For these fractions all I knew was to find the lowest common denominator and after that I wasnt sure what to do
3k/2 - K+3/3 = 8 - k+2/4
3p/4 + p-5/3 = 1/2
u-3/4 -2 = 3u/2 = 2u+1/5
U-5/2 = 2u+5
Try to use parentheses, 'cause it becomes ambiguous.
I assume you mean $\displaystyle \frac32k-\frac{k+3}3=8-\frac{k+2}4$
Multiply both sides by $\displaystyle 12$, this yields
$\displaystyle 6\cdot3k-4(k+3)=96-3(k+2)$
which is easy to take.
The other ones have the same treatment.
Hello, TH1;70157!
Krizalid is absolutely correct . . .
With an equation, we can eliminate the fractions (legally).
Your second problem: .$\displaystyle \frac{3k}{2} - \frac{k+3}{3} \:= \:8 - \frac{k+2}{4}$
The LCD is 12. .Multiply through by 12.
Multiply everything by 12: .$\displaystyle 12\cdot\frac{3k}{2} \,- \,12\cdot\frac{k+3}{3} \;\;=\;\;12\cdot8 \,- \,12\cdot\frac{k+2}{4}$
. . and reduce: .$\displaystyle \not{1}\!\!\!\not{2}^6\cdot\frac{3k}{\not{2}} \,- \,\not{1}\!\!\!\not{2}^4\cdot\frac{k+3}{\not{3}} \;\;=\;\;12\cdot8 \,- \,\not{1}\!\!\!\not{2}^3\cdot\frac{k+2}{\not{4}}$
. . and we have: .$\displaystyle 6(3k) - 4(k+3) \;=\;96 - 3(k+2)$
Can you finish it now?