hlilly2413
  • hlilly2413
How to solve for Y when Y/20+Y * 100 = 40. I haven't done this in so long and am floored as to what to do when the unknown variable appears twice on the same side of the equation. Please help.
Algebra
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SOLVED
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jamiebookeater
  • jamiebookeater
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jdoe0001
  • jdoe0001
\(\bf \cfrac{y}{20}+y\cdot 100=40?\)
phi
  • phi
there are a few ways to tackle this. one way is to multiply both sides (and all terms) by 20 as the first step. can you try doing that ?
hlilly2413
  • hlilly2413
@jdoe0001 it's 20+Y under the Y then * 100 = 40. @phi So 20(y/20+y) and then 20(100) and 40(20)? sooo 20Y over 400+20Y *2000 = 800 would be the new equation?

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phi
  • phi
are you saying the problem is \[ \frac{y}{20+y} \cdot 100 = 40 \] ?
hlilly2413
  • hlilly2413
Yes.
phi
  • phi
in that case, we should start by multiplying both sides by (20+y)
hlilly2413
  • hlilly2413
Alright, I think I understand that by doing so it would cancel out the 20+y on the left. Do I also multiply the 100 by 20+y?
phi
  • phi
when you multiply a fraction times a whole number you do \[ \frac{2}{3} \cdot 4= \frac{2}{3} \cdot \frac{4}{1} = \frac{2\cdot 4}{3\cdot 1} \] you multiply top times top and bottom times bottom so the 100 is actually multiplying the top y
phi
  • phi
you should get \[ 100y= 40(y+20) \]
phi
  • phi
now "distribute the 40" that means 40 times each term inside the parens
hlilly2413
  • hlilly2413
Ok after I asked the question regarding the 100 I did not multiply it by 20 +Y. Thank you for the information there. I did end up getting 100y = 800 + 40Y and can def. take it from there. Thank you very much for your assistance @phi I appreciate it.

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