I need help w/ math hw, (Problem solving basically) |
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 Apr 9 2005, 03:34 PM
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#1
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 Member  Group: Member Posts: 18 Joined: Mar 2005 Member No: 117,664  |
I forgot how to figure this out
 Heres the prob. : What are two whole numbers whose product is 294 and its quiotient is 6. Ive asked a bunch of ppl and they forgot how to. Please help. I dont rlly need answers but can u explain how to solve this equation thingy? THANKS!   down with math!
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Apr 9 2005, 03:38 PM
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#2
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 Another ditch in the road... you keep moving Group: Member Posts: 6,281 Joined: Jan 2005 Member No: 85,152 |
QUOTE What are two whole numbers whose product is 294 and its quiotient is 6. Ive asked a bunch of ppl and they forgot how to. guess and check... try some numbers, and work out if you need to go up or down or what. oh. the answer s 42 and 7. first two i treied, and next time, try in homework help |
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Apr 9 2005, 03:40 PM
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#3
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 Yawn Group: Staff Alumni Posts: 9,530 Joined: Nov 2004 Member No: 65,772 |
I would love to tell you the answer, cept i don't remember how to do those and even if i did, i would probably be telling you the wrong answer since math isn't my thing. There is a homework help section under interests that you can post this stuff in next time
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| *autumn.* |
Apr 9 2005, 05:29 PM
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#4
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yes the answer is 7 and 42, but here is the work involved:
x/y = 6 xy = 294 x = 6y 6y(y)=294 6y^2 = 294 294/6 = y^2 (y squared) 49 = y^2 y = 7. therefore, x = 6 multiplied by 7, which is 42. just algebra 1. |
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| *suddenly she* |
Apr 9 2005, 05:45 PM
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#5
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yeah. substitution is your friend.
basically, write out two equations using the variables, isolate a variable in one equation, and substitute it in the other equation. |
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Apr 10 2005, 07:50 AM
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#6
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 Senior Member Group: Member Posts: 840 Joined: Dec 2004 Member No: 73,927 |
QUOTE(autumn. @ Apr 9 2005, 5:29 PM) yes the answer is 7 and 42, but here is the work involved: x/y = 6 xy = 294 x = 6y 6y(y)=294 6y^2 = 294 294/6 = y^2 (y squared) 49 = y^2 y = 7. therefore, x = 6 multiplied by 7, which is 42. just algebra 1.  Yup, thats how I'd do it. |
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Apr 11 2005, 12:49 AM
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#7
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 白人看不懂 !!!! Group: Member Posts: 3,838 Joined: Aug 2004 Member No: 40,824 |
O GOD I love substitution!! So much better than that ANNOYING elimination.
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| *wind&fire* |
Apr 11 2005, 01:52 AM
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#8
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^ why arent you closing this?
[edit] fageddaboudit... just realised that they might have not seen this |
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May 6 2005, 09:26 PM
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#9
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 Senior Member Group: Member Posts: 2,881 Joined: Apr 2005 Member No: 132,134 |
QUOTE yes the answer is 7 and 42, but here is the work involved: x/y = 6 xy = 294 x = 6y 6y(y)=294 6y^2 = 294 294/6 = y^2 (y squared) 49 = y^2 y = 7. therefore, x = 6 multiplied by 7, which is 42. just algebra 1. the way i would do it. |
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May 7 2005, 09:34 AM
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#10
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 reluctantly gazing Group: Member Posts: 472 Joined: Mar 2005 Member No: 120,555 |
QUOTE(JoeSocks @ Apr 9 2005, 4:34 PM) I forgot how to figure this out Heres the prob. : What are two whole numbers whose product is 294 and its quiotient is 6. Ive asked a bunch of ppl and they forgot how to. Please help. I dont rlly need answers but can u explain how to solve this equation thingy? THANKS! down with math! with only one variable....almost the same thing towards the end, but easier in the beginning for me because i hate using more variables than i need to. Let: x=smaller whole number 294/x=larger whole number so.... (294/x)/x=6 change to multiplication: (294/x) * (1/x) = 6 294/x^2 = 6 multiply both sides by x^2 (x squared) 294 = 6x^2 divide both sides by 6 49 = x^2 they are both perfect squares, find the square root of both: 7 = x plug it into the beginning.. x=smaller whole number ---> 7 294/x=larger whole number ---> 42 |
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