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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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    @Nnesha @e.mccormick @Donblue

  2. e.mccormick
    • one year ago
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    So x is to 3 as x+2 is to 5? Know how to set those up in fractional form?

  3. zzr0ck3r
    • one year ago
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    \(\frac{x}{3}=x+\frac{2}{5}\) 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?

  4. e.mccormick
    • one year ago
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    zz, I think they meant: \(\dfrac{x}{3} = \dfrac{x+2}{5}\)

  5. zzr0ck3r
    • one year ago
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    Was it \(\frac{x}{3}=\frac{x+2}{5}\)?

  6. anonymous
    • one year ago
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    x/3=x+ 2/5 or , 5x=15x+6 or, 15x+6=5x or,10x=6 or,x=6/10=3/5=0.6

  7. anonymous
    • one year ago
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    ye @e.mccormick and @Learner11 i tried that, thats wrong

  8. zzr0ck3r
    • one year ago
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    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 \(x\)?

  9. anonymous
    • one year ago
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    answer choices are -1 , 1 , 2, 3. :P i keep doin this wrong

  10. e.mccormick
    • one year ago
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    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...

  11. anonymous
    • one year ago
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    thank you!

  12. e.mccormick
    • one year ago
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    What steps are you using when you solve the: 5x=3(x+2)

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