Solve For the Unknown Using More Unknowns...?
962761
treat it as a quadratic equation and use the quadratic formula:\n\nfirst subtract A from both sides:\n2pr^2 +2prh - A = 0\n\nthen plug the coefficients: 2p, 2ph, -A into the\nquadratic formula for a, b, c, respectively and simplify:\n\nr = [- b +/- sqrt(b^2 - 4ac)]/(2a)\n\ni'll leave it to you to do the tedious steps to get\nr = [ph +/- sqrt(p^2h^2 + 2pA)]/(2p)
547629
assuming u is an unknown variable y=0
769971
This is solvable, and you wrote it very well!\n\nThe key to setting this up is knowing that for any 2-digit number, you can write it as 10a + b.\n\nSo your problem states that a+b=9 (the sum of the tens digit and ones digit is 9), and it states that a=3b+1 (the tens digit is 1 more than 3 times the ones digit).\n\nSo now you have 2 equations in 2 unknowns, which you can easily solve for:\n\na + b = 9\na = 3b + 1\n\nSubstitute a = 3b + 1 into the first equation to get:\n\na + b = 9\n(3b + 1) + b = 9\n4b + 1 = 9\n4b = 8\nb = 2\n\nNow you can easily solve for a, which equals 7 since a+b=9.\n\nSo your 2-digit number is 72.\n\nCheck: 7+2=9\n7 = 1 + 3*2\n\nIt checks out, so our answer is correct!\n\n~ ♥ ~
626899
solve as many questions as you can
869887
If you have 2 equations, just solve for one variable for one of the equations. Take that equation and plug it into the same variable for the other equation. You will solve for that variable, and plug in that number to any of the 2 equations to solve for the other variable.
846745
Typically you know three out of the four variables and solve for the last variable.\n\nBut sometimes they get tricky by giving you only two but one of the remaining variable is related to one or two of the given variables.
1142710
Use this formula, you can find millions of them\n\nL x W = 2(L + W)\n\nuse any number for L and solve for W
417457
at college level we have different method one of these is this \nwhen solving problem you feel that u required some quantity let them (these are called Unknowns )the no of unknown = no of conditions using these conditions remove all unknowns you will get your answer definitely if it worked ask me for the next \nthanks
1129053
Go to the site below.  It works great, you just put the colors where they belong and click solve, it gives you a step by step on how to solve it.
966925
There are 3 solutions for ABCDEFGHJ -\n\n2 6 7 8 3 5 1 4 9\n2 6 7 9 5 3 1 4 8\n6 7 8 4 1 3 2 5 9
1076564
What do you want to solve for? x? y? or do you want an ordered pair (x,y) that works (integers maybe?)
599042
I'll solve it for you. Send it over.\n\nstill\nshell\ngrill\ncliff\n give us a clue
902349
Hynton has it.  Just remember that you'll be restricted to x > 0 because of the original form of the problem.
201908
Ah, systems.\n\nthe first problem- there are many ways you can choose to do this, but I think we should multiply both sides of the first equation by two so that they positive and negative y's will cancel out.\n\nSo now we have:\n2(2x + y)=2(4)\n3x - 2y = -1\n\nSimplified:\n4x + 2y = 8\n3x - 2y = -1\n\nnow you add down, so you get:\n7x +0y = 7\n7x =7\nx=1\n\nsubstitute 1 for x in any equation:\n2(1)+y = 4   \n2 + y = 4\ny = 2\n\n\n\nfor the second one- lets cancel out the y's again.\n\n5x - 3y = 14\n-3(2x - y) = (6)-3\n\n5x - 3y = 14\n-6x +3y = -18\n\nAdd them together:\n-x = -4\nx = 4\n\nThen solve for y:\n5x - 3y = 14\n5(4) - 3y = 14\n20 -3y = 14\n-3y = -6\ny = 2
78415
You have five different incognitas there:\nTo solve for all the five contemporarily, you need at least other four relations (equations) involving them and no other variable.\n\nOtherwise, you must keep one of them as the variable and the others as non-specified parameters:\n\nAs there are an x and a z, it is possible that both had to be your variables and the others to be parameters: in this case you need one more relation (equation) between x and z.\n\nAnyway: let's solve for x:\n\nTake the non-x terms on the other side of the equation:\n\n45x = 45 z - 6 + k9 + mp\n\nDivide by 45:\n\nx = z + (k9 + mp - 6)/45\n\nYou have a 45°-slope straight in the xz plane, with variable intersection on the z axis depending on k, m and p\n\nHope this helps
36555
There are few ways of solving for the unknown in a quadratic equation.\n\nIn general, a quadratic equation is of the form:\n   a*x^2 + b*x + c = 0, a != 0    --- (1)\nDividing (1) by "a" through out, we get,\n   x^2 + (b/a)*x + (c/a) = 0  --- (2)\n\nOne way of solving is to really guess the solution. It works like this:\nAssume we have (somehow) factorised (2) into a form like:\n   (x - p)(x - q) = 0   --- (3)\n(3) can also be written as (by mutiplying term by term),\n   x^2 - (p+q)*x + p*q = 0   --- (4)\nComparing (2) and (4),\n   p + q = -b/a\n   p*q   = c/a\nSo, if the equation is written like in (2), the sum of roots is negative of x co-efficient, and the product is the constant. A "root" is a value of x that satisfies the given equation (1).\n\nFor e.g, given,\n   x^2 - 6*x + 5 = 0,\nthe sum of roots must be 6 and the product must be 5. So (we guess), the roots must be 1 and 5. Easy!\n\nBut "guessing" is not always possible :(\n\nLook at this equation:\n   x^2 - 6*x + 10 = 0\nThe roots of this are 3 + i and 3 - i. Not easy to guess right?\n\nA generic way of solving is called "completing the square". And it is a very clever piece of work! It works like this:\nEquation (2) again:\n   x^2 + (b/a)*x + (c/a) = 0\nThe aim is to get the equation into a form that looks like:\n   (x + something)^2 = somethingelse\nThen we can solve for x easily.\n\nLet us first get rid of the constant term to the other side:\n   x^2 + (b/a)*x = -(c/a)\nLet us re-write this again like:\n   x^2 + 2*(b/2a)*x = -(c/a) --- (6)\nLook at the left hand side (LHS) of the equation.. it is looking like: x^2 + 2*something*x. If only we have something^2, then we can put it in the form we want. So we'll add that term:\n   x^2 + 2*(b/2a)*x + (b/2a)^2 = -(c/a) + (b/2a)^2\nWe've got it! We now have,\n   ( x + (b/2a) )^2 = -(c/a) + (b/2a)^2\n\nWe can solve for x quite easily from here (left as exercise to the reader).\n\nThe "magic" equation (-b plus or minus sqare-root of etc...) is really derived by completing the square.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.