What's the process of solving equations...?
141310
to solve for X, you have to separate X from everything else. in that problem 2X+3X=5, you would combine the 2X and 3X to get 5X. so the equation end up as 5X=5. from there, divide both side by 5. 5X/5=X and 5/5=1 so X=1.
955298
The KEY to solving any algebra problem, as I tell my students, is to get your unknown variable all by itself on one side of the equal sign.\n\nYou do this by "undoing" everyingthing that's being done to it right now.  And you "undo" the easy stuff first, like addition and subtraction, and then the more complicated stuff like multiplication and division.\n\nSo if you're adding 3 to X, then you would subtract 3.  And remember, what you do to one side of the equal sign, you MUST do to the other, so you would need to subtract 3 from both sides in my example.
316686
the process u follow is correct, but actually it is wrong. fundamentally wat u did is ok, but technically it is wrong. because this becomes a quadratic equation which has more than 1 root. so this equation has more than 1 answers for which it is true.
999333
There are several simple rules for re-arranging and solving equations.  It would be in your best interest to try and learn them, they aren't very hard.
878962
first you get one variable on one side by its self (lets say x), solve for that variable (x), then you put the answer in for "x" and solve for the second variable, Just like you would if you only had one variable to begin with.
993104
take the equation x^2 - 7x + 12 and do the following  (x ) (x) now...see that there is a - and a + in the equation....you need to do this ( x -  ) ( x - ) 'cause you put the minus in the first factor in the second factor you take the - and the + and multiply them....So you get (x- ) (x -) the other 2 terms for each equation comes from the 7x and the 12.  What you do is to find 2 numbers that multiplied gives you 12 and added give you 7....The numbers are 3 and 4 ...Then put them into your factors (x-3)(x-4).\n\nFor the other 2 equations you do the same process\n\n(x^2 - 8x + 7) = (x-  ) (x-  ). In this one, you have to find 2 numbers that multiplied give you 7 and added give you 8. They are 1 and 7. Then (x^2 - 8x + 7) = (x- 1 ) (x- 7 ).\n\nw^2 - 8w + 15 = (w - ) (w- ). As you saw in one and two, three is the same....so the factors are 3 and 5. Then, w^2 - 8w + 15 = (w -3 ) (w-5 )\n\nFor the solve...\n1. x^2 - 6x = 0\n2. x^2 - 2x - 3 = 0\n\nTHe 1 is like this...you have to complete your quadratic equation then you do the following x^2 - 6x = 0 --> x^2 - 6x + 9 =0+ 9...The 9 in the right side was added in order to keep the two sides equal...IN the left side you got a perfect quadratic equation, hence it will be (x-3)^2= 9....Now you pass the 9 of right to the left ...then it will be [(x-3)^2]-9=0\n\n\nThe other equation x^2 - 2x - 3 = 0....This one is quite simple, it is the same as the part 1...The difference is that when you find the factors, they will be 2 numbers that multiplied give you 3 but substracted gives you 2.....The #s are 3 and 1....Here is the thing....when you break the equation into factors, you will have the following (x- ) (x+ )...Since you have a -2 that means that it is the -3 and the +1...Hence your factors will be the following (x-3)(x+1)
594770
Well we are not going to do your homework for you. But we will help you along. why not start out by writing down everything you know. they writing down what you need to find. Then look at your equations and find one that uses what you know to find one of the values you need to find.\n\nSo write it up and put it on here and we will help you along.
159422
Is this a trick question? For a given system of linear equations, there is one solution that will yield a particular answer. There are infinite possible systems that will yield that answer (x=-2). There are also multiple ways to arrive at the same  correct answer for a given system. I think we need more info to be of any real help...
