Confusing math problem?
826747
That means each of the baket cannot contain 8 apples. so all but 1 shall contain 8 and  z in other \n\nso y = 8(x-1) +z where z < 8 = 8x -8 + z\n5x+12 = y\nwe need to solve these\n\n5x+12 = 8x - 8 + z\n\n20 = 3x + z\n\nx =6 means z = 2 y = 42\nx=5 means z =5 y = 37\nx=4 means z = 8 not possible\n\nthere are 6 baskets and 42 appples\n\nor 5 baskets and 37 apples\n\nboth solutions meet above criteria
737332
It's unclear what that equation says.
892068
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
380004
I don't understand.  Can you explain the problem a little better?
987327
Find a number X such that  2X = (X/2) + 99\n\nmultiply both sides by 2 to get  4X = X + 198\nsubtract X from both sides to get 3X = 198\ndivide both sides by 3 to get X = 66\n\ncheck work\n2*66 = 132\n66/2 + 99 = 132
609300
This is more of a word problem or a riddle than a math problem.\n\nThe answer is 7 (SEVEN).  Remove its first letter and it becomes EVEN!
934652
So let's say the problem looks like this\n\n   11\nx 12\n--------\n        <-----the answer from 11*2\n       <------answer from 11*1\n\nfirst hide the 1 from the 12 and work out the problem as 11*2\nthen work on multiplying the 1 with 11 but make sure you start putting your answer down below your first answer and the answer starts in the column that is below the 1 from the 12.\n\nfinally when the two answers are found, you can add them and get your final answer. SOOO here's what it looks like\n\n      11\n   x  12\n--------------\n           22    (this is from 11x2)\n +  11       (this is from 11 x1)\n------------\n     132 = final answer!
85841
How about this? Straight multiplying gives you the following:\n\n18b^3/b^2\n\nb^3 = bxbxb\nb^2 = bxb\n\nYou have 3 bs on top and 2 on bottom. Whenever you have the same number on top and bottom, it cancels out. Since there are 2 bs on bottom, 2 of the ones on the top cancel out as well. This leaves you with 18b.\n\nDid that help at all?
870123
Problem: 17x=51\n\nsolution:  17x=51 (divide 17x  by 17 which will cancel out and will leave x by itself, then do the same for 51. so 51 divided by 17 is equal to 3.) you are left with x on one side and 3 on the other, therefore x=3
795670
Your answer key is wrong, or you are not reading the right answer.\nFrom the data you presented above y = x^3 is OK.
1162175
Question is confusing and vague. Please clarify.
74071
That is right. Just put them in a calculater to check. 2/9=0.22222222 and 8/36=0.2222222. Either they are wrong or you misinterpred the question and it was asking something else. The best way to solve a problem like that is to cross multiply. 2/9=8/? turns into 2*?=8*9.
622612
Joel, you need some help.  This is one of the easiest questions I have ever heard!  You should know this, man!  You're gonna fail your test if you keep cheating on here and don't actually figure out how it works.  Good luck, brotha'\n\nL = -2 or L(x) = -2
912865
bring the a's to one side... u then get 9a = 27 ... divide and answer is a = 3
617282
are u sure its x2? shouldn't it be 2x+10x=(a number) usually there is a number at the end of the equation and u have 2 solve for "x". add details to ur question and then i may be able 2 help u.
515169
u r surely missing some information\nlet us put this in the form of an equation\nlet y=A&B, x=A,B&C\nthen we have,\n275-x=120+110+100-20-30-y+x\nnow we have two variables and one equation\nby the way what dou want to know?\nwhat is the question exactly?
490047
I think a calculator would be great for this one! It sounds perfect for a math wiz to try...Good luck! MayMay