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anonymous
 one year ago
Algebra Question
anonymous
 one year ago
Algebra Question

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If \[\frac{ 3 }{ 4 } = \frac{ 6}{ x } = \frac{ 9 }{ y } \] , then x +y = ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1you can solve for x, and solve for y, by making the following conclusions out of your initial statement (what you wrote above). (Hope you know why they are valid) \(\large\color{black}{ \displaystyle \frac{3}{4}=\frac{6}{x} }\) and \(\large\color{black}{ \displaystyle \frac{3}{4}=\frac{9}{y} }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh so this also follows the transitive property of equality

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, if you have `a=b=c` then you can conclude that: `a=b` `a=c` or `b=c`

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1this way we can solve for x and solve for y. (Have you found any of the unknown variables) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes I have :D After cross multiplying: 3x= 24 and 3y = 26 and then dividing both equations by 3 on both sides gave me x = 8 and y = 12 and therefore x + y = 12 + 8 = 20 Did I do this right or was there a mistake somewhere?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1x=8 is correct, check your y.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.13y=36 yu meant 36 not 26

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1so, y=12, x=8 x+y=8+12=20 RIGHT !!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was like hmm where did you get 27 from haha, thanks :)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Yes, I got confused by 26 there (I thought you meant 27, but the keyboard was off by the first digit.... In any case, you are always welcome:)
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