andriod09
  • andriod09
Need.Math.Help.Very.Badly.In.Comments.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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andriod09
  • andriod09
\[\frac{x}{9}-\frac{4}{3}=x-\frac{5}{6}\] I don't know how to do this, please help me!
andriod09
  • andriod09
@ajprincess @ash2326
whpalmer4
  • whpalmer4
What is the least common multiple of 3, 6, and 9? Multiply the whole equation by that number and you should be in good shape

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whpalmer4
  • whpalmer4
\[\frac{x}{9}-\frac{4}{3}=x-\frac{5}{6}\]Multiply through by 18 \[18*\frac{x}{9}-18*\frac{4}{3}=18*x-18*\frac{5}{6}\]\[2x-24=18x-15\]Collect like terms\[2x-2x-24+15=18x-2x-15+15\]\[-9=16x\]\[x=-\frac{9}{16}\]
andriod09
  • andriod09
Okay, I'm REALLY confused now.
dan815
  • dan815
there less information to take in now >_>
andriod09
  • andriod09
No, just both of you talking at once is really confusing me. One at a time.
andriod09
  • andriod09
And dan, you're the one that really helped me (also for not just giving me the whole problem). so have fun with a medal.
whpalmer4
  • whpalmer4
I chose 18 because it is the least common multiple of 3, 6, 9. To find the LCM, you can either make a list of multiples: 3: 3 6 9 12 15 18 21 24 6: 6 12 18 24 9: 9 18 27 36 18 is the first number that appears in each list Or factor each number: 3: 3 6: 2*3 9: 3*3 = 3^2 Multiply together the highest power encountered of each factor: 2*3^2 = 2*9=18
dan815
  • dan815
lisn to whpalmer he is a good teacher
whpalmer4
  • whpalmer4
what Dan and I are doing is essentially the same thing. he goes around the tree to the left, I go around the tree to the right ;-)
dan815
  • dan815
yes this is your lesson one there are many ways to solve one math question
dan815
  • dan815
just dont get too restricted to one way, understanding all ways will help you get a grasp on math faster
andriod09
  • andriod09
Not when your dyslexic and you can't do variables very well. >.< Thanks to both of you guys at least.
dan815
  • dan815
oh that sucks!!
whpalmer4
  • whpalmer4
the underlying concept we both make use of is that you can multiply a fraction by the same number top and bottom and you don't change its value. for example: \[\frac{1}{2} = \frac{1}{2}*\frac{3}{3} = \frac{3}{6}\]If you divide 1 by 2 on your calculator, you get 0.5. If you divide 3 by 6 on your calculator, you also get 0.5. So, what Dan was doing is multiplying each fraction by another fraction which has the value 1, but converts the denominator to a common denominator so you can add them all together. Here, I'll typeset it: \[\frac{x}{9}-\frac{4}{3}=x-\frac{5}{6}\]If we need to add and subtract those fractions, we need to use a common denominator. As I showed above, 18 is the least common multiple of the denominators in this problem, so we'll make each fraction into one with 18 as the denominator. 9*2 = 18, so the x/9 gets multiplied by 2/2. 3*6=18, so the 4/3 gets multiplied by 6/6. 6*3=18, so the 5/6 gets multiplied by 3/3. Finally, we multiply the solitary\(x\) by 18/18 to give it a common denominator with the rest.
whpalmer4
  • whpalmer4
\[\frac{x}{9}*\frac{2}{2}-\frac{4}{3}*\frac{6}{6}=x*\frac{18}{18}-\frac{5}{6}*\frac{3}{3}\]\[\frac{2x}{18}-\frac{24}{18}=\frac{18x}{18}-\frac{15}{18}\]Now we can add and subtract the fractions and rearrange to get all of the \(x\)'s on one side, and the numbers on the other. Because there's a bigger coefficient of \(x\) on the right, I'll move the other \(x\) of there, and the numbers to the left.\[\frac{2x}{18}-\frac{2x}{18}-\frac{24}{18}+\frac{15}{18}=\frac{18x}{18}-\frac{15}{18}-\frac{2x}{18}+\frac{15}{18}\]\[\frac{-9}{18}=\frac{16x}{18}\]We can cancel the 18's and we have \[-9=16x\]\[x=-\frac{9}{16}\]
whpalmer4
  • whpalmer4
Now, what I did was essentially put everything over a common denominator of 18 and cancel the 18's all in one step. I did that by multiplying each term of the equation by 18 — perfectly legal to do, just need to multiply everything by the same value and the equation holds. \[2+2=4\]Multiply everything by 3\[3*2+3*2=3*4\]\[6+6=12\]Right? When I multiplied through by 18, I made a common denominator and canceled it out, leaving me just the numerators and skipping a bunch of tedious work.
whpalmer4
  • whpalmer4
This is a very handy technique when working with problems involving decimal amounts, too. If you had some equation \[0.23x+0.47y = 3.96\]you could work with it like that, or you could multiply through by 100:\[100*0.23x+100*0.47y = 100*3.96\]\[23x+47y=396\]I don't know about you, but I'd prefer to work with the latter equation! It is exactly equivalent, and you'll get the same answer.
whpalmer4
  • whpalmer4
You could think of that equation as \[\frac{23}{100}x+\frac{47}{100}y = 3+\frac{96}{100}\]and then the multiplication becomes \[\cancel{100}*\frac{23}{\cancel{100}}x+\cancel{100}*\frac{47}{\cancel{100}}y=100*3+\cancel{100}*\frac{96}{\cancel{100}}\]\[23x + 47y = 300+96\]\[23x+47y=396\]
whpalmer4
  • whpalmer4
Does that make sense?

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