[j/2] - [j-3/4] = 9/4
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do u mean $\displaystyle \frac{j}{2} - \frac{j-3}{4}$
Originally Posted by ambergarrett055 [j/2] - [j-3/4] = 9/4 $\displaystyle \frac{j}{2} - \left(j - \frac{3}{4}\right) = \frac{9}{4}$ or $\displaystyle \frac{j}{2} - \frac{j-3}{4} = \frac{9}{4}$ ?
The second one....j-3 over 4. Sorry. I'm new to this.
$\displaystyle \frac{j}{2} - \frac{j-3}{4} = \frac{9}{4}$ multiply every term by 4 to clear the fractions ... $\displaystyle 2j - (j - 3) = 9$ can you finish from here?
Yep. Thanks..I just wanted to make sureI multiplied both sides by four to get rid of the fractions. Thanks!
Originally Posted by ambergarrett055 Yep. Thanks..I just wanted to make sureI multiplied both sides by four to get rid of the fractions. Thanks! next time, show all your work.
I have another equation, too. Do I multiply both sides by 10? [9t + 1] / 5 = [t+8] /10 + [t-6] /10
Originally Posted by ambergarrett055 I have another equation, too. Do I multiply both sides by 10? [9t + 1] / 5 = [t+8] /10 + [t-6] /10 yes
I keep getting t=0 but when I check it it doesn't work out. I'll just keep trying. Thanks!
Originally Posted by ambergarrett055 [9t + 1] / 5 = [t+8] /10 + [t-6] /10 $\displaystyle 2(9t+1) = (t+8) + (t-6)$ $\displaystyle 18t + 2 = 2t + 2$ $\displaystyle 18t = 2t$ $\displaystyle t = 0 $ check ... $\displaystyle \frac{1}{5} = \frac{8}{10} - \frac{6}{10}$ $\displaystyle \frac{1}{5} = \frac{1}{5}$ checks good.
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