What is the subtraction property of equality?
916856
If you have balances of $1800 on two credit cards with no monthly fees and 0%APR, and you make a $500 payment to each of them, then both of those credit cards will have balances of $1300 on them.
576001
By the transitive property, a=c
628648
None. The two numbers have no common factors, bar -1. You mean exactly (with no remainder) not evenly.
15071
-4k-5 <= 3\n-4k <= 8\nk >= 8/(-4) (the sign reverses the inequality)\nk >= -2
855482
You set up two inequalities and solve each inequality separately.\n\nInequality one:  5x-1 <= 14  (drop the absolute value symbols)\n\nInequality two:  5x-1 >= -14  (drop the absolute value symbols, reverse the inequality and make the 14 negative)\n\nBy the way, you have a contradiction in your question.  < is the less than sign, but you refer to it as greater than.  Just a little confusing.\n\nHope this helps.  :)
378323
[It's the commutative property, named after commuters that go back and forth.]\n\n2(8)(b-3-8)(b+1)\n\n2(8)(b-11)(b+1)\n\n2 [8(b-11)] (b+1)  (associative)\n\n2 (8b-88)(b+1)  (distributive)\n\n(8b-88)(2)(b+1)  (commutative)\n\n(8b-88)[2(b+1)]  (associative)\n\n(8b-88)(2b+2) (distributive)\n\n16b^2 + 16b - 176 b - 176 (distributive)\n\n16b^2 - 160b - 176
496089
Adding a negative is the same as subtracting a positive.  Subtracting a negative is the same as adding a positive.
101183
-1<(or equal to)X <(or equal to)6
500151
The distributive property (of multiplication over addition) is an axiom that states that for any numbers a, b, and c, the equality a*(b + c) = a*b + a*c is true. This can be used to distribute or factor (read the equation from right to left. Remember that if a=b, then b=a).\nFor example, 3*1.99 = 3*(2.00 + (-0.01)) = 6.00 + (-0.03) = 6.97.\nLikewise, x^2 + x = x*(x + 1).
792912
Consider if a= 1, b =-1, then abs(a/b) = 1 and a/b = -1\n\n-1 != 1\nabs(a/b) != a/b\n\n(!= means NOT EQUAL TO)
965748
|x - x²| \n= x - x² for x - x² ≥ 0\n= x² - x for x - x² < 0\n\nx - x² = x(1 - x) = 0 when x = 0 and x = 1\n\nSo x - x² ≥ 0 when 0 ≤ x ≤ 1\nand x - x² < 0 when x < 0 and x > 1\n\nSo  dx under the interval [-1,3]\n\n= ∫(x² - x) dx under the interval [-1, 0] +\n∫(x - x²) dx under the interval [0, 1] +\n∫(x² - x) dx under the interval [1, 3]\n\n=  [⅓x³ - ½x² ] [-1, 0] + [½x² - ⅓x³] [0, 1] + [⅓x³ - ½x² ] [1, 3]\n\n= [(⅓0³ - ½0²) - (⅓ (-1)³ - ½ (-1)²) ] + [(½(1)² - ⅓(1)³) - (½0² - ⅓0³)] + [(⅓ 3³ - ½ 3²) - (⅓1³ - ½1²) ]\n\n= [5/6] + [1/6] + [9/2 + 1/6]\n\n= 5 2/3
40850
-3 <4x\n x < -3/4\n solution set : {- infinity ,.............,-3/4}
1033116
A X = D\nX = A^(-1) D. (A inverse * D)\n Do it ur self.
939642
Counterexample.  Let u(x), v(x) be members of {sin x, cos x}.  Both of those are solutions, but you do not get zero by adding one to the other, or by doubling one of them.
