x+y+z=1, x-y+17=0, 2z+15y=0.... x=?
98021
Okay, here we go:\n\nUsing the second equation you can rearrange it to solve that equation for y in terms of x so we get:\n\ny = x + 17  [ I'll call this Equation 4 ]\n\nThen using the third equation you, can obtain z in terms of y:\n\nz = -15/2*y [ Equation 5 ]\n\nSubstituting Equation 4 into Equation 5 we obtain:\n\nz = - 15/2 * ( x + 17 ) [ Equation 6 ]\n\nSo now with Equations 6 and 4 we have y in terms of x and z in terms of x.  We substitute these into Equation 1 to get:\n\nx + ( x + 17 ) + ( -15/2 * (x + 17 ) ) = 1 [Equation 7]\n\nSimplying equation 7 by multiplying things out and combining terms of x we get:\n\n-11/2 * x - 221/2 = 1 [Equation 8]\n\nAnd finally isolating x gives:\n\nx = -223/11 ~= -20.272727272727\n\nNote that we also know y and z in terms of x, so figuring them out is trivial:\n\ny = x + 17 = -233/11 + 17 = -36/11 ~= -3.2727\nz = -15/2 * (x + 17) = -15/2 * ( -233/11 + 17 ) = 270/11 ~= 24.545454\n\nDoes that method make sense?  We used the substitution method to obtain this answer.  Try another one for yourself to make sure you understand!
1021939
2x+y=3..........eq 1         -2x+5y=-9..............eq 2\nAdd eq. 1 and 2\n6y=-6\ny=-1\nPutting this value in eq 1\nx=2\nS.S={ 2, -1}\n\n3x+5y=0............eq 3           2x-5y=-25.................4\nMultiply eq 3 by 2 and eq 4 by 3\nFrom eq 3: 6x+10y=0................eq 5\nFrom eq 4: 6x-15y=-75..............eq 6\nSubtract eq 6 from 5\ny=3\nPut the value of y in eq 3\nx=-5\nS.S={ -5, 3}
801397
equation (2)*2\n6x-4y+8z=12\n9x+4y-z=0\nadding\n15x+7z=12    (4)\n(1)*2\n18x+8y-2z=0\n6x-8y-3z=3\nadding\n24x-5z=3        (5)\n(4)*5        75x+35z=60\n(5)*7      168x+35z=21\nadding     243x=81\nx=1/3\nsubstituting\n8-5z=3\n-5z=-5\nz=1\nsub\n3+4y-1=0\n4y=-2\ny=-1/2\nsolution set{1/3,-1/2,1}
658708
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
806483
Write down all the possible pairs which multiply to 70\nand pick out that pair which sums to 17.\nOr, try to solve the system\nx + y = 17\nxy = 70.\nSo y = x/70,\nx + 70/x = 17.\nNow multiply each term by x and solve the\nresulting quadratic.\nGood luck!
783856
First, you should know that you can multiply any one of these equations through by a nonzero constant, without changing the solution. You can also add or subtract any two of the equations term by term without changing the solution.\n\nWe begin by adding equations 1 and 3 together\nx+y+z = -5\nx-y+4z=12\n---------------\n2x + 5z = 7  (notice that the ys cancel out)\n\nNow we triple equation 1, and subtract equation 2 from it\n\n3x+3y+3z= -15\n2x+3y-2z = 8\n--------------------\nx +5z = -23\n\nNow we have two equations, in which we have eliminated y.\nWe subtract them, and solve for x.\n\n2x + 5z = 7 \nx    +5z = -23\n-----------------\nx = 7 - (-23) = 30\n\nIt is easy to find y and z, now that we know x is 30/\n\nx + 5z = -23\n30 + 5z = -23\n5z = -53\nz = -(53/5) or -10.6\n\nx + y + z = -5\n30 + y + (-53/5) = -5\ny = 53/5 - 30 - 5 = -24.4\n\nSo (x,y,z) = (30,-24.4,-10.6)
424288
1. y2+6y+8 = (y+4)(y+2)\n2. x2+7x+12 = (x+4)(x+3)\n3. z2+2z-24 = (z-4)(z+6)\n4. x2+7x+10 = (x+2)(x+5)
271312
