+0

# Fraction Question

0
554
3
+555

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
+94086
+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
#1
+5

## Now just solve for B. ðŸš²

Guest Jan 19, 2015
#2
+94086
+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
+94086
+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