If 25y = 8x and 125y = 4x - 1, then the value of x is
use substitution so x= 25/8y then plug that into the other equation 125y = 4(25/8y) - 1
er x = 25y/8 then 125y = 4(25y/8)-1
that will give you y.. do vice versa to figure out x
Solve them both for y:\[y=\frac{8x}{25}\]\[y=\frac{4x-1}{125}\] Now that they're both equal to y (and because we know these two equations are talking about the same x and y) you can substitute y for the other equation:\[\frac{8x}{25} = \frac{4x-1}{125}\]Then solve for x:\[(125)(8x) = (4x-1)(25)\]\[1000x = 100x-25\]\[900x = -25\]\[x = -\frac{25}{900} = 0.02777\]
alexray, any ideas on my problems?
used to know how to do your physics problems but it's been a few years
What do you mean? What physics problems?
oh whops, ment feldy. he was looking at it earlier and he had a question about stars orbiting etc. never mind
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