12 divided by 3x-2 + 5 divided by x+4 =1?
827622
X=-9\n\n12/3x - 2 + 5/x + 4 = 1\n12/3x + 5/x  +2= 1\n12/3x + 15/3x = -1\n(12+15)/3x= -1\n27/3x= -1\n3x= -27\nx= -27/3= -9
895527
2x^2-8x+3x-12\n2x(x-4)+3(x-4)\n2x+3 = 0 or x-4= 0\nx = -3/2 or x = 4\n\n2x^2+10x-72\n2x^2+18x-8x-72\n2x(x+9)-8(x+9)\nx = 4 or x = -9\n\nx^2+6x-x-6\nx-1 = 0 or x+6 = 0\nx = 1 or x = -6
775457
multiplying by x\nx^2-5=4x\nx^2-4x-5=0\n(x-5)(x+1)=0\nx=5 or -1
346785
(x^2-5x+4)/(x^2-9) * \n(x^2+7x+12)/(x^2+3x-28)=\n\n[(x^2-5x+4)*(x^2+7x+12)]/\n[(x^2-9)*(x^2+3x-28)]=\n\n[(x^2-4x-x+4)*(x^2+4x+3x+12)]/\n[(x-3)(x+3)*(x^2+7x-4x-28)]=\n\n[(x(x-4)-(x-4))*(x(x+4+3(x+4)]/\n[(x-3)(x+3)*(x(x+7)-4(x+7)]=\n\n[(x-4)(x-1)*(x+4)(x+3)]/\n[(x-3)(x+3)*(x+7)*(x-4)]=\n\n[(x-1)(x+4)]/[(x-3)(x+7)]
292806
first... need to get all the like terms to one side of the equation\n\n-20=3(4x-5)-4(4x+1)\n\ndistribute and simplify\n\n-20=12x-15-16x-4 (distrubting a -4)\n-20=-4x-19\n-1=-4x\n(1/4)=x
231477
1/3 x 1/4 = (1x1)/(3x4) = 1/12
389436
5x4+4+4+5+4/2\n=20+4+4+5+2\n=35
560796
x = 4/5 + 1 3/8\nx = 4/5 + 11/8\nx = 32/40 + 55/40\nx = 87/40\nx = 2 7/40
516994
The trick is to substitute a known true equation. I used 2+3=5, but any true equation will work.\n\nSubstitututing in the first equation you get\n4(2) - 3(2) + 4(3) - 3(3) = 4(5) - 3(5)\n\nwhich is 8 - 6 + 12 - 9 = 20 - 15\n\nor 5 = 5.\n\n\nIn the second equation you get\n4(2) + 4(3) - 4(5) = 3(2) + 3(3) - 3(5)\n\nwhich is 8 + 12 - 20 = 6 + 9 -15\n\nor 0 = 0.\n\n\nHence, a + b - c = 0.  In the next equation, you are dividing by zero, which is not permissible in math.  If a +  b - c had not been equal to zero, then this would have been possible.
925101
The number is 4(x+3), or 4x+12
881909
No, I don't think you are right.  I assume this is a system of simultaneous equations and you're trying to find a solution set.\n\n4x - 12y = 5\n-x + 3y = -1\n\nFirst, multiply the second equation by 4 on both sides, giving\n\n4x - 12y = 5\n-4x + 12y = -4\n\nNext, add the two equations, giving\n\n4x-4x +12y -12y = 5 -4\n0x + 0y = 1\n\nThere are no solutions (x,y) for which this is true, so the solution set is the empty set.\n\nTaking the solution set you've determined, (5,0), putting that into the first equation gives\n4*5 -12 * 0 = 5\n20 -0 = 5 or 20 = 5, which is not true, so (5,0) cannot possibly be a solution.\n\nYou've made several errors, as follows:\n\n4x-12y=5  OK\n-x+3y= -1 OK\n\n-x=3y+-1 NO!, should be -x = -3y + -1 or x = 3y +1\n4(3y+-1)-12y=5 OK\n12y+-4-12=5 NO!, should be 12y -4 -12y = 5 or -4 = 5 \n0y=1 I have no idea how you got this\ny=0 NO!, if 0y =1, y is not 0, there is no y such that 0y =1\n-x=12y+5  I have no idea how you got this\n-x=12(0)+5 I understand how you got this, but y does not equal 0\n-x=0+5 OK\n-x=-5 previously you had -x = 0 + 5 , which was wrong but if it was right, you should get x = -5, not -x = -5\n\nOMG
720107
4x -12y =5...........(1)\n-x +3y = -1..........(2)\n\nMultiplying eq.(2) by 4 and add with eq.(1)\n(4x -12y =5) + (-4x +12y = -4)=\n0 = 1\nwhich is invalid.
1109407
-4(3-x) = 2(x+6) \n-12+4X=2X+12 use distributive property\n-12+4x-2x=2x-2x+12 move the variable x onto 1 side \n-12+12+2x=12+12 isolate the variable x\n2x=24\n(2x/2)=24/2\nx=12\n\nCheater!
406780
3x=2x+5 \n\nyou have to isolate x \n\nmake x on one side of the equal sign\n\n3x=2x+5 \nsubtract 2x from both sides\nx=5
889666
5(x-1)=2(x+3)\n5x-5=2x+6\nadding -2x+5\n3x=11\nx=11/3 or 3 2/3\notherwise\n5(x-1)=2x+3\n5x-5=2x+3\n3x=8\nx=8/3\n=2 2/3
147408
(15-5) divided [ (12 divided by 2 *2) -2]\n\n= (15-5)/[12/(2*2) - 2]\n= 10/[12/(2*2) - 2]\n= 10/[12/4 - 2]\n= 10/[3 - 2]\n= 10/1\n= 10
560506
that works... so does this:\n\n-2 x -1 x 1 x 2 x 3 = 12\n\n-2*-1=2\n2*1=2\n2*2=4\n4*3=12