extremely easy, need assistance? Medal! so I'm in college algebra now, and I seem to have forgotten some basics. Can someone help me solve for y in this equation x=9y+7/7y-2
HI!!
Hi!
this is not really that easy so it is easy to forget
lol okay, thanks for the reassurance.
\[x=\frac{9y+7}{7y-2}\] right?
yup!
first multiply both sides by \(7y-2\) to clear the fraction
should you always do the denominator first?
you get \[x(7y-2)=9y+7\] then multiply out on the right
you gotta get rid of the denominator to solve, yes
okay, and that's what I got also.
when you multiply out you get \[7xy-2x=9y+7\]
so 7yx-2x=9y+7
right
now put all the stuff with \(y\) on one side of the equal sign, (say the left) everything else on the other
okay, one second
I got 5x-7=9y/y
no not quite
how do i get the y alone then?
that's where I always mess it up.
we are not there yet, hold on
we are good to here \[7xy-2x=9y+7\] now you want all the y terms on the left, subtract \(9y\) from both sides, which does not really mean "subtract" it just means "write it" \]7xy-9y-2x=7\]
oops \[7xy-9y-2x=7\] that's better
oh... I went about it the wrong way...
then add \(2x\) to both sides what do you get?
7yx-9y=2x+7
ok good so we have all the y terms on the left, all else on the right that is correct because we want to get y by itself now factor out the y from the left do you know what i mean by that?
yeah.
y(7x-9)=2x+7
good now one step and you are done, namely divide by \(7x-9\) to get \(y\)
and then divide by the 7x-9
yup `
okay, got it. Thank you so much. lol that was much more difficult than I remember learning in 6th grade. I caught on to bad math habits and it shows in how I try to do things now ):
not terrifically hard, but not really that easy take several steps
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Thanks again! have a great day!
you too
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