• Seats: 1b + 2t + 1u = 220
← 2 equations
We still don’t have enough information. What else can we count?
Count pedals: 440 pedals
• Pedals: 2b + 4t + 2u = 440
← 3 equations
Is this enough info? Why or why not? Can we see any other info?
Subtract seat equation from wheel equation to get
1b + 0t + 0u = 120
⇒ b = 120
← 4 equations
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Sub back in to get
2t + 1u = 100
← system of 5 equations
Parametrize solution:
t=p
u = 100 − 2p
⇒ inf. many solutions
What if someone told you b = 105?
⇒ no solutions
Need something else to distinguish the numbers of each type.
Count handlebars: 200 handlebars
• Handlebars: 1b + 2t + 0u = 200
b = 120
t = 40 ⇒ unique solution
u = 20
Main points:
• Systems of equations can have any number of equations. (In this example we found some
useful equations and some redundant ones.)
• Some have:
– infinitely many solutions (e.g. when we had the first 5 equations)
– no solutions (e.g. when we were told b=105)
– unique solution (e.g. after we counted the handlebars)
Section 1.2
Exercise # 8 a)
Find values for a and b such that the system has infinitely many solutions, no solutions or unique
solution.
x − 2y = 1
ax + by = 5
Solution:
1 −2 1 R2 −aR1
1
−2
−−−−−→
a b
5
0 b + 2a
Case 1: If b + 2a 6= 0 then,
1
R
1 −2 1
1 0
R1 +2R2
b+2a 2
−−−−→
−−−−→
5−a
0 1
0 1
b+2a
2
1
5−a →
1 + 2 5−a
1 0
b+2a
→
5−a
0 1
b+2a
b+10
b+2a
5−a
b+2a
x=
y=
b+10
b+2a
5−a
b+2a
and we get a unique solution
Case 2: If b + 2a = 0 then,
→
1
1 −2 0 0 5−a
and this provides two subcases.
a). 5 − a = 0 ⇒ a = 5. Hence b = −10
→
1 −2 1
0 0 0
x = 1 + 2t
y=t
Infinite solutions
b). 5 − a 6= 0 ⇒ a 6= 5.
→
1 −2 1
0 0 some number
No solutions
Notes originally typed up by Olga Krylova. Edited by Geoff Williams.
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