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Having trouble with fractions
- Thread starter smooty21
- Start date
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.
\(\displaystyle \L\\\frac{1}{5}(3x+4)=\frac{1}{3}(2x-8)\)
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.
Another way is multiply each side by the lcd 15:smooty21 said:1/5 (3X + 4) = 1/3 (2X - 8)
3(3x + 4) = 5(2x - 8)
9x + 12 = 10x - 40
10x - 9x = 12 + 40
x = 52