Find x in this proportion. x|3 = x+2|5 (they are fractions. Please show answer and how you got it!)

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Find x in this proportion. x|3 = x+2|5 (they are fractions. Please show answer and how you got it!)

Mathematics
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@Nnesha @e.mccormick @Donblue
So x is to 3 as x+2 is to 5? Know how to set those up in fractional form?
\(\frac{x}{3}=x+\frac{2}{5}\) I think the easiest thing would be to multiply everything by the LCM of the denominators, which is 15 \(15\frac{x}{3}=15x+15\frac{2}{5}\) \(5x=15x+6\) Can you solve now?

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Other answers:

zz, I think they meant: \(\dfrac{x}{3} = \dfrac{x+2}{5}\)
Was it \(\frac{x}{3}=\frac{x+2}{5}\)?
x/3=x+ 2/5 or , 5x=15x+6 or, 15x+6=5x or,10x=6 or,x=6/10=3/5=0.6
ye @e.mccormick and @Learner11 i tried that, thats wrong
ahh, try the same thing with that! \[\frac{x}{3}=\frac{x+2}{5}\] \[15*\frac{x}{3}=15*\frac{x+2}{5}\] \[5x=3(x+2)\] now solve for \(x\)?
answer choices are -1 , 1 , 2, 3. :P i keep doin this wrong
As a general rule: \(\dfrac{a}{b}=\dfrac{c}{d}\) type proportions can alsways use cross multiplication to move forward. Cross multiplication basically does what xx did. He just multiplied hte bottoms (3 and 5) first. But if you don't multiply them out, then you can cancel quickly: \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\dfrac{a}{b}\times \dfrac{bd}{1}=\dfrac{c}{d}\times \dfrac{bd}{1}\) \(\dfrac{a}{\cancel{b}}\times \dfrac{\cancel{b}d}{1}=\dfrac{c}{\cancel{d}}\times \dfrac{b\cancel{d}}{1}\) \(\dfrac{a}{1}\times \dfrac{d}{1}=\dfrac{c}{1}\times \dfrac{b}{1}\) \(a\times d=c\times b\) \(ad=cb\) That is a general way to start. Then solve for the unknown...
thank you!
What steps are you using when you solve the: 5x=3(x+2)

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