How many solutions are there to solve a system of linear equations using substitution and get 7 = y ?
138047
just 1...that being 7
260573
Well, we know that y = 7, so just plug that into the first equation and solve for x.\n\nx + 7 = 5\nx = 5 - 7\nx = -2\n\nANSWER : (-2, 7)
753930
The most intuitive way to answer is to use Gauss-Jordan elimination, as described below.\n\n\nw + x + y + z = 24\n3w + 3x + 3y + 3z = 72\n3w - 2x - 3y + 3z  = -9\n5x + 6y= 81\n\nw + x + y + z = 24\n5w + 5x + 5y + 5z = 120\n5w + 3x - 2y - 2z  = 4\n2x + 7y + 7z = 116\n\nw + x + y + z = 24\n2w + 2x + 2y + 2z = 48\n2w + 4x + 4y - 5z  = 22\n-2x - 2y + 7z = 26\n\n2x + 7y + 7z = 116\n-2x - 2y + 7z = 26\n\n4x + 9y = 90\n5x + 6y= 81\n\n20x + 45y = 450\n20x + 24y = 324\n21y = 126\n\ny = 6\n\n4x + 9y = 90\n4x + 54= 90\n4x  = 36\nx = 9\n\n2x + 7y + 7z = 116\n18 + 42 + 7z = 116\n7z = 56\nz = 8\n\nw + x + y + z = 24\nw + 9 + 6 + 8 = 24\nw = 1\n{w,x,y,z} = {1,9,6,8}\n\nAnother method is to put the coefficiants into a matrix, and put it into reduced-row echelon form.\n\n|1 1   1  1  24|\n|3  -2 -3 3 -9|\n|5  3  -2 -2 4|\n|2  4  4  -5 22|\n\nIn RREF...\n|1 0 0 0 1|\n|0 1 0 0 9|\n|0 0 1 0 6|\n|0 0 0 1 8|\n\nThese are all new valid equations, just in a much nicer form. Specially, 1w+0x+0y+0z=1, etc.\nThis tells us that w=1, x=9, y=6, and z=8.\n\nAlso, you could use Cramer's rule with these equations and get the same answer, but I will not demonstrate that one.
201908
Ah, systems.\n\nthe first problem- there are many ways you can choose to do this, but I think we should multiply both sides of the first equation by two so that they positive and negative y's will cancel out.\n\nSo now we have:\n2(2x + y)=2(4)\n3x - 2y = -1\n\nSimplified:\n4x + 2y = 8\n3x - 2y = -1\n\nnow you add down, so you get:\n7x +0y = 7\n7x =7\nx=1\n\nsubstitute 1 for x in any equation:\n2(1)+y = 4   \n2 + y = 4\ny = 2\n\n\n\nfor the second one- lets cancel out the y's again.\n\n5x - 3y = 14\n-3(2x - y) = (6)-3\n\n5x - 3y = 14\n-6x +3y = -18\n\nAdd them together:\n-x = -4\nx = 4\n\nThen solve for y:\n5x - 3y = 14\n5(4) - 3y = 14\n20 -3y = 14\n-3y = -6\ny = 2
282293
The techniques used to solve systems of linear equations (substitution and elimination) tend to do a pretty good job of solving quadratic systems.\n\nThis page does a pretty good job of demonstrating how:
819984
restate the first equation in terms of y:\n\ny = 12 - x\n\nNow substitute this in for y in the second equation:\n\n2x² + 3(12 - x)² = 7x(12 - x)\n\nNow, multiply it out and solve for x:\n\n2x² + 3(144 - 24x + x²) = 84x - 7x²\n2x² + 3x² + 7x² - 72x - 84x + 432 = 0\n12x² - 156x + 432 = 0\n12(x² - 13x + 36) = 0\nx² - 13x + 36 = 0\n(x - 9)(x - 4) = 0\n\nx = 9 or 4\n\nSo, {x,y} = {9,3} or {4,8}
1137447
2 linear equations\n\n X  + 3Y - 1 = 0\n4X + 7Y + 5 = 0\n\nFROM (1)   4X + 12Y - 4 = 0\nFROM (2)   4X +  7Y + 5 = 0\n\nSUBTRACT\n                          5Y - 9 = 0\n\n                          5Y = 9\n\n                            Y = 1.8\n \n  FROM (1)      X + 5.4 - 1 = 0\n\n                        X + 4.4 =   0\n\n                        X =  -4.4
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}
250421
Solutions do exist, it’s just possible you got your numbers mixed up (it happens to the best of us)\n\n2x-3y+z=5  --------(1)\n\nx+4y-2z= -3 --------(2)\n\n4x-2y+3z= 6 ---------(3)\n\nConsider (1) . let’s make z the subject of the formula to eliminate it\n\n2x-3y+z= 5\n\nTherefore z= 5+3y-2x    [by rearranging]\n\nYou can substitute z=5+3y-2x in (2) and (3), I’ll put it in (3) first, the order doesn’t matter\n\nTherefore 4x-2y+3(5+3y-2x)= 6\n\n4x-2y+15+9y-6x= 6\n\nSimplifying this you get 2x-7y= 9\n\nPut z=5+3y-2x in (2)\n\nSo  x+4y-2(5+3y-2x)= -3\n\nx+4y-10-6y+4x+-3\n\nsimplifying, you get 5x-2y= 7\n\nso now we have two new equations in terms of x and y, I’ll label them (4) and (5)\n\n2x-7y= 9 ---------(4)\n\n5x-2y= 7 ---------(5)\n\nWe have to solve these simultaneously now, consider (5)\n\n5x-2y= 7\n\nMaking x the subject of the formula we get\n\nx= (7+2y)/5\n\nput this new equation in (4)\n\n2* (7+2y)/5- 7y=9\n\nSimplifying this you get 14-45= 31y therefore y= -1\n\nNow we use ‘backward substitution’ to find x and z, first x\n\nFrom (4) we know 2x-7(-1)=9 therefore x=1\n\nNow put x and y in (1) to find z\n\n2(1)-3(-1)+z= 5\n\nFrom this you find z is 0\n\nTherefore x=1, y= -1 and z=0\n\nP.S: I'm not sure what you mean by the addition method, I used both elimination and substitution methods here, however I hope this helps!
