Asked by Kat

I've been trying to figure this one out all day. I have to use either substitution or elimination to solve a system for how many cookies and brownies two kids sold at a bake sale.
The system is:
c+b=600
$.35c+$.75b=$360
(obivously, c is the amount of cookies they sold,[35 cents each] and b is the amount of brownies they sold[75 cents each)
If you could help me out that'd be great. Thanks.
Answered by bobpursley
multiply the top equation by .35
.35c+.35b=600
.35c+.75b=360

subtract the bottom equation from the top.

.35b-.75b=240
solve for b.
Answered by bobpursley
as you probably will find out, this leads to a negative price for the brownies, a most unusual situation. They must be really bad, to pay folks to take them. Are you certain the equations are correct?
Answered by Kat
absolutely
Answered by bobpursley
Ok, I found the error I made. I didn't multipy the 600 by .35

Here is is.
35c+.35b=210
.35c+.75b=360

subtract the second fromfirst.
-.40b=-150
b= 3.75

check that carefully, I am tired.
Answered by Kat
it's a little off, but it's close enough for me. Thank you so much!
Answered by nattily
I got it. bobpursley was REALLY close for b. the only difference is that there's no decimal. also, c=225. just use the substitution method and you should be fine. If you're tired Kat, especially since you've been staring at it a while, i suggest forgetting about it now and getting some rest before you look at it again. As for bobpursley, i'm sure that she's quite grateful that anyone even attempted to solve it. and you were EXTREMELY close.
Answered by Anonymous
Actually, the answer is b=107 c=492
bobpursley made a typo. It is not .75, it is 1.75.
So 0.35b-1.75b=-1.4b
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