Intermediate Algebra: solve 2k + ar + r - 3y for r
- Thread starter mikey13
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Re: here it is right
Get all the 'r's together.
Subtract 'r'
2k+ar - r = -3y
Get the 'r's alone -- only terms with 'r' are allowed.
Subtract '2k'
ar - r = -3y - 2k
Get only one 'r'
Use the distributive property (sometimes called "factoring out")
(a-1)*r = -3y - 2k
Get 'r' REALLY alone.
Divide by (a-1)
\(\displaystyle \L\,r = \frac{-3y - 2k}{a-1}\)
The secret here, if there is one, is just to go one step at a time. No need to see the end before you get started.
In this case, you may wish to rewrite it a bit.
\(\displaystyle \L\,r = \frac{3y + 2k}{1-a}\)
It's a useful exercise to track down how that happened and how it's exactly the same.
2k + ar = r - 3ymikey13 said:
Get all the 'r's together.
Subtract 'r'
2k+ar - r = -3y
Get the 'r's alone -- only terms with 'r' are allowed.
Subtract '2k'
ar - r = -3y - 2k
Get only one 'r'
Use the distributive property (sometimes called "factoring out")
(a-1)*r = -3y - 2k
Get 'r' REALLY alone.
Divide by (a-1)
\(\displaystyle \L\,r = \frac{-3y - 2k}{a-1}\)
The secret here, if there is one, is just to go one step at a time. No need to see the end before you get started.
In this case, you may wish to rewrite it a bit.
\(\displaystyle \L\,r = \frac{3y + 2k}{1-a}\)
It's a useful exercise to track down how that happened and how it's exactly the same.