tag:blogger.com,1999:blog-2217022227008948602.post132813691403218908..comments2023-10-29T06:43:10.927-07:00Comments on The Puzzle Page: I Want Candy!Oosterwalhttp://www.blogger.com/profile/14514993727423832898noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-2217022227008948602.post-2776872457823443822010-02-25T15:16:48.776-08:002010-02-25T15:16:48.776-08:00I am 13 in algebra. We have to figure this problem...I am 13 in algebra. We have to figure this problem out. But using substitution. I have no idea how to do it!!!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2217022227008948602.post-64391604760384723002009-11-09T21:40:03.407-08:002009-11-09T21:40:03.407-08:00TELL ME THE ANSWER PLEASETELL ME THE ANSWER PLEASEAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2217022227008948602.post-10440593832528933602009-02-14T13:49:00.000-08:002009-02-14T13:49:00.000-08:00J = jellybeansC = chocolateA = CaramelL = licorice...J = jellybeans<BR/>C = chocolate<BR/>A = Caramel<BR/>L = licorice<BR/><BR/>J + 2C = $2<BR/>4A + J = $3<BR/>3L + J + A = $1.50<BR/><BR/>AND HOW ON EARTH DO THEY DO THE NEXT BIT?????Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2217022227008948602.post-78370073803118083732009-02-04T06:17:00.000-08:002009-02-04T06:17:00.000-08:00There is more to this than meets the eye. With fo...There is more to this than meets the eye. <BR/>With four unknowns and only three equations there are several possibilities:<BR/>1) There could be no solution<BR/>2) There could be a set of solutions that forms a line in the 4-dimensional space.<BR/>3) There could be a set of solutions that forms a plane or hyper-plane.<BR/><BR/>There <B>cannot</B> be a set of solutions that includes only a single point. Hence, there is not enough information to compute the price of the four types of candy.<BR/><BR/>However, it turns out that all of solutions of these three equations satisfy the equation:<BR/>J+C+L+A = $2 as Morris Tau showed.Ken Stanleyhttps://www.blogger.com/profile/02500250199126923469noreply@blogger.comtag:blogger.com,1999:blog-2217022227008948602.post-38021461863075290182009-01-31T14:26:00.000-08:002009-01-31T14:26:00.000-08:00So what is the answerSo what is the answerAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2217022227008948602.post-66229829403165000282009-01-10T17:10:00.000-08:002009-01-10T17:10:00.000-08:00Where is this place? I want some free jellybeans!...Where is this place? I want some free jellybeans!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2217022227008948602.post-19660905964708917322008-03-16T21:34:00.000-07:002008-03-16T21:34:00.000-07:00J = jellybeansC = ChocolateA = CaramelL = Licorice...J = jellybeans<BR/>C = Chocolate<BR/>A = Caramel<BR/>L = Licorice<BR/><BR/>From the problem you can write these three equations:<BR/>1) J + 2C = $2<BR/>2) 4A + J = $3<BR/>3) 3L + J + A = $1.50<BR/><BR/>Note you don't need to solve for J,C,A,L just the sum of them.<BR/><BR/>Rewriting the above 3 equations:<BR/>3 x (J + 2C) = 3 x $2<BR/>4A + J = $3<BR/>2 x (3L + J + A) = 2 x $1.50<BR/><BR/>Add up all three and you get:<BR/>6J + 6C + 6A + 6L = $12<BR/>or<BR/>J + C + A + L = $2Anonymousnoreply@blogger.com