695856
I'll try and help you with this one and give you details on how to solve others.\n\nTo solve by substitution you must first solve for a single variable.  Take the first equation, for example and solve for x.\nx - y = 3\n\nAdd y to both sides:\nx = y + 3\n\nNow substitute this equation for x into the second equation:\n3x - 2y = 3\n3 (y+3) - 2y = 3\n\nMultiply the 3 through:\n3y + 9 - 2y = 3\n\nCombine like terms:\n(3y - 2y) + 9 = 3\ny + 9 = 3\n\nSubtract 9 from both sides:\ny = 3 - 9\ny = -6\n\nNow solve for x by substituting this back into your original equation:\nx - y = 3\nx - (-6) = 3\nx + 6 = 3\nx = 3 - 6\nx = -3\n\nSo your solution is x = -3, y = -6.\n\nIt's a good idea to double check your answer:\nx - y =? 3\n-3 - (-6) =? 3\n-3 + 6 =? 3\n3 = 3 <-- check\n\n3x - 2y = 3\n3(-3) - 2(-6) =? 3\n-9 - (-12) =? 3\n-9 + 12 =? 3\n3 = 3 <-- check\n\nNow to answer the second part of your question.\n\nIf you get to a solution with actual values (like we did) then you have an *independent* system of equations.\n\nIf you get to a solution like y = y or x = x or n = n, then you have a *dependent* system of equations.  An example of this would be:\nx = y + 1\n2x = 2y + 2\n\nHere you can see that the second equation is just double the first, so it doesn't give you any more information.  If you substitute x = y + 1 into the second equation you get:\n2(y + 1) = 2y + 2\n2y + 2 = 2y + 2\n2 = 2 <-- dependent system (x or y can be anything and the system is true)\n\nThe last case would be an *inconsistent* system.  This is when you get to an answer like 2 = 7.\nx = y + 1\n2x = 2y - 7\n\nHere when you substitute in x = y + 1, you get:\n2(y+1) = 2y - 7\n2y + 2 = 2y - 7\n2 = -7 <-- inconsistent system (x or y can be anything and the system will never be true).\n\nSo in summary, with your example:\nx = -3 \ny = -6\nAnd since you have a result, the system is independent.
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.
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
447443
Ok...you have to solve two equations\n\nFIRST ONE\n\nif (8-4x) < 5\n\nthen (8-4x-5) < 0 (by subtracting 5 from both sides of the equation)\n\nthus (3-4x) < 0 (obviously)\n\nso 3 < 4x (by adding 4x to both sides of the equation)\n\nso 4x > 3\n\nso......... x > 3/4\n\nSECOND ONE\n\nif -(8-4x) < 5\n\nthen -8+4x < 5 (obviously...see that -*- = +)\n\nthen -8+4x-5 < 0 (by subtracting 5 from both sides of the equation)\n\nso -13+4x < 0 (obviously)\n\nand 4x < 13 (by adding 13 to both sides of the equation)\n\nso......x < 13/4\n\nFINAL ANSWER IS.......(3/4 < x < 13/4)
681497
Solve meaning???\n\nploting?\nfinding the focus etc.?\n\nbe specific
826575
Use one equation to substitute into the other,then solve for a one variable problem.  Use your new information (the solved variable) to solve for the second variable.\n\nFor the first set:\n\n-3x - y = 2 rearranged becomes:\ny = -3x -2\n\nUsing this "definition" of y, substitute it into the second equation as such:\n\n-6x + 2(-3x -2) = -2\n-6x - 6x - 4 = -2\n-12x = 2\nx = -1/6\n\nSubstitute -(1/6) for x in either equation to solve for y:\n\n-3(-1/6) - y = 2\ny = -3/2\n\nDo the same for the second pair of equations to make sure you got it (the answers for x and y are x=9/7, y = 1/7).
768295
Using an equation is the lazy, mechanical way to do it. If you want to understand what's going on, factor it if possible.\n\nIf you *really* want to know what's going on, derive the quadratic formula yourself. It isn't as hard as it looks, and it can be done on one page.
291468
Oh yes i have cracked it at last....\n\nsolution...\n\nLet x=tan-1 (y) (i mean tan inverse)\nthen tan x = y,and cot x = 1/y (coz tan-1 (x) = cot-1 (1/x))\n\nthen ur given equation's Left hand side part (LHS) will be\n\n(1/y)/(1-y) + (y)/(1-(1/y))\n\n= 1/(y-y2)+ y/(1-(1/y))\n\n= 1/(y-y2) + y2/(y-1)\n\n= y2(y-1) - 1/y(y-1)\n\n= y3/y(y-1)\n\nI hope u understood till this step.\n\nNow we have (a3-b3) = (a-b) (a2+ab+b2)\n\nthen the LHS becomes...\n\n(y-1)(y2+y+1)/y(y-1)\n\n= (y2+y+1)/y\n\n= y+1+(1/y)\n\nsubstitute back y=tanx, 1/y = cot x and leave 1 as it is...\n\nU have LHS = 1 + tan x + cot x...Thats ur RHS !!!
537199
got to math.com, click on calculators and click on solve equations.....it'll do the math for ya