This document explains how to solve a word problem about the number of keychains Susan, Amy, and Mary have using algebra. It introduces letting Y represent Susan's keychains, writes an equation for the total keychains as 22Y + 34, solves to find Y = 6, and uses that value to determine each person's keychains. The conclusion compares this algebraic method to modeling and explains algebra is more efficient and better for harder problems.

1. Algebra method
In this question, we only need to use one symbol
to represent a value.

2. Solving
 So, Let Y be the number of
keychains Susan has.
 Total number of keychains they have
would be Y(Susan)+20Y(Amy) +
Y+34(Mary), which gives us 22Y + 34.

3. Solving
 Since the number of keychains they
have altogether is 166, 22Y+34=166
 22Y=166-34
 22Y=132
 Y=6

4. Solving
 Now, we simply use the value of Y to
find out how many keychains each of
them have.
 Amy has 20Y keychains, so Amy has
20 x Y= 120 keychains
 Susan has Y keychains, so Susan has
6 keychains
 Mary has Y+34 keychains, so Mary
has 6+34=40 keychains

5. Conclusion
 As you can see, this method is much
faster than the modelling method
 This is because you don’t have to
spend time to slowly draw out the
units

6. Conclusion
 Also, the modelling method and
Algebra method do have some
similarities, but the Algebra method is
more efficient
 Algebra is also better for solving
tougher problems as compared to
modelling (not very obvious in this
example as the question was fairly
simple)