have a problem I keep getting the wrong answer

Well, what answer do you get?
Why notr show us how you are working the problem?
 
LN said:
7[2-(3+4r)]-2 = -9+2(1-15r)
Click to expand...

If you had posted your work and the wrong answer, we could much more quickly see just what the problem is. Without that, it's a guessing game.

The most likely problem is misinterpretation of items in and outside parentheses. For example, on the left side, what is that "-2"? Is that supposed to be an exponent, or a -2 tagged onto the LN7[2-(3+4r)]?
 
Hello, LN!

If you don't show us your work, we can't tell you what you're doing wrong ... duh!

The right answer is r = 1\displaystyle r\,=\,1 . . . but, of course, you know that, right?

I'll do the whole problem for you
    \displaystyle \;\; and you'll answer, "Oh, I forgot the minus.
    \displaystyle \;\; or worse . . . "I was looking at the wrong answer ...hahaha!"

Okay, I'll do it for you . . . just this once.


We are given: \(\displaystyle \L\;7[2\,-\,(3\,+\,4r)]\,-\,2\;=\;-9\,+\,2(1\,-\,15r)\)

The left side is:  7[2 − 3 − 4r] − 2  =  7[−1 − 4r] − 2  =  −7 − 28r − 2  =  −28r − 9\displaystyle \:7[2\,-\,3\,-\,4r]\,-\,2\;=\;7[-1\,-\,4r]\,-\,2\;=\;-7\,-\,28r\,-\,2\;=\;-28r\,-\,9

The right side is:  −9 + 2 − 30r  =  −7 − 30r\displaystyle \:-9\,+\,2\,-\,30r\;=\;-7\,-\,30r


The equation becomes: \(\displaystyle \L\;-28r\,-\,9\;=\;-7\,-\,30r\)

Can you finish it now?
 
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