what is the answer to this math problem: y +23az/35+{3/7[2/7]}=47azy-23{7-8[97+5/7]}   what do, z,and y equal?
108506
y=az+(-6/49)\n(6/49)a+ya=z\n\nam i right or close i hope
784552
z/0.23-0.11=0.17\n=>z /0.23=0.17+0.11=0.28\n=>z=0.28X0.23 =0.0644=0.1 (rounded to the nearest tenth)\nTherefore,(D) is correct
850927
9x/5 + 5y/4 = 47/10\n2x/9 +3y/8 = 5/36\n\n36x+25y=94\n16x+27y=10\n\ny=(94-36x)/25\n400x+2538-972x=250\n\ny=-2\nx=4\n\nfor another one use the same method!
1143016
y=kx\n44=k(4)\n44/4=k(4)/4\nk=11\nConstant is 11...easy algebra
961056
5z-(3+z)>=2(3z+1)\n5z-3-z>=6z+2\n4z-3>= 6z+2\n-3-2>=6z-4z\n-5>=2z\n-2.5>=z\nz<=-2.5
685350
4 x (7 - 6) / 3 = 4/3\n9 - (38 / (7 + 12)) = 7\n2 5/8 - (3 / (4 - 1)) + 1 / 2 = 2 1/8\n6 + (15 / 3) - (2 x 5) = 1\n18 / (9 - 6) + 8 = 14 \nMath Rulez!!!!
431382
let the 2 numbers be x and y :\n\nx - y=23----A\nxy = -120 ---B\nx = -120/y\n\nsubstituting x in A:\n-120/y - y = 23 \n-120/y - y2/y = 23\n-120/y - 23 = y2/y\n-120/y - 23/y = y2/y \n-120-23/y = y\n-120 - 23 = y2\n0 = y2 + 23 + 120\nsolving the quadratic equation by the formula;\ny = -b (sq.root) b2-4ac/ 2a\n\ny= -8 and x is 15\nand then 15-(-8) = 23 \n         15x(-8) = -120
941575
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.
696503
It's perfectly doable. You have to use those tricks where you multiply through and combine the equations to eliminate variables.\nLike here:\n\n       3x + 5y +2z = 17\n (-2)(x + y + z     =   0)\n------------------------------\n      x   + 3y          =   17\n\n\nYou do that a few times, try to get two of the variables in terms of  the other variable, and plug back into one of the equations they give you. \nMy answer was x = -67, y = 28, z = 39.
631059
3y+5=y-25\n2y+5=-25\n2y=-30\ny=-15\n\nSee, you just use inverse operations.\n\n4n-5=2n+3\n2n-5=3\n2n=8\nn=4\n\nSo your answers are y=-15 and n=4. I hope that helped!
817577
you are going to have to use SUBSTITUTION. meaning place the equations one above the other. See what you can cross out... i.e.\n\n1) TAKE OUT THE X'S BY MULTIPLYING THE FIRST EQUATION BY 5 AND THE SECOND EQUATION BY 3. YOU WILL GET...\n\n15x+20y+4z=35.\n-15x+9y+24z= -93\n\nthus, you can cross out the x's and get the equations:\n\n20y+4z=35\n9y+24z= -93, and now you must cross out the z's by multiplying the top equation by -6...\n\n-120y-24z=-210\n9y+24z= -93\n\nand, so on....
702715
I know this can be done with matrix operations, but as I understand it, it represents the following system of equations which can be solved using junior algebra methods:\n\n-10x + 11y - z = -3\n14x  +  9y + 8z = 2\n12x  - 2y - 11z = 3\n\nAdding 8*first equation to second, gives \n-66x + 97y = -22\nSubtracting 11*first from third gives\n122x - 123y = 36, and I begin to see why one might try something different!\n\nThese last two give \nx = 393/1858\ny = -77/929  and then sub in the first to get\nz = -50/929\n\nPersonally, I don't believe it, but that's what I got.  How about you?
975557
5x-3y=-12\n-3y=-5x-12\nmultiplying by -1\n3y=5x+12\nm=5/3\nb=12/3=4\n\n2.y=3x+5\n\n3.y+5/5+5=x+3/4+3\ny+5/10=x+3/7\nmultiplying by 70\n7(y+5)=10(x+3)\n7y+35=10x+30\nadding -7y\n35=10x-7y+30\nadding -35\n10x-7y-5=0
995498
5 - y = 6y - [4 + (1/7)(21y -14)]\nDistribute the 1/7\n5 - y = 6y - [4 + 3y - 2]\n5 - y = 6y - [3y + 2]\n5 - y = 3y - 2\n-4y = -7\ny = 7/4
902088
Well done! Not one is wrong. The answer to the third one is y = 11. If y = 11/12, you'd end up with 11=132!
132964
You are solving for k so it you need to rewrite the problem to look like this k=z/(xy) where x=3, y=8, z=6 substitute the letters for numbers so k=6/(3*8) k=6/24 or k=1/4 to check just solve for any other letter such as z by using what you know of k and you should come out with z=6
973910
17.\n5x+7y=-1  (1)\n-2y+3z=9  (2)\n7x-z=27    (3)\n(3)*3\n21x-3z=81\n-2y+3z=9\nadding\n21x-2y=90 *7\n147x-14y=630\n10x+14y=-2\n157x=628\nx=4\nsub in (1)\ny=-3\nsubin (2)\nz=1\n\n\n18.\nr-s+3t=-8    (1)     \n2s-t=15       (2)\n3r+2t=-7      (3)\n\n(2)*2  4s-2t=30\n          3r+2t=-7\nadding\n4s+3r=23\n(2)*3  6s-3t=45\n       r-s-3t=-8\nadding\n5s+r=37\n4s+3r=23\n15s+3r=111\nsubtracting\n11s=88\ns=8\nsub in (2)\n16-t=15\nt=1\nsub in (3)\n3r+2=-7\nr=-3