Algebra: turning word problem into equation
I think the final equation should end up looking like this, for example: (1870)(4.0)=(680)(p) Solve for the unknown (p). However the question I am stuck on has three unknowns. How do I turn this into an equation? Thanks.
Here is what I have so far: PS=Pulley Small; PL=Pulley Large
PL*256=PS*X This is how the final equation looks on the other problems.
2PS=PL
PL=256rpm
X=PSrpm
Solve for PS:
2PS*256=PS*X
-PS -PS
PS*256=X
PS/256=X/256
PS=X/256
Solve for PL:
2PS=PL
2*(X/256)=PL
R03.jpg
Here is what I have so far: PS=Pulley Small; PL=Pulley Large
PL*256=PS*X This is how the final equation looks on the other problems.
2PS=PL
PL=256rpm
X=PSrpm
Solve for PS:
2PS*256=PS*X
-PS -PS
PS*256=X
PS/256=X/256
PS=X/256
Solve for PL:
2PS=PL
2*(X/256)=PL
R03.jpg
1. one pulley is twice as large in diameter as a second pulley
2. the circumference of a circle is pi * diameter
The simple answer is that, since the diameter the first is twice that of the second pulley, one revolution of the larger pulley will cause two revolutions of the smaller one!ASKER
The simple answer is that, since the diameter the first is twice that of the second pulley, one revolution of the larger pulley will cause two revolutions of the smaller one!
I that's true then the smaller pulley should be 512rpm. I unfortunately do not have the answer in the back of the book to this question.
HainKurt: Is your answer almost correct? Thanks for your response.
I that's true then the smaller pulley should be 512rpm. I unfortunately do not have the answer in the back of the book to this question.
HainKurt: Is your answer almost correct? Thanks for your response.
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SOLUTION
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ASKER
Thank you all for helping me learn how to put this together. Little by little I will learn this.
"final equation should end up looking like this, for example: (1870)(4.0)=(680)(p) "
-
The goal of algebra is NOT to make one equation look like another. It is to use the rules of algebra to solve the equation.
"However the question I am stuck on has three unknowns"
What are your three?
PS, PL, RPM of PS
What is known? RPM of PL The ratio PL/PS (=2)
what else do you know? eq of circle cir = pi * diam
In one minute PL lays down 256 * pi * PL worth of rope
in one minute PS lays down x * pi * PS worth of rope both these lengths are the same thus
x * pi * PS = 256 * pi * PL
or
x = 256 * PL/PS but PL/PS = 2 so
x = 256 * 2 which is 512 as paulsauve: said
-
The goal of algebra is NOT to make one equation look like another. It is to use the rules of algebra to solve the equation.
"However the question I am stuck on has three unknowns"
What are your three?
PS, PL, RPM of PS
What is known? RPM of PL The ratio PL/PS (=2)
what else do you know? eq of circle cir = pi * diam
In one minute PL lays down 256 * pi * PL worth of rope
in one minute PS lays down x * pi * PS worth of rope both these lengths are the same thus
x * pi * PS = 256 * pi * PL
or
x = 256 * PL/PS but PL/PS = 2 so
x = 256 * 2 which is 512 as paulsauve: said
ASKER
Thanks again. It seems to help to know things like (cir = pi * diam). I think that's from geometry.
PL=256rpm
X=PSrpm
2PS=256rpm =>
PS=128rpm
X=128rpm