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

#1**+5 ****Start by multiplying both sides by (x+2), giving 2x-1 = A(x+2)+B **

**Now, multiply out and equate x terms on the left to x terms on the right, and we have A=2, and also -1=2A+B **

**Now just solve for B. ðŸš²**

Guest Jan 19, 2015

#2**+10 **

Best Answer

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}$$

Don't give anon negative points for trying to help you!!!

Melody
Jan 19, 2015

#3**+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