what is the solutions of the syst of equations. x+y=8 & x-y=-4, the ordrd pair is ?
952487
x+y=8\nx-y=-4\n______\n2x=4 (because +y and -y cancel out)\nx=2\nthen plug 2 in for x in one of the equations\n2+y=8\nsolve for y\ny=6\nordered pair is (2, 6)
897159
x + y = 2- - - - - - -Equation 1\ny - x = 2- - - - - - -Equation 2\n- - - - - - - -\n\nRearrange equation 2 to read\n\n-x + y = 2\n\n- - - - - - - -\n\nx + y = 2\n-x + y = 2\n- - - - - - - -\n\n2y = 4\n\n2y/2 = 4/2\n\ny = 2\n\nThe answer is y = 2\n\nInsert the y value into equation \n\n- - - - - - - - - - - - - - - - - - - - - \n\nx + y = 2\n\nx + 2 = 2\n\nx + 2 - 2 = 2 - 2\n\nx = 0\n\nThe answer is x = 0\n\nInsert the x value into equation 1\n\n- - - - - - - - - - - - - - - - - - - - - -\n\nCheck for equation 1\n\nx + y = 2\n\n0 + 2 = 2\n\n2 = 2\n\n- - - - - - - - -\n\nCheck for equation 2\n\ny - x = 2\n\n2 - 0 = 2\n\n2 = 2\n- - - - - - -\n\nThe solution set is { 0 , 2 }\n\n- - - - - - -  -s-
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
951394
x+y=4\nx-y=8.5\n\nJust add the 2 equations\n\nit will result to this one:\n\n2x-0y=12.5\n2x=12.5\nx=6.25\n\nusing x, we could get the value of y.just choose from the 2 given equations where you  could substitute x.\n\nx+y=4\n6.25+y=4\ny= -2.25
469518
There is an obvious 1-dimensional continuum of solutions to this equation: the line y=x.\nHowever, there are a continuum of more solutions, most of which are nontrivial:\nNote that log(x^y) = y*log(x). Thus log(x)/log(y) = x/y. This implies that y = x^(y/x) and similarly x = y^(x/y). This is the general form of the solution where the obvious solution is the restriction x/y = 1. \nIf we let x/y = 2, we get 2y = y^2, or y(y - 2) = 0 which gives nondegenerate solution point (4, 2) as mentioned previously. Setting x/y = 3 gives us 3y = y^3 or y^2 - 3 = 0, which gives the nontrivial solution point (3*Sqrt[3], Sqrt[3]) and so on.
369017
Set one equation equal to one of the variables, then insert it into the other equation and solve for the second variable.  then insert that variable into the original equation.\n\nx+y=5 or x=5-y\nx+2y=6\n\nput x=5-y into the second equation\n\n5-y+2y=6 \n\nnow solve for y\n\ny=1\n\nput this back into one of the original equations\n\nx+1=5\n\nx=4
694839
(y+4)(y+4)-2y -11=0\ny^2 +8y + 16 -2y -11 =0\ny^2 +6y +5=0\n(y+5)(y+1)=0\ny=-5, x=-1 , and   y=-1, x=3\n\nCheck with y=-5\n(-1)^2 -2(-5) = 1+10=11  OK\n\nCheck with y=-1\n3^2 -2(-1) = 9+2=11  OK
201908
Ah, systems.\n\nthe first problem- there are many ways you can choose to do this, but I think we should multiply both sides of the first equation by two so that they positive and negative y's will cancel out.\n\nSo now we have:\n2(2x + y)=2(4)\n3x - 2y = -1\n\nSimplified:\n4x + 2y = 8\n3x - 2y = -1\n\nnow you add down, so you get:\n7x +0y = 7\n7x =7\nx=1\n\nsubstitute 1 for x in any equation:\n2(1)+y = 4   \n2 + y = 4\ny = 2\n\n\n\nfor the second one- lets cancel out the y's again.\n\n5x - 3y = 14\n-3(2x - y) = (6)-3\n\n5x - 3y = 14\n-6x +3y = -18\n\nAdd them together:\n-x = -4\nx = 4\n\nThen solve for y:\n5x - 3y = 14\n5(4) - 3y = 14\n20 -3y = 14\n-3y = -6\ny = 2
748264
You didn't give any full ordered pairs. All you have to do to see if the ordered pairs are solutions is put the first number in for X and the second in for Y.  If your X multiplied by 3 plus your Y multiplied by 4 is 12, then it is an ordered pair.  For example, if you ordered pair was (4/3, 2); 3(3/4) + 4(2) = 4+8 =12.  But if you had (3, 4); 3(3) + 4(4) = 9+16 =25, so that one would not be a solution.
949784
y= (x^2 - 1)^2\nx^2-1=+/-√y\nx^2=1+/-√y\nx=+/-√(1+/-√y)
685420
x-y = 4\nx = 4 + y\nx-4 = y\n\nSo all pairs of the for (x, x-4).
940135
Use factoring.\nx2-2x-8=0\n(x-4)(x+2)=0\nso\nx-4=0 and x+2=0\nsolve\nx=4 and x=-2
747583
y= √(1/(x² + 1) – 2) \n\nWell 1/(x² + 1) – 2 ≥ 0\nie 1/(x² + 1) ≥ 2\nSince x² + 1 ≥ 0 for all x\nOn taking reciprocals\nx² + 1 ≤ ½\nx² ≤ -½ This is not possible as x² ≥ 0 for all x \nSo there is no domain\n----------------------- -----------------------\nWell that makes sense as  x² + 1 ≥ 1\nSo 1/(x² + 1) ≤ 1\nSo 1/(x² + 1) - 2 ≤ -1\n\nSo taking square roots gives no real solutions.
892609
y=2x-8\nx=2y-8\nadding 8\n2y=x+8\ndividing by 2\ny=(x+8)/2
1083422
-2d^2+8d=0\n\nsimplifying the equation d^2-4d=0  \nfurther simplifying    d(d-4)=0        d=0   or d-4= 0   d=4\n\n0 and 4 are the possible roots\n\nthe second answer is wrong here my proof:\n\n-b^2+/-sqrt b^2-4ac/2a\nthe correct equation is\n\nx= (-b+/-(sqrt b^2-4ac))/2a \n\nx= (-8+/- sqrt 8^2-4(-2)(0)))/2(-2)\n\nx= (-8+/- sqrt 64-0)/-4\n\nx= (-8+/- 8)/-4\n\nx= (-8+8)/-4 or x= (-8-8)/-4\n\nx= 0/-4 --> x=0 or x= (-16)/-4 ----> x=4
561927
Y - Y1 = M(X - X1)\n\nY - 4 = -2(x + 1)\nY - 4 = -2x - 2\n\nY = -2x + 2
704540
Add the two equations together: (2x+y)+(x-y) = 1+8, so 3x = 9, x = 3.  Then 2*3+y = 1 so y = -5.\n\nSteve