y= Whole question under square roots. x/x-5?
830542
y = sqrt[x/(x-5)]\n\nStart by getting a few points:\nf(0) = 0  (it passes through the origin)\nf(5) is undetermined (divide by zero)\n\nSee what happens when 0 < x < 5.  For x = 3, x/(x-5) is negative, therefore the function is undefined over that range.\n\nFor x > 5, x/(x-5) is positive, so the function exists.  When x gets very large, x >> 5 (e.g., x = 1 million), x/(x-5) approaches one, so f(x) approaches sqrt(1) = 1 from above.  Therefore, y = 1 is an asymptote, but f(x) is always larger than one for very large x.\n\nNow look at x < 0.  x/(x-5) is always positive, and it's always less than one.  A good point to use is:\n\nf(-4) = sqrt[-4/(-4-5)] = sqrt(4/9) = 2/3\n\nand another good point is\n\nf(9) = sqrt[9/(9-5)] = sqrt(9/4) = 3/2\n\nWe know enough to plot the graph:\n\n1.  Draw the axes; draw a dotted-line asymptote at y=1; and draw another dotted-line asymptote at x=5.\n\n2.  Let's do negative x first.  Start at the origin, draw the curve through (-4, 2/3), and continue the line out to the left, using y=1 as an asymptote.\n\n3.  The function is undefined for 0 < x <= 5.\n\n4.  For x>5, start the curve way up on the x=5 asymptote, bring it down through (9, 3/2), and continue it out to the right, using y=1 as an asymptote.
736379
for this just substitute for x and y in eqn and see if LHS=RHS\nLHS=4\nRHS=(-1/2)(-3)-5/2\n=3/2-5/2\n=-2/2\n=-1\nnot = RHS\nso pt. not on line
525953
X = 24/Y  Just start plugging in Y and getting X, there are lots of possibilities, unless you restrict it to just whole numbers.
399118
y-int= (0,-5)\nx-int= (-1.2,0)
757845
x  y\n0 0\n1 -1/5\n-5 1
1108240
2(x-5)^2 = 3\ndivide by 2:  (x - 5)^2 = 3/2\nsquare root: (x - 5) = +/- root(6)/2\nadd 5: x = 5 +/- root(6)/2
680375
actually, you only found where y=x, but since x and y are squared, x=-y is also valid\n\nsolve\n3 +4/x =y\nx=+-y\n\nx=y\n3y+4-y^2=0\n(y-1)(y-4)=0\ny=1,4\n\nx=-y\n3y-4-y^2=0\nsolve using quadratic formula - will give you two more answers\n(y+)(y-)=0
925669
You need to use the quadratic equation...\n\nhttp://mathworld.wolfram.com/QuadraticEquation.html\n\nOR\n\nx (x - 5) = 0
612878
xy = 4 then y = x/4\n\nx^2 +  4y^2 = 20\nx^2 + 4 x^2/16 = 20\nx^2 + x^2/4 = 20\n4x^2 + x^2 = 80\n5x^2 = 80\nx^2 = 80/5\nx^2 = 16\n\ntherefor\nx = 4\nthen \ny=1
77583
add 5 to both sides:    x^2 = 5\ntake the sqrt of both sides:  x= + or - sqrt of 5
576742
ANSWER: y = 2; x = -1/5\n\nSOLUTION:\n\n1) First solve for y.  You can do this by adding the two equations together:\n\n5x - 2y = -5\ny - 5x = 3\n\nNotice that you have a positive 5x on the first equation, and a minus 5x on the second equation, so they both cancel out and you can re-write the equation as follows:\n\n- 2y = -5\n   y = 3\n\nAdd up these 2 equations and you get: \n\n- 1y = -2   or y =  2\n\n2) Then solve for x by replacing y with 2.  You can use any of the 2 equations from the problem.\n\n5x - 2(2) = -5\n\n5x - 4 = -5\n\n5x = -5 + 4\n\n5x = -1\n\nx = -1/5
875144
you have to take the square root of each side, then just simplyfy\n\nx = 1.2472191289247
421674
y = (x mod 5) + 1.\n\nThough that does seem a bit like cheating :)
256804
5x^2  -10 x +5\ncan be factorised into (x-1)(5x-5)\nsimplified to 5(x-1)(x-1)\n\nthus square root of given sum becomes square root of \n5(x-1)^2\n\nwhich is (x-1)*5^1/2
193592
The answer is a.  Look at the equations and solve for y.  In the first one you can do it with just addition and multiplication, but for the second two you'd need to take a square root.  Multiplication and addition don't change the function at all, but for the square root, you'd need to use the square root function which only gives you the positive half, or it wouldn't be a function.
436432
X = (0*Y) -5\nY = 0 * X\n\nbut basically just write X = -5 and thats it
762368
NB! sqrt(x)=x^½ \n2*(2+1)^½ / -5*(2+3)^½<=>\n( 2*(3)^½*(5)^½ )/(-5*5)<=>\n( -2*(3*5)^½ )/25<=>\n( -2(15)^½ )/25<=>\n-2( 3*5/5*5*5*5)^½<=>\n-2(3/5*5*5)^½= -sqrt(3/5)*2/5\n=========================