# Math Help - Gr11 Fractions

1. ## Gr11 Fractions

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

2. Originally Posted by TH1
3k/2 - K+3/3 = 8 - k+2/4
Try to use parentheses, 'cause it becomes ambiguous.

I assume you mean $\frac32k-\frac{k+3}3=8-\frac{k+2}4$

Multiply both sides by $12$, this yields

$6\cdot3k-4(k+3)=96-3(k+2)$

which is easy to take.

The other ones have the same treatment.

3. Hello, TH1;70157!

Krizalid is absolutely correct . . .

With an equation, we can eliminate the fractions (legally).

Your second problem: . $\frac{3k}{2} - \frac{k+3}{3} \:= \:8 - \frac{k+2}{4}$

The LCD is 12. .Multiply through by 12.

Multiply everything by 12: . $12\cdot\frac{3k}{2} \,- \,12\cdot\frac{k+3}{3} \;\;=\;\;12\cdot8 \,- \,12\cdot\frac{k+2}{4}$

. . and reduce: . $\not{1}\!\!\!\not{2}^6\cdot\frac{3k}{\not{2}} \,- \,\not{1}\!\!\!\not{2}^4\cdot\frac{k+3}{\not{3}} \;\;=\;\;12\cdot8 \,- \,\not{1}\!\!\!\not{2}^3\cdot\frac{k+2}{\not{4}}$

. . and we have: . $6(3k) - 4(k+3) \;=\;96 - 3(k+2)$

Can you finish it now?