SAT MATH LEVEL 2
\[\frac{ 6 }{ x } + \frac{ 3 }{ y }=4 \]
and
\[\frac{ 3 }{ x }+\frac{ 6 }{ y }=\frac{ 7 }{2 }\]
What is the value of xy? not x and y only xy
for the first one: 6(1/x + 1/2y)=2/3
you're solving for xy right
exactly
lets try this first equation --> y = 4/3 -2x and will input that in to the bottom equation to get 3/x + 6/(4/3-2x) = 2/3 and solve for x
ok
= 7/2 not 2/3
I don't understand a part
3/x + 6/(4/3-2x) = 7/2 what is that 6 doing there?
are this two separate questions or are the equation part of the same question?
the equations are part of the same question. I believe we have to use the method of substitution. I am not sure how.
give me sec
ok
lets start from scratch, I messed up. the first equation is y = 4/3 -2x ---> we subtracted 6/x out and then divided by 3
now where back to the same equation the 6 is there because in your second equation y is under a 6
wait a second, for the first equation, it must be 1/y = 4/3 - 2x
you wrote y
divide by one again
how is it possible? 1/1 . 1/y = 4/3 - 2x 1 at the bottom cancels out the 1 at the top. so 1/y = 4/3 - 2x
lets find a different way then
I believe we have to divide both equations by a constant term which would be 3
(6/x+3/y=4) -2(3/x+6/y =7/2) this will elemenate the x value
Ok wait, this is the answer key: the answer: 2/x and 1/y = 4/3 1/x + 2/y = 7/6 y = 3 and x = 2 xy=6
follow me we're on the right track
Ok you have the answer. I know that at the begining we have to divide both equations by the value of 3. show me how you get y=3 and x=2
so from the equation you should get 1/y = 1/3 so then multiply by y and you'll have 1= y/3 so then multiply by 3 which will give you y = 3
how did you get 1/y = 1/3
did you divide by the 9
-9/y = -3 ---> 1/y = 1/3 --> y = 3
are saying all this from the answer key I provided you?
or your own way
my own, I followed the equation
still gives the same answer
you followed the equation from the answer key or the original equation?
original equation
Ok, let me put all the steps into one. If I have further equestions. I will tag you here. Thanks a lot :)
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