SOLUTION: Solve for x. 7[2-(3+4r]=-9+2(1-5r) 7[-1+2r] = -9+2.1 2.-15r -7+2r -7 -30r +7 +7 2r = 0 -30r +30r +30r 32r = 0 I have no idea if I am even clos

Algebra ->  Equations -> SOLUTION: Solve for x. 7[2-(3+4r]=-9+2(1-5r) 7[-1+2r] = -9+2.1 2.-15r -7+2r -7 -30r +7 +7 2r = 0 -30r +30r +30r 32r = 0 I have no idea if I am even clos      Log On


   



Question 269920: Solve for x.
7[2-(3+4r]=-9+2(1-5r)
7[-1+2r] = -9+2.1 2.-15r
-7+2r -7 -30r
+7 +7
2r = 0 -30r
+30r +30r
32r = 0
I have no idea if I am even close with this one. The brackets are throwing me off I think.

Found 3 solutions by drk, checkley77, unlockmath:
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
we start with
(i) 7%282-%283%2B4r%29%29=-9%2B2%281-5r%29
first 2-(3+4r) is -1 - 4r
we get
(ii) 7%28-1-4r%29+=+-9%2B2%281-5r%29
distribute the 7 and then the 2 to get
(iii) -7+-+28r+=+-9+%2B+2+-+10r
simplify the right to get
(iv) -7+-+28r+=+-7+-+10r
the -7's cancel so we get
(v) -28r+=+-10r
adding 28r we get
(vi) 18r+=+0
and then r = 0.

Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x.
I don't see any (x) so I'll solve for r.
7[2-(3+4r)]=-9+2(1-5r) {I've added a ) after the 4r.}
7[2-3-4r]=-9+2-10r
7[-1-4r]=-7-10r
-7-28r=-7-10r
-28r+10r=-7+7
-18r=0
r=0 ans.
Proof:
7[2-(3+4*0)]=-9+2(1-5*0)
7[2-3+0]=-9+2*1
7*-1=-9+2
-7=-7

Answer by unlockmath(1688) About Me  (Show Source):
You can put this solution on YOUR website!
Hello,
Let's take it step by step: (Note: we'll be solving for "r" not "x")
7[2-(3+4r]=-9+2(1-5r)
7[-1-4r]=-9+2(1-5r) Notice this step was corrected.
7[-1-4r] = -9+2-10r This step is corrected also.
-7-28r= -7 -10r From this step add 7 to both sides to get:
-28r=-10r Now add 10 r to both sides to get:
-18r=0 Divide both sides by -18 to result in:
r=0
Plug 0 into the original equation to see if it works.
Make more sense?
RJ
There's a book I wrote that might help:
www.math-unlock.com