you are trying to explain to a classmate how equations can arise with infinitely many solutions and with no solutions. give such an explanation. does having no solution mean that 0 is a solution? expain.
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Infinite solutions? Take the equation X = Y\nWhen X = 1, Y = 1, When X = 2, Y = 2 etc. you have as many solutions as you wish. There is no single answer!\n\nSingle solution  5(X+3) = 25\nremove brackets  5X+15  = 25\ntake 15 from each    5X = 10\nsoluution             X = 2     ONLY one answer.\n\nNo solution      X+1  =  X+5\nNo answer will satisfy both sides of this equation so there is no solution.
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When Graphed\n\nEquations can be any line\nLinear Equations form only straight lines\nFunction is only a function if you don't have 2 identical x values.
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One of the solutions that can be applied () personal development classes.\n\nSince "of the solutions" and "personal development" are details, are descriptions  of "one" and "classes" respectively, just remove it to test the (is) and the (are).\n\nOne [] that can be applied IS [] classes.\nOne [] that can be applied are [] classes.\n\nOne [] that can be applied is....\nOne is singular. One solution.\n\nIf it was "Two that can be appllied .... ARE." Because two is not singular.\nOne is.\nTwo are.
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To me it means an unfixed variable can be a wide range of numbers, and to solve for it, you are looking for the most optimal value.\n\nX=X+1, to me means, the value of X will be changed each cycle by adding 1 to it.  That will change the solution to the equation to which X is a part.\n\nYour math teacher uncle probably doesn't have experience with cycles, and saw X=X+1 as an illogical equation.
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No no solution since n>2\n\nPierre de Fermat died in 1665. Today we think of Fermat as a number theorist, in fact as perhaps the most famous number theorist who ever lived. It is therefore surprising to find that Fermat was in fact a lawyer and only an amateur mathematician. Also surprising is the fact that he published only one mathematical paper in his life, and that was an anonymous article written as an appendix to a colleague's book.\n\nThere is a statue of Fermat and his muse in his home town of Toulouse:\n(Click it to see a larger version)\n\n\nBecause Fermat refused to publish his work, his friends feared that it would soon be forgotten unless something was done about it. His son, Samuel undertook the task of collecting Fermat's letters and other mathematical papers, comments written in books, etc. with the object of publishing his father's mathematical ideas. In this way the famous 'Last theorem' came to be published. It was found by Samuel written as a marginal note in his father's copy of Diophantus's Arithmetica. \n\nFermat's Last Theorem states that \n\nxn + yn = zn \n\nhas no non-zero integer solutions for x, y and z when n > 2. Fermat wrote \n\nI have discovered a truly remarkable proof which this margin is too small to contain.
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Here's what you should think: 3 times a number that has to be zero or higher, then having another 3 added to it, cannot possibly equal 1 (it must be at least 3).  So there is no solution.\n\nYou did say it was an inequality, though.  If the = was really > or >=, the answer is all real numbers.  If it was < or <= the answer is the empty set.
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Your statement translates to the following algebraic equation.\na+b=a-b=a*b\nSolving a+b=a-b, we get b = 0; \nSubstituting b=0 and solving a+b=a*b, we get a = 0;\nSo the two numbers are zeroes.
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The answer to that is a line.  If you want a unique solution to a two-variable problem, you need two equations.
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Consider a = 1, b = 0, c = 1.  Clearly, b^2-4ac = -4 < 0.  This corresponds to the equation x^2 + 1 = 0.  But x^2 is never negative, and 1 more than that is always positive, so this has no solutions.\n\nIf you want only non-imaginaries, there are no solutions when b^2 < 4ac and only one solution when b^2 = 4ac.\n\nYou clarify: "that does not suffice because every quadratic is meant to give probable solutions because all polynomials and quotients of polynomials are rational, and rational equations don't give imaginary solutions."\n\nRational equations do indeed give complex solutions, many of them nonreal.  The example I gave above (x^2 + 1 = 0) is an example.  The algebraic closure of the reals is the complex numbers; the algebraic closure of the rationals is the algebraic numbers.  If you want all quadratics to give two solutions (counting multiplicity), you'll need to allow imaginaries.
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Hello Rajeash\nYour additional question "\nProve that when x is even then it is impossible to write the above equation For n>2.\n"\nIsnt this what thye good o'l  Sir Fermat already has stated even for odd x ?\n\n\na^(n) - b^(n) = x^(n) \nsuppose that n is not prime, thus n = p*q, with p a prime\nthen we can write\n(a^q)^p -(b^q)^p = (x^q)^p \nso we can limit our problem to n a prime. Fermat stated his statement for p>2 thus we only have to investigate the odd primes.I do not agrede with You Rashesj that we have to investigate 4 as a separate case, because 4 = 2^2 we can write the equation as (a^2)^2 - etc , Do you agree with this ?\n\nYou also say that we can assume that if we have a solution then x will be an odd prime. If a,b,x have a common factor then we can divide that factor out. So we can assumet that they dont have a common factor. If x is even then a and b should be odd.
