if y = xz  find a value for y given that x= 3 and z = 10?
405853
y=xz\n\nplug in the numbers tht are given.\n\nyou are given x=3, so x in the equation becomes 3. \ny=(3)z\n\nz=10, so zin the same equation becomes 10.\nNow it's y=(3) x (10)\n\nWhen multiplied, you get 30.
373486
10y - 3 = -10 + 6y\n    + 3 =  +3 <-- subtract 3 from both sides  \n-------------------\n10y = -7 + 6y\n-6y =    - 6y <-- subtract 6y from both sides\n---------------\n4y = -7\n--   --\n4  =  4 <-- divide both sides by 4\n\ny = -7/4
785524
y = 3x-5\n\nx^2 +(3x-5)^2 = 25\nx^2 + 9x^2 -30x + 25 = 25\n10x^2 -30x = 0\n10x(x-3) = 0\n\nso x = 0 or x = 3\n\nWhen x = 0 y = -5\n\nWhen x = 3  y = 4
1169467
y = x^2 + 5x - 3\n2y=3x - 2\n----------------\n2y = 2x^2 + 10x - 6\n2y = 3x - 2\n-----------------------\n    0 = 2x^2 + 7x - 4\n=>2x^2 + 7x - 4 =0\n=>2x^2 + 8x - x - 4 = 0\n=>2x(x + 4) -1(x+4) = 0\n=> (2x - 1)(x + 4) = 0\ntherefore 2x - 1 = 0 => x = 1/2 or \nx + 4 = 0 or x = -4\nx = - 4 or 1/2  \ngot it?
833533
Go for calculator.\npress  number (z), press x^y, press 5 \nthen press x^y then press 0.3\ngotcha, u will get the answer
239864
Solve for y in the second equation, then plug that into the first equation and solve for z.  After you find the value of z, plug that into the other equation.\nMake sure you always check your answer!  To do that just plug in the values for y and z into each equation to see if you get a true statement.  If you don't then your answer is wrong or else you made a mistake when checking the answer.
902917
y=1/x^2-3 => y+3 = 1/x^2 => x= +/-sqrt(1/(y+3))
4959
=8(y^3)z + 12(y^2)(z^2) - 20y^2
1142827
You don't give a feasible region. However, only one of your answers actually satisfies z=6x+10y. That should solve your problem.
961063
let # of large ones = x and # of small ones = y\n\nthen x = 30+3y\nand x=6y\n \n=> 6y = 30 + 3y\n=> 3y = 30 \n=> y = 10\n\nand x = 6*10 = 60
738586
Friend, \n            let the number be y\n                        3*y=9\n                            y=9/3\n                            y=3\n           So the number is 3
677039
(1) { -2x-y-z=0\n(2) { -x-2y-z=0\n(3) { -x-y-2z=0\n\nLet's look at (2):\n-x-2y-z=0 => x=-2y-z\nNow at (3):\n-x-y-2z=0 => x=-y-2z\nNow we can see that\n-2y-z=-y-2z => *** y=z ***\n\nNow let's put these values into (1):\n-2x-y-z=0 => -2x-y-y=0 => -2x-2y=0 => 2x=-2y => *** x=-y ***\n\nFor now, we know that x=-y=-z\nLet's put all of what we know into (2), and express everything using 'x':\n-x-2y-z=0 => -x-(-2x)-(-x)=0 => -x+2x+x=0 => 2x=0 => *** x=0 ***\n\nThus, x=0 => y=-0=0 => z=0\nSo the only solution is (0,0,0).\n\nGood luck!
750664
Let,the number be x\nkTherefore according to the problem\n10-2x=x+3\n=>-2x-x=3-10\n=>-3x=-7\nx=-7/-3=7/3=2 1/3\nTherefore the number is 2 1/3
951523
vertical: x=3\nhorizontal: y=5\n`
865074
To start off, you are given a simple equation to find x with respect to y and z. \n\nx = 2y - 3z +5\n\nThe next step is to substitute into the other two equations.\n\n3(2y - 3z +5) -3y + z = 9\n6y - 9z + 15 - 3y + z = 9\n3y - 8z = -6\n\n5(2y - 3z + 5) + y - 3z = 3\n10y - 15z + 25 +y - 3z = 3\n11y - 12z = -22\n\nNow what I would do is take the first new equation and solve for y:\ny = (8/3)z - 2\n\nAnd then substitute it into the second equation:\n\n11((8/3)z - 2) - 12z = -22\n(88/3)z - 22 - 12z = -22\nz = 0\n\nInsert this into the previous equation:\ny = (8/3)(0) - 2\ny = -2\n\nAnd then finally into the very first equation:\nx - 2(-2) + 3(0) = 5\nx + 4 = 5\nx = 1\n\nNow, to be sure, we check our answer with the other two equations:\n\n3(1) -3(-2) + 0 = 9\n3 + 6 = 9\n\n5(1) + (-2) - 3(0) = 3\n5 - 2 = 3\n\nEverything checks out, so:\nx = 1\ny = -2\nz = 0
794793
x = 3. Try entering it on your calculator.
840475
x^9- y^3z^3=(x³-yz)³ \n                  =x^9-3*x^6*yz+3x³y²z²-y³z³