Let the two numbers be x,y \n\nTwo numbers multiplied together is - 15\n==> xy = -15 --> (1)\n\nTwo numbers added together is 2\n==> x + y = 2 --> (2)\n\nUsing (2)\nx + y = 2 ==> x = 2 - y --> (2A)\n\nSubstituting the value of "x" from (2A) in (1) we get\n\n(2 - y)y = -15\n==> 2y - y^2 = - 15\n==> y^2 - 2y - 15 = 0\n==> y^2 - 5y + 3y - 15 = 0  \n==> y(y - 5) + 3(y - 5) = 0  \n==> (y + 3)(y - 5)= 0 \n==> (y + 3) = 0 (Or) (y - 5)= 0 \n==> y = - 3 (Or) y = 5 \n\n-------------------------------------------------------\n[y^2 - 2y - 15 = 0\n(Coefficient of y^2)*(Constant) = (1)(-15) = -15\n\nFactorize -15 in such a way that the sum of the factors is equal to the coefficient of y\n\n-15 = - 1 * + 15 [(-1) + (+ 15) = +14] Na\n-15 = + 1 * - 15 [(+1) + (- 15) = -14] Na\n-15 = + 3 * - 5 [(+3) + (- 5) = -2] Ha ... ha ... ha\n-15 = - 3 * + 5 [(-3) + (+ 5) = +2] Na\n--------------------------------------------------\n(y + 3)(y - 5)= 0 \nProduct of two factors is zero\n==> \nOne of the factors should be equal to zero for the product to be equal to zero.\n==> (y + 3) = 0 (Or) (y - 5)= 0 \n-------------------------------------------------------\n\nNote: \n(2) and (2A) are two forms of the same equation.  Therefore, when the value of "x" as in (2A) should not be substituted in (2) as it amounts to using the same equation for solving for values of two variables which would not give you a result.... try to confirm this..\n\n\nhttp://www.schoolingkids.com/\n\nhttp://www.futureaccountant.com/
344034
1.y+2\n2. x+3\n3. z+6\n4. i wasnt sure what it was equal to, so i assued 0... making the answer (x+2)(x+5) however,if it is x2+7x=10...then this answer's wrong...
146086
2x+y=6 => y=6-2x \n3x+y=1 => 3x+(6-2x)=1 => x+6=1 => x=-5\ny=6-(2*-5) => y=16\n\nx=-5\ny=16
1061869
5y+2x=-4--------(1)\ny=3x-7-----------(2)\nSubstitute the value of y from (2) into (1), we get\n5(3x-7)+2x=4\n15x-35+2x=4\n17x-35=4\n17x=39\nx=39/17\nnow, y=3(39/17)-7=-2/17
555353
The answer is C.  a=0.\n\nD = sqrt (x2-x1)^2 + (y2-y1)^2\n17 = (1-2)^2 + (4-a)^2\n17 = 1 + (4-a)^2\n16 = (4-a)^2\n4 = 4-a, so a = 0
1011851
2x + y=14 and x - y= 1\nhence x=1+ y now substitute that in the 1st eq 2(1+y)+y=14\n2+3y=14\n3y=12\ny=4 now find out value of x you will get lhs = rhs simillary you can solve the second one...
983933
(x+y+z)(x+y+z) = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 1\n\n(1-x^2 - y^2 - z^2) = 2xy + 2xz + 2yz\n\n(1 - x^2 - y^2 - z^2) \n------------------------ = xy + xz + yz\n.............2\n\nClearly the left hand side is less than 1/2
10079
X+Y+Z=24\nY=X+2\nZ=2*X-2\n\nX+(X+2)+(2*X-2)=24\nCollect terms: 4*X=24; X=6\nY=6+2; Y=8\nZ=2*6-2; Z=10\n\n6 cm, 8 cm, 10 cm
955620
2x+2y+3z=-1 .....(1)\n3x-1y-2z=18 .......(2)\n-3x+1y-1z=-9 ......(3)\n\n(1)+(2)x2\n2x+2y+3z=-1\n6x-2y-4z=36\n-------------------+\n8x-z=35 .............(4)\n\n(2)+(3)\n3x-1y-2z=18\n-3x+1y-1z=-9\n---------------------+\n-3z=9\nz=-3\n\nsubstitute this value of z to (4)\n8x-z=35\n8x-(-3)=35\n8x+3=35\n8x=32\nx=4\n\nsubstitute the value of x and z to (1)\n2x+2y+3z=-1\n2(4)+2y+3(-3)=-1\n8+2y-9=-1\n2y-1=-1\n2y=0\ny=0\n\n(x,y,z)=(4,0,-3)\n\nhope this is clear enough
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.