How many solutions are there to solve a system of linear equations using substitution and get - 8 = - 8 ?
98555
If the system has no answer never you get -8=-8.\nIf the system has only one answer, never you get -8=-8.\nIf the system has more than one solution, that is the equations are dependent, again never reach -8=-8.\n\nBut for every kind of known equation, -8=-8 do not contradict with the equation, that is you can write it as a new equation and make a new system of equations, then you get from this system -8=-8.\nAnother way is this:\nSuppose first equation is f(x1,x2,...)=0 :\n\n----> -f(x1,x2,...)=0 \nNow add up these two equations: f-f=0 ---> 0=0\n\nNow add -8 to both sides of last equation, you get -8=-8
704540
Add the two equations together: (2x+y)+(x-y) = 1+8, so 3x = 9, x = 3.  Then 2*3+y = 1 so y = -5.\n\nSteve
133517
the solutions is 7,8,3,2 or 7,6,5,2. The problem works with both sets of numbers.
966925
There are 3 solutions for ABCDEFGHJ -\n\n2 6 7 8 3 5 1 4 9\n2 6 7 9 5 3 1 4 8\n6 7 8 4 1 3 2 5 9
427860
There is an infinite number of solutions, \nthe set of solutions is :\n{(k,-k+12):k belongs to the set of real numbers}
1187525
well start by multiplying things out\n\n-8x+12=-8x+5\n-8x+8x+12=5\n0+12=5\nno solution?\n\n2x+3y=6\n2x=6-3y\nx=6/2-3y/2\nx=3-(3/2)y
919399
6(x+4)-2x=-8\n6x+24-2x=-8\n6x-2x+24+8=0\n4x+32=0\n4x=-32\nx=-32/4\nx=-8
599219
you have to solve the ones in the brackets first. so in this instance, 4(-1) = -4.\n\nand then, you solve -4 - 3 = -7
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
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}
5554
Solving the problem stepwise -\n(x - 8 ) ( x - 2 )\n= x(x-2) - 8(x-2)\n=x^2-2x-8x+2\n=x^2-10x+2\n\nThese two sites will help you for the concepts.
447914
-4(2x+8)=32\n-8x-32=32\n-8x=64\nx=-8\n\nEDIT: The guy above missed a sign.  Insert -8 into the original equation to see if it checks out (it does).
789649
We'll use the "addition" (or "elimination") method, but if we were to use the substitution method, we'd get the same result.\n\nIf you multiply both sides of the first equation by 3, you'll get:\n\n9x - 9y = 27\n\nAdding the two equations together then produces:\n\n9x - 9y - 9x + 9y = 27 - 25\n\nWhen you combine like terms, you get\n\n0 = 2\n\nNow, obviously, that's not true, which means this is an "inconsistent system" -- a system without any solutions.\n\nHope that helps!
940135
Use factoring.\nx2-2x-8=0\n(x-4)(x+2)=0\nso\nx-4=0 and x+2=0\nsolve\nx=4 and x=-2
896553
4x^2+8x-x-2\n(4x-1)(x+2)
107281
Okay, I see your two equations are 5x + 5y = -30 and \n2x - y = 15.  I'm not certain if these are right so I'll explain how to solve systems of equations in general.\nThe way we solve a system of equations by substitution is to solve one of the equations for one of the variables.  We then take what that variable is equal to and substitute it into the equation we did not yet touch.  We then solve the resulting equation.  \nThe explanation is confusing so I will provide an example.\n4x + y = 9\n-x + 3y = 1\nSo I'll solve the second equation for x like so:\n-x + 3y = 1\n-x = 1 - 3y\nx = 3y -1\nThen, I substitute 3y - 1 into the first equation for x, since this expression is equal to x.\n4x + y = 9\n4(3y-1) + y = 9\n12y - 4 + y = 9\nAdding four to both sides and combining like terms I get:\n13y = 13\nTherefore y = 1.\nWe can then solve for x using one of the original equations, we'll pick 4x + y = 9.\n4x + y = 9, and we know y=1 so:\n4x + 1 = 9\nSubtracting 1 and solving gives us\n4x = 8, and therefore x = 2.\nYou can use this same process to solve your problem.  I suggest starting by solving the second equation for y.\nHope this helps, please remember to come back and choose a best answer!  :-)
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?