# Algebra Ratio and Proportion Inverse Variation

All the equations accept the one I am stuck on 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 two unknowns.

Here is what I did, but I don't think it's correct much less how to solve it. The book answer is 10cm; 17cm. How do I turn this into an equation and solve? Thanks.

PL=Pulley large, PS=Pulley small.

(80)(PL)+7=(136)(PS)
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Commented:
you need 2 equation
1. 80PL=186PS
2. PL=PS+7
Now you have two equations for 2 unknows values
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80xPL = 136xPS

PL=PS+7

Now evaluate by substitution
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Author Commented:
80PL=136PS
PL=PS+7

80PL/80=136PS/80
PL=136PS/80

PL=PS+7
136PS/80=PS+7
(136PS/80)-7=PS-7
(PS)(136PS/80)-7=PS(PS)
136/80-7=PS
1.7-7=PS
-5.3=PS

80PL=136PS
80PL=136(-5.3)
80PL/80=-720.8/80
PL=-9.01

The answer should be 10 and 17. What am I doing wrong? Thanks.
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Commented:
you have to do the following:
first the original equations
1. 80PL=136PS
2. PL=PS+7
replace 2 in 1:
80*(PS+7)=136*PS
80*PS + 560 = 136*PS
PS = 560/56 = 10
then replace PS value in equation 2
PL=10+7 = 17

Thats it!
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Commented:
80*PL=136*PS
PL=PS+7

Stay away from division until the end:

80*(PS+7) = 136*PS   ==>   80*PS + 560 = 136*PS    ==>   56*PS = 560
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Author Commented:
1. 80PL=136PS
2. PL=PS+7
replace 2 in 1:
80*(PS+7)=136*PS
80*PS + 560 = 136*PS
PS = 560/56 = 10

I think you skipped a step. How did you go from
80*PS + 560 = 136*PS
PS = 560/56 = 10

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Commented:
This is the step
80*PS + 560 = 136*PS
560 = 136*PS - 80*PS
560 = 56*PS
PS = 560/56
PS = 10
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Author Commented:
I thought you could only divide because it is the opposite of multiply. Thanks. I learned something here.

80*PS + 560 = 136*PS

80PS/80 + 560 = 136PS/80

PS + 560 = 136PS/80

PS -560 = 136PS/80-560

PS = 136PS/80-560
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Commented:
sorry I really don't undertand why your are dividing PS by 80 in both sides.
For me is better not to divide until the end.
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Author Commented:
When I see for example:
10x = 800

It tells me to divide
10x/10 = 800/10  x = 80.
That is what I am used to doing.

I guess what I learned here is that when there are two of the same variable on both sides of the equation, then I can subtract or add one side over to the other.

For example:
15y*6 = 800 + 40y
15y-15y*6 = 800 + 40y-15y
6 = 800 + 24y
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