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)
R02.jpg
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)
R02.jpg
80xPL = 136xPS
PL=PS+7
Now evaluate by substitution
PL=PS+7
Now evaluate by substitution
ASKER
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.
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.
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!
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!
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
PL=PS+7
Stay away from division until the end:
80*(PS+7) = 136*PS ==> 80*PS + 560 = 136*PS ==> 56*PS = 560
ASKER
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
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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ASKER
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
80*PS + 560 = 136*PS
80PS/80 + 560 = 136PS/80
PS + 560 = 136PS/80
PS -560 = 136PS/80-560
PS = 136PS/80-560
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.
For me is better not to divide until the end.
ASKER
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
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
1. 80PL=186PS
2. PL=PS+7
Now you have two equations for 2 unknows values