Find the distances btw the parallel planes 15x+6y-6z=7 and 5x+2y-2z =6?
601325
I'll get you started by answering the first question.\n\nSay that you have two parallel planes in the form\n\nax+by+cz = d1\nax+by+cz = d2\n\nThe vector normal to these planes is N = (a,b,c) i.e., if you draw pictures of these planes then the vector N defines a line that goes perpendicularly through them both.  This line intersects the planes at the points\n\nP1 = d1/(a^2 + b^2 + c^2)*N\nP2 = d2/(a^2 + b^2 + c^2)*N\n\nThe distance between the two planes is simply the distance between these points:\n\ndist = | P1-P2 | = |d1-d2|/sqrt{a^2 + b^2 + c^2}\n\nFor your equations, I'm going to multiply the second equation by 3:\n\n15x+6y-6z=7\n15x+6y-6z=18\n\nThe distance between these planes is\n\ndist = |18-7|/sqrt{15^2 + 6^2 + 6^2} = √33/9
708030
5x + 7 = 6x\nSubtract 5x from each side\n7 = x\nAnswer: x = 7
893261
15a = 3(3a+8)\n15a=9a+24\n15a-9a=24\n6a=24\na=4
860215
6x - 2y = 8 dividing both sides by 2 we get => 3x - y = 4 -----(1)\n3x + 7y = - 4 -----(2)\n(1) - (2) => 3x - y - 3x - 7y = 4 - (-4)\n-8y = 8 and y= -1\nso x = 1
261684
5x-2y =7\n5x = 7+2y\n x = (7+2y)/5 -- substitute into\n-16-2x+y =7\n-16 -2(7+2y/5)+y = 7\n-16 -14-4y/5 +y = 7\n-16(5) -14-4y +y(5) = 7(5)\n-80 -14 -4y +5y = 35\ny = 35+80+14\n  = 129\n\nsubsitute y=129 into 5x-2y =7,\n5x -2(129) = 7\n5x -258 = 7\n5x =265\nx = 265/5\n  =53\n\nThus,x= 53; y=129
797240
I'll get you started.\n\n1) x + y = 6, so x = 6 - y\n\nx + z = 8, so since x = 6 - y, (6 - y) + z = 8,  y = 6 - 8 + z\n\ny + z = 2, so (6 - 8 + z) + z = 2, and z + z = 2 - 6 + 8, so \n\n2z = 4, and then z = 2\n\nnow put that number back into the other formulas, \n\ny + z = 2, so y + 2 = 2, so y = 0\n\nthen x + y = 6, so x + 0 = 6, so x = 6\n\n\n2) 2x + 4y = 9 & -2x + y = 5\n\nso 2x = 9 - 4y, so x = (9 - 4y) / 2\n\nsubstitute x, so -2((9 - 4y) / 2) + y = 5, so -9 + 4y + y = 5\n\nso 5y = 5 + 9, 5y = 14, y = 14/5 = 2.8\n\n2x + 4(2.8) = 9,   so 2x + 11.2 = 9,   so 2x = 9 - 11.2\n\nthen x = (9 - 11.2) / 2,  so x = -1.1, \n\ncheck 2x + 4y = 9, 2(-1.1) + 4(2.8) = 9\n\n-2.2 + 11.2 = 9,   9 = 9
791030
3x – 2y = 6\n3x = 6 + 2y\nx = 2 + 2/3y
1001971
[(16x*2-25) / (24x*2-30x)] / [(8x*2-2x-15) / (6x*2-7x-3)]\n= [(16x*2-25) / (24x*2-30x)] [(6x*2-7x-3) / (8x*2-2x-15)\n= [(16x*2-25) (6x*2-7x-30] / [(24x*2-30x) (8x*2-2x-15)]\n= [(4x-5) (4x+5) (3x+1) (2x-3)] / [(6x) (4x-5) (4x+5) (2x-3)]\n= (3x+1) / 6x
143223
a) (6x-7y)(6x+7y)\nb) y(4xy-7)\nc) (x-8)(x+3)
469648
2x - 6y = 37\n-6y = -2x + 37\ny = (2/6)x - (37/6)\ny = (1/3)x - (37/6)
1078373
2z+7<z+10\nyou need to get like terms together so first...\n2z+7+-7<z+10-7\n2z<z+3\nnow get z over to the  2z\n2z-z<z-z+3\nz<3\n\ncheck ya later ♥
941984
z = 6 is a plane parallel to the x-y plane through 6 on the z-axis\n\n(x+2)² + (y-2)² = 2² is a cylinder with axis through (-2, 2) on the x-y plane parallel to the z-axis with a radius of 2\n\nSo what you have described is a circular cross-section of that cylinder cut off by the plane z = 6\n\nParametrically x = -2 + 2cosθ, y = 2 + 2sinθ, z = 6\n\nReasons: \nx = -2 + 2cosθ →x+2 = 2cosθ, so (x + 2)² = 2²cos²θ\ny = 2 + 2sinθ → y - 2 = 2sinθ, so (y - 2)² = 2²sin²θ\n\nWhence (x + 2)² + (y - 2)² = 2²(cos²θ + sin²θ) = 2²\n\nAnd z = 6 is not variable
991735
easy\n6x+9=2x+7    by puting variables on same side\n6x-2x=7-9\n4x=-2\nx=- 0.5    (minus half)\n\n#2\n15x + 30= 19x +11\n 30-11= 19x-15x\n4x=19\nx=19/4=4.75
142158
Use the FOIL method, since this is really (6x+5)(6x+5)\n\nFirst terms (6x)(6x)=36x^2 +\nOutside terms (6x)(5)=30x +\nInside terms (5)(6x)=30x +\nLast terms (5)(5)=25\n\nFor a total of  36x^2 + 60x + 25
919291
I will show one example on how to do it and then you take that example and apply it to the other two problems:\n\nOn the first problem, start with these two equations:\n2x-y+z=15\nx-y+2z=12\n\nTake the second equation and multiply it by -2:\n2x-y+z=15\n-2x+2y-4z=-24\ny-3z=-9\n\nLet's focus on the other part:\nx-y+2z=12\nx+2y-z=-3\n\nMultiply the second equation by -1:\nx-y+2z=12\n-x-2y+z=3\n-3y+3z=15\n\nLet's put those two equations together:\ny-3z=-9\n-3y+3z=15\n-2y=6\ny=-3\n\nPlug it into a y-value:\n-3-3z=-9\n-3z=-6\nz=2\n\nPut the y and z value into this equation:\nx-(-3)+2(2)=12\nx+3+4=12\nx+7=12\nx=5\n\n(5,-3,2)\n\nCheck:\n5-(-3)+2(2)=12\n5+3+4=12\n8+4=12\n12=12\n\nI hope that helps!\n\nP.S. I checked with the original equations and they equaled out also. You can check the answers with a graphing calculator. Whoever is giving the people for thumbs down for their answers, whether it is right or wrong, is being very unkind and rude for doing so. I am sure that I am not only person who does not appreciate the thumbs down given too.
172253
6x= -15-3\n6x= -18\nx= -18/6\nx= -3
1142827
You don't give a feasible region. However, only one of your answers actually satisfies z=6x+10y. That should solve your problem.