anonymous
  • anonymous
What is the value of x in the proportion. x/2=16x-3/20 1. 2 2. -3/2 3. 1/4 4. 1/2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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whpalmer4
  • whpalmer4
\[\frac{x}{2} = \frac{16x-3}{20}\] is that how the problem reads?
anonymous
  • anonymous
yes
whpalmer4
  • whpalmer4
Do you know about cross-multiplication?

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anonymous
  • anonymous
yes
whpalmer4
  • whpalmer4
Good. Cross-multiply and solve the resulting equation for \(x\)
anonymous
  • anonymous
actually i am confused on how to cross multiply on this can you explain?
whpalmer4
  • whpalmer4
Okay, \[\frac{x}{2} = \frac{16x-3}{20}\] multiply the numerator of one by the denominator of the other \[x*20 = 2*(16x-3)\]\[20x = 32x - 6\]
whpalmer4
  • whpalmer4
What you are effectively doing is making a common denominator of 1
whpalmer4
  • whpalmer4
Can you solve \[20x = 32x - 6\] for \(x\)?
anonymous
  • anonymous
x=1/2
whpalmer4
  • whpalmer4
let's try it in the original equation and make sure it works: \[\frac{\frac{1}{2}}{2} = \frac{16(\frac{1}{2})-3}{20}\]\[\frac{1}{4} = \frac{8-3}{20}\]\[\frac{1}{4} = \frac{5}{20} = \frac{1}{4}\checkmark\]
anonymous
  • anonymous
Okay i get it now thanks!
anonymous
  • anonymous
Can you help me with one more?
whpalmer4
  • whpalmer4
We could also solve like this: \[\frac{x}{2} = \frac{16x-3}{20}\]Multiply both sides by 20\[20*\frac{x}{2} = 20*\frac{16x-3}{20}\]\[10x = 16x-3\]\[10x-10x+3 = 16x-10x+3-3\]\[3=6x\]\[x=\frac{1}{2}\]
anonymous
  • anonymous
Wait so 1/2 is the answer or 1/4?
whpalmer4
  • whpalmer4
the answer is x = 1/2. what I did was plug the value we got for x back into the original equation to make sure that it made a true statement. Let's say that I made a mistake somewhere and decided that \(x=2\) is the right answer. Well, when I plug that back into the original, I get: \[\frac{2}{2} = \frac{16(2)-3}{20}\]\[1 = \frac{29}{20}\]and that is not true, so that means that \(x=2\) is NOT the correct answer.
anonymous
  • anonymous
Ohhh okay thank you so much. I am gonna open another question can you help me?
whpalmer4
  • whpalmer4
sure, just "tag" me by putting "@whpalmer4" in a response and I'll get a notification

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