+0

# Fraction Question

0
277
3
+524

How would I express (2x - 1) / (x + 2) as A + B / (x+2), where A and B are integers?

I can do 2x / (x + 2) - 1 / (x+2) of course, but that doesn't work...

Any ideas?

EDIT: Sorry if I wasn't clear, but it's NOT:

$${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}} = {\mathtt{A}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{B}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}$$

But instead I need to change:

$${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}$$

Into something that looks like:

$${\mathtt{A}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{B}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}$$

Thank you!

Will85237  Jan 19, 2015

#2
+91412
+10

I think anon's logic should work but I would do it much more simply.

$$\frac{2x-1 }{ x+2}\\\\ =\frac{2(x+2)-1-4}{x+2}\\\\ =\frac{2(x+2)-5}{x+2}\\\\ =\frac{2(x+2)}{x+2}+\frac{-5}{x+2}\\\\ =2+\frac{-5}{x+2}$$

Melody  Jan 19, 2015
Sort:

#1
+5

## Now just solve for B. ðŸš²

Guest Jan 19, 2015
#2
+91412
+10

I think anon's logic should work but I would do it much more simply.

$$\frac{2x-1 }{ x+2}\\\\ =\frac{2(x+2)-1-4}{x+2}\\\\ =\frac{2(x+2)-5}{x+2}\\\\ =\frac{2(x+2)}{x+2}+\frac{-5}{x+2}\\\\ =2+\frac{-5}{x+2}$$

Melody  Jan 19, 2015
#3
+91412
+5

YES anon understood what you wanted and gave you a correct answer!

I shall show you

$$\\\frac{2x-1}{x+2}=A+\frac{B}{x+2}\\\\ multiply both sides by (x+2)\\ 2x-1=A(x+2)+B\\ 2x-1=Ax+2A+B\\ 2x-1=Ax+(2A+B)\\  equating co-efficients\\ 2=A\\ -1=2A+B\\ -1=4+B\\ B=-5\\ so\\ \frac{2x-1}{x+2}=2+\frac{-5}{x+2}\\\\$$

so maybe you owe anon an appology

Melody  Jan 19, 2015

### 31 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details