If you are over 45, you can remember problem like this from Algebra I.
Dave can do a job in 5 hours and his
friend Moreno can do it in 4 hours.
How long would it take if they did
it together?
Not only do I remember doing them, but when I began my career as a teacher in
1969 the same text was being used, and I found myself sending my class home
with those same problems. However, realizing they were all the same, except
that they might have a twist, Dave got to work one hour late, or something.
I thought there must be a better way, and that is what I asked my class.
This is why. Think of a circle. Do we find the Area of one circle differently
then in another. NO! A circle is a circle.
We have a FORMULA: RT = 1
This means the RATE/HOUR x TIME WORKED = 1
1 MEANS THE ENTIRE JOB
DAVE above does 1/5 of the job per hour because he does the job in 5 hours.
MORENO does ¼ of the job per hour because he does the job in 4 hours.
1/5 x 5 = 1 and ¼ x 4 = 1
Now we are simply combining their work. R₁T + R₂T =1
The 1 and 2 in the formula are subscripts, and are just used as counters
As I stood before my classes that 1st year I said,
"Lets do it in general, using variables"
(1/a) (x) + (1/b)(x) = 1
Here a and b stand for the times each spent working alone, and x is the
combined timed. Now we solve.
x/a + x/b = 1 Our common denominator is ab
(bx + ax)/ab = 1/1 Now use the rules of proportions.
(bx + ax) = ab Factor x out of both terms on left
x (a + b) = ab Divide by sum (a + b)
x = ab/(a + b)
Now think of our friends Dave and Moreno
Dave = 5 hours
Moreno = 4 hours
Thus if we use our new formula: x = (5)(4)/(5 + 4)
x = 20/9 hours or 2 and 2/9 hours
With the same Algebraic tools we can create a formula for one of them
being 1 hour late. With these formulas we can solve a difficult Algebra I
problems as easily as we use A = bh for a Rectangle.
Add this formula to you list. But I suggest deriving it in an Algebra I
class can inspire students to feel better about story problems, and look for
new strategies to solve them.
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