can you find values for x,y,z?
217117
It's a Fermat Last Theorem
395793
Use one of them to get x to y relation and substitute x or y in the other. This way you can find the one unknown and after substituting it with it's known value in the x/y relation you will find the other one.\n\nThere are other ways for dealing with system equation, but this one suits your system best.\n\nTry to make it yourself. This is the best way to learn.
520758
You have z=2, and y-z=-3, so if you put the value of z in there you should have \ny - 2 = -3\nso y is -1\nNow, with the value of y, you can replace it in the x-y=4 equation\nx-(-1)=4\nso x=3\n\nYou were on the right track, you just stopped a bit short of the finish line.
80080
Sorry, but I think you need to do this yourself.  Here are some guidelines though; give them a try.\n1.  Make a line drawing so you can see the info you're given and the path to the required answer.\n2.  Make a list of ALL the related formulae you were given.  Check each one against the given data; if you find one in which you have all the variables EXCEPT one, substitute and solve for the unknown.\n3.  Sometimes, you may have to use one formula to find a variable and THEN use that in ANOTHER formula to arrive at your ansver.\n\nGood luck!!
375161
x=2y\n\n-2 -4\n0 0\n1 2\n3 6
318586
this one isn't possible..wait...i think i got it....\n\nxa+yb+xz+yz\nxa+yb+z(x+y)\n\nif a typo, and a = b then:\nxa+ya+z(x+y)\na(x+y)+z(x+y)\n(a+z)(x+y)\n\notherwise, not possible unless, use the math trick:\nxa+yb+z(x+y)+ya-ya+xb-xb\nxa+xb+ya+yb+z(x+y)-xb-ya\n(x+y)(a+b)+z(x+y)-xb-ya\n(x+y)(a+b+z)-xb-ya
484092
Yes, It is possible if we use x,y,z plane in plotting graphs. When using this method we should be very precise.\n\n3D model gives very spectacular graphs. There is algebraic expression for many 3D shapes e.g sphere, cones
244252
a. multiply y and z to get x\nb. mulitply y and y and z to get x\nc. multiply y and z and z to get x\nd. multiply y and y and z and add y to get x.\nf. multiply y by the quantity of 3 times y minus 2 times y.  Then take the square root of that answer to get x.
153865
x + y + z = t\n\nx +x% + y + y% + z + z% =\n[(100 + 1)( x + y + z)]% =\n101t% = t + t%. So\nx +x% + y + y% + z + z% = = t + t% or\nx% + y% + z% = t%
442548
okay well\n\n1) combine two to figure out the first variable\nx-2y+z=4\n-x+2y+3z=-4\nso 4z=0; z=0\nthen figure out the next variable with a different set of equations\nlike and multiply so the ys will cancel [take Zs out cause zero]\nx-2y+z=4\n2(2x+y)=6\n\ntherefore\nx-2y=4\n4x+2y=6\n5x=10\nx=2\n\nthen substitute for y\n\n-x+2y=-4\n-2+2y=-4\n2y=-2\ny=-1\n\nso y=-1, x=2, z=0\n\n\ncontinue the process with the rest.
496263
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
521843
Right away I can see that I can eliminate y from equations 1 and 2 by adding them together.\nEquation 1 + Equation 2 gives\n3x - 2z = 1\n\nI also noticed that I can elminate y again by adding equations 2 and 3\nEquation 2 + Equation 3 gives\n5x + z = -20\n\nI now have two equations and two variables.\n3x - 2z = 1 and\n5x + z = -20\nSolving the second equation for z gives me\nz = -5x - 20\nPlug that into the first one \n3x - 2(-5x-20) = 1\n3x + 10x + 40 = 1\n13x = -39\nx = -3\nSince z = -5x - 20 that means\nz = -5(-3) - 20 = 15 - 20 = -5\nNow replace x with -3 and z with -5 in one of the original equations to find y.\nUsing x + y + z = -6 gives\n-3 + y + -5 = -6\ny - 8 = -6\ny = 2\n\nSolution: x = -3, y = 2 and z = -5 or (-3,2,-5)\n\nYou can check your answer by plugging these values into the other two equations to see if the answer makes sense.\n2(-3) - 2 - 3(-5) = 7\n-6 - 2 + 15 = 7\n7 = 7 True\n\n3(-3) + 2 + 4(-5) = -27\n-9 + 2 - 20 = -27\n-27 = -27 True\nThis guarantees that the solution (-3,2,-5) is correct.
256400
1. k = -9\n2. s = -15\n3. y = 8\n4. p = 7\n5. m = -5\n6. z = 4\n7. b = -9\n8. c = -40\n9. d = 12\n10. a = -42
428621
is this for a algebra class? calculs class? matrix methods?\n\nTry this.\n\nWe want to simplify the complicated list of 3 constraints to two simple constraints: a range for y and a range for x.\n\nThe lower bound is simple: constraint III says all x and y are greater than or equal to zero.\n\nFind a range for y. Algebra can be done on these eqations the same as with an equal sign as long as the alligator is in the correct direction (it is). So eliminate x from the equations by using, "Constraint II minus Constraint I"; or;\n\n  x+3y<_6\n-(x+ y<_2)\n\n=2y<_4....... or, y<_2. Therefore, 0<_y<_2.\n\nSimliarly for range x, eliminate y by "Three times Constraint I plus Constraint II.", or:\n\n3*(x-y<_2) =  3x- 3y<_6\n                    +( x+3y<_6)\n                    = 4x<_12.... or, x<_3. Therefore, 0<_x<_3.\n\nNow we simply have an equation of three variables with two constraints.\n\nz=3x+2y;  0<_y<_2\n                 0<_x<_3\n\nI'm not sure what you want from here, but I think I understand.\n\nThe graph will be a plane in the x,y,z coordinate system.\n\nThe maximum z is achieved by using maximum values of x and y.\n\nThe corners will be defined by the extreme values of x and y with the correseponding z.\n\nFor instance, with (x,y,z)...\n\n(0,2,z1)\n(3,0,z2)\n(0,0,z3)\n(3,2,z4)\n\nThis will create a "polygon...?" in the x,y,z plane with the corners at the locations shown above.\n\nThis might be able to be done more simply by using matrices but I can't see an obvious way.
400703
1) (a - b)(x - y)\n2) (x^9 + y^3)(x^9 - y^3)\n3) (6x + 5)(x + 8)\n4) (x - 9)(x - 7)
470251
[(x1+x2)/2, (y1+y2)/2]\n(1,8)
183805
x=0,2,3,4 and y=1,-1,0,3