# Strategy for Simplfication?

#### TheJason

Staff member
Moderator
Simplify

@MarkFL @harpazo this seems like this might go on forever.

$$\displaystyle \dfrac{3}{5b} - \dfrac{2}{4b} + \dfrac{1}{7c} + \dfrac{7}{21c}$$

$$\displaystyle \dfrac{-2}{4b} + \dfrac{1}{7c}$$

$$\displaystyle \dfrac{(-2)(7c) + (4b)(1)}{(4b)(7c)} = \dfrac{(-2)(7c)}{(4b)(7c)} + \dfrac{(4b)(1)}{(4b)(7c)}$$

Cancel out stuff.

$$\displaystyle \dfrac{-2}{4b} + \dfrac{1}{7c}$$

$$\displaystyle \dfrac{3}{5b} -\dfrac{2}{4b} + \dfrac{1}{7c}+ \dfrac{7}{21c}$$

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
I noticed you were ending up where you started on several threads. I think what is intended here is to combine all terms...

$$\displaystyle \frac{3}{5b}-\frac{2}{4b}+\frac{1}{7c}+\dfrac{7}{21c}=$$

$$\displaystyle \frac{3}{5b}\cdot\frac{3\cdot4\cdot7c}{3\cdot4\cdot7c}-\frac{2}{4b}\cdot\frac{3\cdot5\cdot7c}{3\cdot5\cdot7c}+\frac{1}{7c}\cdot\frac{3\cdot4\cdot5b}{3\cdot4\cdot5b}+\dfrac{7}{21c}\cdot\frac{4\cdot5b}{4\cdot5b}=$$

$$\displaystyle \frac{42c+200b}{420bc}=\frac{100b+21c}{210bc}$$

#### harpazo

##### Pure Mathematics
Banned
Simplify

@MarkFL @harpazo this seems like this might go on forever.

$$\displaystyle \dfrac{3}{5b} - \dfrac{2}{4b} + \dfrac{1}{7c} + \dfrac{7}{21c}$$

$$\displaystyle \dfrac{-2}{4b} + \dfrac{1}{7c}$$

$$\displaystyle \dfrac{(-2)(7c) + (4b)(1)}{(4b)(7c)} = \dfrac{(-2)(7c)}{(4b)(7c)} + \dfrac{(4b)(1)}{(4b)(7c)}$$

Cancel out stuff.

$$\displaystyle \dfrac{-2}{4b} + \dfrac{1}{7c}$$

$$\displaystyle \dfrac{3}{5b} -\dfrac{2}{4b} + \dfrac{1}{7c}+ \dfrac{7}{21c}$$
What is the LCD? When you figure that out, then multiply every term by the LCD.

Staff member