Algebra Question

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Algebra Question

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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If \[\frac{ 3 }{ 4 } = \frac{ 6}{ x } = \frac{ 9 }{ y } \] , then x +y = ?
you 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 \(\large\color{black}{ \displaystyle \frac{3}{4}=\frac{9}{y} }\)

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oh so this also follows the transitive property of equality
yes, if you have `a=b=c` then you can conclude that: `a=b` `a=c` or `b=c`
this way we can solve for x and solve for y. (Have you found any of the unknown variables) ?
Yes 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?
you mea 3y=27?
x=8 is correct, check your y.
yes
hmm
3y=36 yu meant 36 not 26
so, y=12, x=8 x+y=8+12=20 RIGHT !!
OH haha
I was like hmm where did you get 27 from haha, thanks :)
Yes, 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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