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Zero of zero is undefined because of the definition of division and some basic field axioms.  It's not zero and it's not infinity, it's UNDEFINED.  I can prove it.\n\nThe multiplicative inverse of n is defined as the number 1/n such that n*(1/n)=1.  It's a property of zero that anything times zero is zero.  Therefore for any 1/n, 0*(1/n)=0, and since zero cannot equal one, zero has no multiplicative inverse.\n\nDivision by n is defined as multiplication by the multiplicative inverse of n.  Since zero has no multiplicative inverse, you cannot divide by zero.\n\nAll you people who said that anything divide by zero is infinity are just flat out wrong.  If n is not equal to zero, the limit as x approaches zero for the positive side of n/x is infinity, but that does not mean that n/0 is infinity.  Also, note that if you come at it from the negative side the limit is negative infinity.  n/0 is undefined because zero has no mutiplicative inverse so the expression 1/0 is mathematically meaningless  Also, the limit of 0/x as x approaches zero would be zero, not infinity.\n\nWhoever it was who did the calculator thing, calculators are a far cry from a mathemtical proof.   There are plenty of arithmetic problems that make perfect sense but can't be done on a calculator.  In fact, there are infinetly many of them.  And there are some properties of multiplication and addition that can be proven mathematically but are not true on a calculator.  But you and the first poster get major props for being the only people to answer this question and demonstrate anything resembing an basic understand of elementary arithmetic.\n\n______\nH2++ cannot exist because it would have to have a covalent bond but would have no electrons.  Two hydrogen nuclei can't form a diatomic molecule.  O-- is not going to exist in water because the hyrogen atom would have to dissociate from the water molecule one at a time, and OH- is more stable than O-- and H+.  It could in theory exist as 2H+ and O--, but OH- is a much happier molecule than O--.\n\nNothing can go faster than the speed of light because as an object approaches the speed of light it's mass becomes infinite, and so the force required to cause it to accelerator more becomes infinite.  You can't apply infinite force to something.  Normally I'd do out the math to show that, but it's 3:30 in the morning so I doubt I can make it coherent.\n\nTensor calculus is too complicated to explain at 3:30 in the morning.  You can read about it here: http://en.wikipedia.org/wiki/Tensor_calculus
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No solution.\n\nDo (equation 3) minus 2*(equation 1)...\nyou get\ny + z = 13\nwhich contradicts equation 2.  Either you have a mistake, or the answer is no solution.\n\nHope this helps,\nGod bless,\nShane
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Any non-zero number divided by a limit approaching 0 WILL go to infinity:  n/(lim->0) is inifinity so n/0 is infinity.\nBut 0/0 is UNDEFINED.
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The same equation, in this case, cannot have two different answers. So there is no solution to this system (to understand it, try graphing it)
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We say that a it is indeterminate because if we have two functions, f and g such that:\n\n   lim (x->oo) f(x) = infinity\n   lim (x->oo) g(x) = 0\n\nthen we know nothing about:\n\n   lim (x->oo) f(x)^g(x)\n\nThat is, this third limit can be any value.\n\nFor example, given a real number A>=1, if we let f(x) = A^x and g(x) = 1/x, then obviously, f(x) converges to infinity, g(x) converges to 0, and f(x)^g(x) converges to A.\n\nIt's important to note that mathematicians aren't trying to be difficult, but often use terms carefully to have a precise meaning.  The phrase 'indeterminate' is different from 'undefined.'\n\nFor the person who says infinity to any power is indeterminate, you are wrong.  The form:\n\n    infinity^R\n\nis determinate if R is a non-zero real number.  If R is negative, then this 'form' is always zero.  If R is positive, this form is always infinity.  This means that if we have two functions f and g so that:\n\n   f(x) -> infinity\n   g(x) -> R\n\nthen f(x)^g(x) -> 0 if R<0, infinity if R>0, and indeterminate if R=0.
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if we speak of numbers and not of limits, the answer is no without exceptions: 0 is defined in group theory as the null element, so it is BY DEFINITION the element which multiplied by any other gives itself: always 0.\nInstead, what we usually assume as always true, but is not true in some finite algebraic models, is that if A*B=0 then either A=0 or B=0. This is not always true in all algebraic models (for exemple in modular algebra).
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they are both the same line. It is called a dependent solution. Or a coincident line. The line would start at (0,2), and then move up two spaces, and one to the right before making another dot. repeat proccess about 2 or 3 times.