Show work!!! :D Medal may be rewarded! :O 5. (5 over y - 5) = (y over y - 5) - 1 6. (1 over 3x - 2) + (5 over x) = 0
5. \[\frac{ 5 }{ y - 5 } = \frac{ y }{ y - 5 } - 1\] 6.\[\frac{ 1 }{ 3x - 2 } + \frac{ 5 }{ x } = 0\]
@Firejay5 Are you ready for my explaination?? :)
yea
Ok so first I want to know if you know how to subtract fractions and numbers?
1. I suck fraction so don't even try to ask me 2. I am partially good with numbers
Ok that's ok I can help :) When we want to subtract fractions and numbers we need a common denominator. In this question lets work with the left side ok
I mean the right side Sorry :P
So lets begin :) This is what the equation is \[\frac{ y }{ y-5 }-\frac{ 1 }{ 1 }\] Do you agree??
I guess
Why aren't you confident??
I thought it would be y - 5
What do you mean? All I did to the right side of the equation was put 1 over 1 because all numbers are over 1. Are you still confused?
see before you can add or subtract the denominator must be the same and all of the denominators are y - 5, so it should be 1 over y - 5
You're on the right track but not quite. In order to have the same denominator you have to multiply both the numerator and denominator by \[y-5\] so it would be \[\frac{ y }{ y-5 }-\frac{ y-5 }{ y-5 }\] Does that make sense?
see that's the point I was making
Ok so can we proceed?? :P
yes
Ok so now we have the same denominator we can evaluate the numerators together \[\frac{ y-(y-5) }{ y-5 }\] does this make sense?
yes
Ok so now we see that there is a negative sign in front of the (y-5) so that negative will affect everything in the bracket. So... \[\frac{ y-y+5 }{ y-5 }\] Do you agree??
yes
Ok Can you take it from here then?
so do we have so far?: \[\frac{ 5 }{ y - 5 } = \frac{ y - y + 5 }{ y - 5 }\]
Yes but we are solving the right side in order to prove both sides. So finish the right side
Do I cross multiply or what???
No on the right side y-y+5 has two common terms which can be simplified
what 2 common terms
Y and Y so they cancel out
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