how many real solutions does this nonlinear system have?
361972
if you add your two equations you get 3y^2=27\n\nthat is y^2 = 9\n\ni.e y = ± 3\n\nthen since x^2 = 13 - y^2\n           x^2 = 4\n           x = ± 2\n\nhence there are four soltions\nx=2 y=3, x=-2 y=3, x=2 y=-3, x=-2 y=-3
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
195201
i think u typed the question wrong, can u please correct it and if u have 2 unknowns, please give 2 equations, else u cant get a definite solution
801480
4 is the answer I get
759105
False... y has to be 2...\n(3,2) is a solution, but not (3,1).
739088
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.
920768
One solution is\n0, 2, 4, 6, 8, 10, 12, 1, 3, 5, 7, 9, 11\nBut there is a huge variety of solutions\n\nSorry Chris I did not copy so here is another\n\n0, 11, 5, 2, 9, 3, 1, 4, 10, 7, 12, 6, 8
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
593616
It's not hard if you draw the graph of the functions y=x and y=5*sin(Pi*x), and see how many times they cross.\n\nNow, you know that sin(pi*x) is always less than 1, so there are only solutions when -5<x<5.  And you know that at x=5, sin(pi*x) =0. (Same at x=-5)\n\nIf you plot the graph of sin(pi*x), you see that it's positive if 0<x<1, 2<x<3, and 4<x<5.  Since the sine curve goes up and down, you can see that there are two solutions in each interval.  Counting these up gives 6 solutions in each interval, or 6 solutions.  Now, look at x<0-- there are 6 solutions there also, but the x=0 is included in both sets, so you have to ignore one of them-- there are 11 solutions to the equation x=5sin(pi*x)
296877
It has a solution, indeed. It actually has two solutions, -3 and 7. To find it out, set each pair of nomials, (e.g. (x + 3)) to zero. So it would be (x+3)=0. Then solve for x. Do the same for the other nomial, and there you have it.
1066425
A system of linear equation has a unique solution if and only if the determinate of the coefficients matrix is not equal to zero. Therefore, for your system\n\nx+5y=10\n-2x-10y= -20\n\nThe determinate is\n|   1     5    |\n|  -2   -10  |\n\ndet= 1*(-10) – 5*(-2) = 0\n\nThen the system is linear dependency and has infinite number of solutions.
15744
Classes of Solutions to Mathieu's Equation\n\n  Floquet's theorem asserts that any solution to the Mathieu equation (Eq. gif) is of the form:[4,29,28]\n\n \n\nwhere is either real or imaginary and is a -periodic functiongif of . This assertion implies that solutions of Eq. gif can be separated into the product of a term of the form and a series of harmonics, , each of which is -periodic in . For values of with a non-zero imaginary component and , Eq. gif indicates that:\n\nThus, , yields an unbounded solution to the Mathieu equation. On the other hand, values of which are purely real yield bounded, oscillatory solutions for .\n\nThere are two types of bounded solutions to Mathieu's equation: functions of fractional order and functions of integral order. Functions of fractional order are obtained for . The oscillatory periods of the frequencies associated with fractional-order Mathieu functions are non-integer multiples of . The special case of corresponds to an -periodic (in terms of ) solution. These -periodic solutions are Mathieu functions of integral order.\n\nAll bounded solutions to the Mathieu equation--- those of fractional as well as integral order--- are described (in accordance with Floquet theory) by an infinite series of harmonic oscillations whose amplitudes decrease with increasing frequency:\n\n \n\nThe argument of in Eq. gif represents harmonics of frequency , while the term represents a (secular) frequency which is a function of the trap specifications (or, equivalently, of the Mathieu parameters a, q). The Mathieu parameters and also determine the coefficients , which approach zero as n increases.[36]
274355
Depends on the system you are trying to solve. check your limits of integration. from -infinity to 0 plus 0 to infinity.
610726
there is a solution, all x,y that satisfy this equation:\ny = 4 - 2x\n\nwhich is a straight line with intercepts (0,4) & (2,0)\n\nIt has an infinite amount of solutions. To be able to solve for actual numbers, you need as many equations as you have variables.
800072
It has one real solution.  Irrational roots are always in groups of two, so there could have been one or three real solutions.\n\nThe real solution is -0.531447857.  Unfortunately, there is not any easy way to find this one other than guess and check.\n\nYou will also note that plotting the line x^3-4x^2+7x+5=y is always increasing.  (The first derivative is always positive).  This tells you that you are guaranteed to only have one real solution.\n\nAnd 2.26... doesn't work
979866
If I am reading this as you have submitted, then, the solution of x=2secx is transcendental, and needs to be solved graphically giving a set of values that satisfy it; you plug that set of values for x that satisfy it into y=3+2tanx, and that gives the solution set for y.
1047731
Assume dc/dz is constant, n(t=T)=-DA(dc/dz) T\nBut in reality dc/dz often changes.  Normally dc/dz should decrease with time.  If it is a close system, it will reach the equilibrium state.