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Algebra 16 Online
OpenStudy (anonymous):

285p+15f=1275 270p+30f=1350 f represents the # of calories in gram of fat and p represents the # of calories how many calories are in a gram of fat ?

OpenStudy (anonymous):

please someone help me

OpenStudy (anonymous):

solve the system of equations

OpenStudy (anonymous):

ok im going 2 try

OpenStudy (anonymous):

When I solved the system I got p=4 and f=9. I'm not sure how to apply that exactly. Does it mean that 9 fat grams is 4 calories...so 1 fat gram is 2.25 calories. sorry...I can't be of any more help than that

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

doing a little more work, I've determined there are 9 calories in 1 gram of fat

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

zorry but i try to get p and f but i couldnt can you tell me how you got it

OpenStudy (anonymous):

OK

OpenStudy (anonymous):

285p+15f=1275 -2(285+15f=1275) -570p-30f=-2550 270p+30f=1350 270p+30f=1350 __________________

OpenStudy (anonymous):

multiply the top equation by (-270/285) then add the two equations, which will eliminate the p terms and give you a simple equation for f.

OpenStudy (anonymous):

sorry...hit wrong button...simplifying the part with the dotted line, I get -300p=-1200 solve for p and you get 4 I multiplied by -2 above so I could solve the system of equations by elimination. Since I know p, and can find f by substituting 4 into the given equations. When I did that, I got f=9.

OpenStudy (anonymous):

What douglas...said is also a way to solve. I didn't want to work with fractions, so i decided to eliminate the other variable.

OpenStudy (anonymous):

thank you that help me alot thanks both of you

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