how to solve absolute value equations?
413384
If you had something like\n\n| x - 3 | = 14\n\nthat really means\n\nx - 3 = 14\nOR\n-(x - 3) = 14\n\nSolving your original equation means breaking it down into these two scenarios and solving them individually.  So in this example, x = 17 from the first one, and x = -11 from the second one.\n\nBecause of the absolute value, expect to have double the usual amount of answers.  In other words, for a linear equation, normally you'd expect to have one answer.  If it's a linear absolute value, you'll have at most 2 (this doesn't mean you'll always have 2 - you might have 2 0's for an answer, as in |x/15| = 0).  A quadratic yields at most 2 answers, but a quadratic absolute will yield at most 4.
317809
absolute values are important in higher levels of analysis when using wide ranges of numbers, also it help when we try to have equations of only real numbers make sense sometimes, i.e. various sorts of root equations
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
697874
Multiply or divide all components in one question to set the variables in the opposite questions to cancel out if you added the two problems together by combining like terms. For example:\n2x + 3y = -13  and 4x + y = -1\n\nTo keep this simple I will multiply all the components of the 2x + 3y = -13 equation by negative 2, resulting in:\n\n-4x - 6y = 26\n\nThen add the new equation to the other equation 4x + y = -1\n\n-4x -6y = 26\n4x + y = -1\n----------------\n0x -5y = 25  which simplifies to y = -5\n\nInsert the value found for y back into one of the original eqations and solve for x\n\n2x + 3(-5) = -13\n\n2x + -15 = -13\n\n2x = 2\n\nx = 1\n\nYour end result is x =1 and y = -5
408462
to solve such a system in general, what you do is this: take the first equation, solve it for x treating y as you would a constant. In your case, this is x = 4 + y.\n\nThen, take your expression for x and substitute it everywhere you see x in the second equation; in your case you get 3 *(4+y) + 2y = 7.\n\nSolve this now 1-variable equation for y, once you find y (in this case, -1).\n\nSubstitute the value of y into either equation, and you get the value of x (3 in this case).
811215
If the value in an absolute value is positive, it remains the same. If it is negative, it is multiplied by -1.\n\nEither a<x, a = x, or a > x. (I think this is called trichotomy?)\n\nIf a = x, then both sides equal 0.\n\nIf a > x, then a-x is positive, and x - a is negative, so...\na - x =  -1(x - a)\n\na - x If a > x, then x-a is positive, and a - x is negative, so...\n-1(a - x) =  x - a\nx - a = x - a\n\na - x = a - x\n\nQED
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
967368
Just treat all other letters as constants. You can multiply, divide, add and subtract both sides of an equation and maintain the equality.
259003
Use the quadratic formula.  It works everytime!\nx = (-b plus or minus sprt(b^2 - 4ac))/2\nwhere\nax^2 + bx + c = 0\nSome of these will factor, however not all of them will.
490656
Make Y=0\nyou have to have m and x though to solve b\nExample:\nY=mx+b\n0=3(4)+b\n0=12+b\n -12 on both sides of equal sign\n-12=b\nhope this helped!
764542
This is an impossible figure as by your own figures ΔABC is equilateral with AB = 2AL = 10\nBC = 2BD = 10 and\nAC = 2AE = 10\nso AT = 2/3 * BT = 2/3* 5√3\n= (10√3)/3\n≠10
901005
Use the following equation and solve:\n\nP1V1/T1=P2V2/T2\n\nT2=P2V2T1/P1V1\n\nalso note the significant figures!!!!
366751
make x the width, and solve the two equations (the one given, and the one assumed about the rectangle where l*w=A)
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
634747
What you want is express \n\nv = a*v1 + b*v2 + c*v3 \n\nWhere a, b and c are, of course, scalars. The coordinate vector will then be (a, b, c).\n\nNow how do we do this: \n\n(2, -1, 3) = a(1, 0, 0) + b(2, 2, 0) + c(3, 3, 3) \n\nEither you just look at it and try to guess what a b and c can be, or you write this up as a set of 3 equations in a,b and c: \n\n2 = 1a + 2b + 3c\n-1 = 0a + 2b + 3c\n3 = -0a + 0b + 3c \n\n...and then solve it: \n\nFrom the last equation we see that c = 1, then using the second equation, -1 = 2b + 3  <=>    -4 = 2b  <=>  -2 = b\n\nAnd substituting into the first, \n\n2 = 1a + 2(-2) + 3(1) = a - 4 + 3 = a - 1 \n\nthus 3 = a. \n\nSo the coordinate vector is (3, -2, 1) and with these numbers the criteria are (supposedly) satisfied. :)
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.
247965
infinite values for both,\nits caus u cant solve two variables with just one equation