-2x-5y=-15\n-6x-15y=-45\n\noption are 5,1\n15/2+5/2y,y\n15/2-5/2y,y\nor no solution
202248
Gauss Jordan elimination has to do with matrix operations.  Take your coefficients and constants and put them in an augmented matrix (basically a matrix with an extra column for the constants):\n\n| -2 -5 | -15 |\n| -6 -15 | -45 |\n\nUsing elementary row operations, try to get this into row reduced echelon form (all 1's on the diagonal):\n\nR1/(-2) -> R1\n| 1 5/2 | 15/2 |\n| -6 -15 | -45 |\n\nR2 - (-6)R1 -> R2\n| 1 5/2 | 15/2 |\n| 0 0 | 0 |\n\nFrom here, we see that we can't get this into RRE, and from this we can say that there is no solution.\n\nThis was seen in the previous answers in solving it as a system of equations versus using a matrix.  From inspecting the equations as a system, you can see that these are actually the same line - the second is the first multiplied by 3.  Since they are the same line, there are actually infinite solutions.\n\n(If these were parallel lines, then you would have no solution.  Remember that a "solution" is a point of intersection in this case.  If they are parallel, they never intersect, and thus the system would have no solution.  But if they overlap, they always intersect, and thus the system would have infinite solutions.)\n\nsomeguy's explanation in turning it into a parameter is a legitimate approach if you _have_ to have an answer.  The answer he came up with works, regardless of which equation you started with.\n\nSo from this, we have three answers:  no solution (from G-J elimination), infinite solutions (from inspecting the lines graphically/analytically), and (15/2 - 5y/2, y) (from turning it into a parameter solution).  Take your pick based on whatever method you are required to use for this particular problem.
403904
(y-11) -(y+6)=6y\ny - 11 - y - 6 = 6y\ny - y - 6 - 11 = 6y\n0 - 17 = 6y\n6y = -17\ny = -17/6\nThe answer is D) -17 over 6
513715
put -2y=5-3x call it equation {1}\ncall 4x-5y=86 {2}\n\nmultiply all of the terms i {1} by 5\nmultiply all ot the terms in {2) by 2\nthis gives -10y=25-15x {3} &  8x-10y=172 {4}\nyou've got -10y in terms of x in {3}, so just put that into {4}  8x+25-15x=172, -7x=157, \nx = -172/7 sub in to {1} to get y \nman that was ablast from the past, haven't done that for years.
431136
Multiply equation 1 by 2 so that when adding both equation, the x variable is cancelled,\n\n6x + 12y = 16\n-6x + 3y = 2\n0 + 15y = 18\ny = 6/5\n\nSubstituting y in equation1\n3x + 36/5 = 8\n3x = 4/5\nx = 4/15\n\nTherefore  x = 4/15 , y = 6/5
661010
i) 8x -10x-15-1+6x = 8+4=6x\n   8x-6x+6x-10x = 8+4+15+1\n   -2x = 28\n    x = -14\n\nii) 15x-12-14x+21 = 3x+3+5\n     15x-3x-14x = 3+5+12-21\n      -2x = -1\n         x = 1/2\n\npeice of cake :-)
490934
1/2y= -3x+7\ny= -6x+14
66101
2x - 3y = 5        \n2x^2 + 3y^2 = 35 \n\n2x = 5 + 3y\nx = (5 + 3y)/2\n\n\n2 [(5 + 3y)/2]^2 = 5 + 3y\n2 [(25+30y+9y^2)/4] = 5 + 3y\n25+30y+9y^2 = 10+6y\n9y^2-24y+15=0\n(3y-3)(3y-5)=0\n\ny=1; y=5/3\n\nif y=1\n2x - 3(1) = 5\nx = 4\n\nif y = 5/3\n2x - 3(5/3) = 5\nx = 5
481034
First, I multiplied the first equation by 2 and the second one by -5 in order to make it easy for me to cancel out both x's.\n2(-5x+7y=-40)\n-5(-2x+4y=-16)\n\n-10x+14y=-80\n10x+-20y=80\n\nNow, you can actually cancel out the x's and the constants at the end.\n\n-6y=0\ny=0\n\nNow you plug y=0 into one of the equations to find x.\n\n-5x + 7(0) = -40\n-5x +0 = -40\n-5x = -40\nx = 8\n\n(0,8) is the answer.