459734
The goal is to get the term containing x (or other unknown) on one side and the numbers (and other unknowns) on the other.\n\nYou do this by performing the same operation on both sides at once. That way, the equation remains an equality. If you did different operations on the two sides, the two sides would no longer be equal.\n\nHere is an example: -4(2x+8)=32\n\nDivide both sides by -4. The minus four on the left becomes (-4/-4) or 1. It kind of disappears. And dividing the 32 on the right by -4, changes the 32 to -8. However, the left and right sides are still equal because you performed the same operation on both sides.\n\nAnd here is the rest:\n\n2x + 8 = -8\nAdd -8: 2x + 8 - 8 = -8 -8\n2x = -16\nDivide by 2: 2x/2 = -16/2\nx = -8
365867
2x - (7 + x) <= x\n2x - 7 - x <= x\nx - 7 <= x\n-7 <= 0\n\nSince we're left with a statement with no variables, and the statement is true, x = all real numbers.
493260
Knowing what a proof is, is the first step in answering this question. From what I am familiar with...\n\nA proof is: a set of logical steps acquired through deductive (therefore, not making any giant leaps in logic, unless by definition), and hence, empirically (from the evidence provided) resulting in a direct equivalence (being, among other types of equivalence, but primarily, in permutation,  multiplicative/additively & negatively/positively & even/odd... meta-mathematically) of states, that's shortest distance is (in absolute terms), either infinity, zero, and/or, also, one.\n\nReally, the attempted 'proof' of 2 + 2 = 5 is based on a distorted type of Trigonometry, which was in essence the source of today's Calculus (just try to draw Tangent or Secant without running into the idea of Calculus' derivative & integral, respectively), and actually is the result of any additive equavalence of any two numbers' to being alike to any number, (because measuring hypotenuse of a given sides is essentially multiplicative, hence partially irrational). \n\n(Which makes me wonder... is there a 2 * 2 = 5 equivalent? and the answer is a resounding, yes! But first the 'proof' as written by Charles Seife.)\n\nLet a & b each be equal to 1. Since a ^ b are equal,\n\nb^2 = ab ...(eq.1)\n\nSince a equals itself, it is obvious that \n\na^2 = a^2  ...(eq.2)\n\nSubtract equation 1 from equation 2. This yeilds\n\n(a^2) - (b^2) = (a^2)-ab ...(eq. 3)\n\nWe can factor both sides of the equation; (a^2)-ab equals a(a-b). Likewise, (a^2)-(b^2) equals (a + b)(a - b) (Nothing fishy is going on here. Ths statement is perfectly true. Plug in numbers and see for yourself!) Substituting into the equation 3 , we get\n\n(a+b)(a-b) = a (a-b) ...(eq.5)\n\nSo far, so good. Now divide both sides of the equation by (a-b) and we get\n\na + b = a  ...(eq.5)\n\nb = 0 ...(eq.6)\n\nBut we set b to 1 at the very beginning of this proof, so this means that\n\n1 = 0 ...(eq.7)\n\n ...Anyways, getting that far gives us the jist of the proof, later in the proof, Charles Seife goes on to prove that Winston Churchill was a carrot! if you want to know how that is possible, I recommend you read the book.\n\nFrom equation 7, add a number to either side and get it equal to any other number, one greater than itself.\n\nMultiplying equation 7 after adding to it, and one can get: any number is equal to any other number. \n\nHence, conceptually, any number is equal to zero, and, theoretically, that includes infinity. But that's also the reason why when you divide by zero, it is 'Undefined.' Which, consequentially, is what is happening in this equation... just subsistute 1 into equation 3 and one will see that we are dividing by zero in equation 5. \n\nThis is what lead to the invention of calculus. Really, from here this segways into Hilbert Space... but that is best left for another entry, hopefully, on the actual subject of quantazation.\n\nThat's all I have time for... \n\nTHIS PROOF IS BY DEFINITION INCORRECT, but it provides a good tool as of why we define things in mathematics the way we do.\n\nA good question to ask from here would be (based on my previous tangent):\n\nDoes 1/3 plus 1/3 plus 1/3 = 1?\nOr, does it equal just zero point nine repeating?