Associated Topics || Dr. Math Home || Search Dr. Math

### Solving for a Variable in One Equation

```
Date: 3/18/96 at 20:52:4
From: Anonymous

and solve for the indicated variable:

(2r+3)(2r-1) = -(3r+1)

Thank you,
Robert
```

```
Date: 3/20/96 at 10:56:55
From: Doctor Patrick

Hi!

To solve for r we are going to have to set one side of the equation
to 0.

To do this we must add the -(3r+1) to both sides.  This gives us
(2r+3)(2r-1) + (3r+1) = 0 . Unfortunately, we can't do anything with
the problem until we clear the parentheses.  Once we do this we get
(4r^2 + 4r - 3) + (3r+1) = 0.  Remove the parentheses:
4r^2 + 4r - 3 + 3r + 1 = 0.
Now we add like terms and get 4r^2 + 7r - 2 = 0.

Now all that is left to do is to factor the problem and solve each half
separately. I'll let you try this part on your own, but write back if
you need more help.

Good luck!

-Doctor Patrick,  The Math Forum

```
Associated Topics:
Middle School Equations

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search