487338
Don't panic, you can do this...\n\nJust start by distributing the numbers through the parentheses:\n5(2x-3)=2(-3x-2)+5\n10x - 15 = -6x -4 + 5\n\nNow group letters on one side and numbers on the other, start by adding 15 to both sides:\n10x = -6x - 4 + 5 + 15\n10x = -6x + 16\n\nNow add 6x to both sides:\n10x + 6x = 16\n16x = 16\n\nNow divide both sides by 16:\nx = 1\n\nIf you did it correctly, you should be able to plug x = 1 back into the equation and get a valid statement:\n\n5(2(1) - 3) = 2(-3(1) - 2) + 5\n5(2 - 3) = 2(-3 - 2) + 5\n5(-1) = 2(-5) + 5\n-5 = -10 + 5\n-5 = -5 <-- check\n\nSo x = 1 is the correct answer.
202693
3y+2x=12\n2x=-3y+12\nx= -3/2y+6\nx=-3/2(x-1)+6\n =-3/2x-(-3/2)+6\n = -3/2x+15/2\n5/2x=15/2\nx= 3\n\ny= -3/2y+6-1\n5/2y= 5\ny= 2
37394
y= 2x + 11 and y=-x + 5 \n\n2x + 11 = -x + 5\n3x + 11 = 5\n3x = -6\nx = -2\n\ny = -(-2) + 5\ny = 2 + 5\ny = 7\n\n(-2, 7)
677040
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
404453
why not, you just plug y in the 1st equation:\n5x + 3y = -19   & y=-2x-8\n5x + 3(- 2x - 8) = -19\nand solve for x\n5x - 6 x - 24 = -19\n-x - 24 = -19\n-x = 24 - 19 = 5\nx = -5\ny = -2x-8\ny = -2(-5) - 8\ny = 10 - 8 = 2\n\nx = -5 , y = 2
501183
3x + 2y - 2z\n\n3(5) + 2(2) - 2(1)\n\n15 + 4 - 2\n\n19 - 2\n\n17\n\nThe answer is 17\n\n- - - - - - -s-
357953
1st one:\n\n5x-2y=-5; flip 2y to other side\n5x=2y-5; divide by 5\nx=2/5y-1; put this into 2nd equation for x\ny-5(2/5y-1)=3\ny-2y+5=3\n-y=3-5\n-y=-2\ny=2\n\nput this into 2nd equation for y\n2-5x=3\n-5x=1\nx=-1/5\n\n2nd one\n8x-4y=16; divide by 4\n2x-y=4; flip 2x to other side\n-y=4-2x; multiply by -1\ny=2x-4; lo and behold this is exactly the same as the 2nd equation - can't be solved with substitution, equations 1 and 2 are the same\n\n3rd one\n4x-12y=5; divide by 4\nx-3y=5/4; flip 3y to other side\nx=5/4+3y; substitute this into x in 2nd equation\n-(5/4+3y)+3y=-1; simplify\n-5/4-3y+3y=-1; cancel out the -3y and + 3y\n-5/4=-1; well, how can you have -5/4 ever equal to -1, equations are not the solvable (unless either of the numbers equals infinity)
350298
2x -  4y = -5\n3x + 4y = -15\n\n5x = -20\n\nx = -20/5                                       \n\nx = -4\n\n*************************\n\n2(-4) - 4y = -5\n\n-8 - 4y = -5\n\n- 4y = -5 + 8\n\n- 4y = 3              \n\ny = -3/4\n\n**************************\n2 (-4) - 4 (-3/4) = -5                                        \n\n-8 - (-3) = -5           \n\n&\n\n3 (-4) + 4 (-3/4) = -15                           \n\n-12 + (-3) = -15
616457
x-(-2 -7)=-1\nx+2x+14=-1\n3x=-15\n\nx = -5\n\ny=-x-7\ny=-(-5) - 7\ny=5-7\n\ny = -2