Decide rather or not the ordered pair  is a solution to the equation.  Explain?
472975
Replace x with 4 and y with 5.  If the resulting statement is true, then it is a solution.  If it is not true, then it is not a solution.\n5x +3y = 35\n5(4) + 3(5) = 35\n20 + 15 = 35\n35 = 35\nThis is true, so (4,5) is a solution.\n(There is no need to solve for y.  You can solve for y, but this just adds additional work.)
55379
I would say, \n\nonly in the case when n=0 or 1, then this equation stands good! If n is not equal to zero or one, then everytime the answer would definitely be above 1 (even when u use negative integers)
118612
Kerplode is right - but that's kind of complicated.  Basically you solve both in exactly the same way.  The only difference is that instead of an = you have an > or <.  \n\nIt just looks different, there really is no difference in how you solve them.
948825
First answerer correct.  \n\nBut the second set is a function, because there are no repeated x values.  If you have a component of the first set corresponding to two different components of the second, then it isn't a function, e.g. in the second set, if the next pair were (15, 5) then it wouldn't be a function because of the double result for 15.  In the first set, if your misprinted pair were (1,2) then it wouldn't be a function, same reason.
747036
0.12 \nP(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) - P(A)P(B|A) = 0.1 + 0.1 - 0.1(0.8)
2302
(1) = x\n(2) = / \n(3) = x
150385
f(x)=Int(f'(y), 0, x)-f(0) => yes, they're opposite operation, but if x has a given value, e.g. x=2, 3...no, they're not, b/c in that case, the RHS of the previous egality is a real number, not a function
374365
it will be choice (c)\n(x+1)(x-1)\nx^2+x-x-1\nx^2-1
3037
This is not presented as an equation, so the goal appears to be to combine the rational expression --> x+10 \nThat's all you can do with the question as it is written.
367272
Suppose the number of combinations of k objects out of n objects without respect to order chosen is C(n, k). If we wanted to count order, we would multiply each combination by the number of ways it could be permuted. Thus P(k)*C(n, k) is the number of ways one could choose k people out of n people with regard to order. \nBut we can easily count this number another way. The number of possible choices for the first person is n, for the second person n-1, and so on until we reach n - k + 1. Thus, we get the equation P(k)*C(n, k) = P(n, k) = n*(n - 1)*...*(n - k + 1) = n!/(n - k)!. Solving for C(n, k), we get C(n, k) = n!/[P(k)(n - k)!] = n!/[k!(n - k)!]. \nNow you should be able to derive it whenever you get confused.
694639
From the second equation:\n\np/(1+p)=r\np=r(p+1)\np=rp+r\np-rp=r\np(1-r)=r\np=r/(1-r) which is not equivalent to p=(r-1)/r
676814
First equation: sqrt[x] + y = a\nor sqrt[x] = a - y\nSquaring both sides\nx = (a - y)^2__________(1)\n\nSecond equation: x + sqrt[y] = b\nor sqrt[y] = b - x\nSquaring both sides\ny = (b - x)^2___________(2)\n\nSubstituting 'x' from eqn. (1) in eqn. (2)\ny = [b - (a - y)^2]^2\n  = [b - a^2 + 2*a*y - y^2]^2\n\nSo it is clear that you have to solve a 4-th order equation.\nIt is really tough.
245200
Proof by contradiction: Assume A or B.\n\nIf not B then not A\nalso not C (because of not B) so neither A nor C which contradicts our assumption. Therefore B.
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
196036
If you divide 2 to the second equation, you will notice that it is identical to the first one, so they are dependent to each other, and there are infinitely many solutions to this problem.\n\n^_^
837785
hi\np=(qm)/n\ny=x+(b/m) or x+(bq/np)
364344
emmmmmm\n\nd. its not a fuction of x in terms of y explicitly. It needs to be rearranged