Remember some basic rules
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and
![x^ \frac{m}{n}= \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7Bm%7D%7Bn%7D%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%20%20)
so
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1/2-2/7=
7/14-4/14=
3/14
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Let y represent Tim's age and let x represent Chris's age.
y=3x
y=x+14
Set the two equations equal to each other:
3x=x+14
Subtract the x from the one side to get it on the other side.
2x=14
Divide by 2 on both sides
14/2=7
x=7
Plug it back in and see if it works:
y=3(7)=21
Other equation:
7+14=21
x=7
y=21
There is one clock that shows the right time so we do not have to worry about the one which is always correct.
Talking about the second clock that loses a minutes in every 24 hours (or in a day), so after 60 days (since it has lost 60 minutes because it is losing 1 minute everyday) it will show 11:00 a.m when it is exactly the noon.
So this way, in total it will take
days before it shows the correct noon.
Now, the third clock gains a minute every 24 hours (or in a day) , after 60 days (when it has gained 60 minutes or a complete hour) it will show 1:00 p.m when it is exactly the noon.
This way, it will take
days (since it has gained a minute everyday) when it shows the correct noon.
Therefore, it will take 1440 days before all the three clocks show the correct time again.
14 = 2•7 and 35 = 5•7, so the GCF of 14k and 35 is 7.
14k + 35 = 2•7k + 5•7 = 7 (2k + 5)
(Remember that the distributive property says a (b + c) = ab + ac.)