Having trouble with fractions

smooty21

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Joined
Jun 13, 2007
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2
1/5 (3X + 4) = 1/3 (2X - 8)


If someone could show me the correct way to do these I would appreciate very much. Thank you!!!
 

galactus

Super Moderator
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Sep 28, 2005
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7,216
I suppose this is what you have:

\(\displaystyle \L\\\frac{1}{5}(3x+4)=\frac{1}{3}(2x-8)\)

I will show you this one and hopefully it'll help you with others. Okey-doke?.

Distribute:

\(\displaystyle \L\\\frac{3}{5}x+\frac{4}{5}=\frac{2}{3}x-\frac{8}{3}\)

Now, get x's on one side and constants on the other by subtracting \(\displaystyle \frac{2}{3}x\) from both sides and subtracting \(\displaystyle \frac{4}{5}\) from both sides:

This gives:

\(\displaystyle \L\\\frac{-1}{15}x=\frac{-52}{15}\)

Multiply both sides by -15:

\(\displaystyle \L\\x=52\)

See?. Practice a few more and you'll be doing it blindfolded.
 

smooty21

New member
Joined
Jun 13, 2007
Messages
2
Thank you very much. Your'e a life saver.
 

Denis

Senior Member
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Feb 17, 2004
Messages
1,465
smooty21 said:
1/5 (3X + 4) = 1/3 (2X - 8)
Another way is multiply each side by the lcd 15:
3(3x + 4) = 5(2x - 8)
9x + 12 = 10x - 40
10x - 9x = 12 + 40
x = 52
 
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