792411
put y = 7 and x =4 and solve for c\nwe get 7 = 16+c or c = -9
701694
7x + y = 1 (subtract 7x from both sides)\ny = 1 - 7x\n\n3x + 2y = -9 (substitute in for y)\n3x + 2(1 - 7x) = -9\n3x + 2 - 14x = -9 (subtract 2 from both sides)\n-11x = -11\nx = 1\n\ny = 1 - 7x (substitute in for x)\ny = 1 - 7(1)\ny = 1 - 7\ny = -6\n\nAnswer: (1, -6)
220478
(x^2y - xy^2) = xy*(x-y) = 7*5 = 35
1116483
7y = x + 23\n(x - 2)^2 + y^2 = 25\n\n7y = x + 23\nx = 7y - 23\n\n(x - 2)^2 + y^2 = 25\n((7y - 23) - 2)^2 + y^2 = 25\n(7y - 23 - 2)^2 + y^2 = 25\n(7y - 25)^2 + y^2 = 25\n((7y - 25)(7y - 25)) + y^2 = 25\n(49y^2 - 175y - 175y + 625) + y^2 = 25\n49y^2 - 350y + 625 + y^2 = 25\n50y^2 - 350y + 600 = 0\n50(y^2 - 7y + 12) = 0\ny^2 - 7y + 12 = 0\n(y - 4)(y - 3) = 0\ny = 4 or 3\n\nx = 7y - 23\nx = 7(4) - 23\nx = 28 - 23\nx = 5\n\nor\n\nx = 7(3) - 23\nx = 21 - 23\nx = -2\n\nANS : (3,-2) and (5,4)
119155
you're given these two equations:\n\n2x+4y=16\n8y+4x=32\n\nthe second equation can also be:\n\n8y+4x=32\n4x+8y=32\n2x+4y=16, which woah - is the same as the first equation!  in other words, the two equations are the same and, when graphed, are the same lines.  that means any point along the line is a solution.  thus, there are an infinite number of solutions.  one POSSIBLE solution is shown below:\n\n2x+4y=16\nx+2y=8, choose any "x" since this equation will work for all "x"s; say, let x=2.\n2+2y=8\n2y=6\ny=3, therefore x=2,y=3 is A solution.  plugging these back in gives you:\n\n2x+4y=16 ==> 2(2)+4(3)=16 ==> 4+12=16 ==> 16=16\n8y+4x=32 ==> 8(3)+4(2)=32 ==> 24+8=32 ==> 32=32\n\nta-da!
900362
To explain this properly, we need an example so let's use the following 2 equations to explain the substitution method\n\n3x+2y=12 \n3x+y=9 \n\nFor ease, we would refer to the two equations as equations (1) and (2) respectively\n\n3x+2y=12 ... (1)\n3x+y=9 ... (2)\n\nFirst of all, to find x and y, we need 2 equations because there are 2 unknowns; x and y. If there is one unknown, we need only one equation to find it, likewise if there are 3 unknowns; x, y, z, we need 3 equations to solve it.\n\nNow, we need to make one of the unknowns the subject of one equation and substitute it in the other to find the solutions. When an unknown is the subject of an equation, it means that unknown stands on one side of the equality sign.\n\n3x+2y=12 ... (1)\n3x+y=9 ... (2)\n\nLooking at the two equations, in which one do you think that we can easily make an unknown (x or y) the subject. You are right! In equation (2) we can easily make y the subject of the equation because it already stands on it's own, it does not carry any co-efficients(whole numbers that stand in front of unknowns).\n\nSo making y the subject of the equation (2), we have:\n\ny = 9 -3x ... (3)\n\nWhen 3x crosses to the other side of the equality sign, it becomes -3x. We call this equation (3) for easey reference.\n\nNow, substitute the value of y in (3) in equation (1). The rule here is, if you use equation (1) to get equation (3), then you MUST substitute in equation (2). Likewise, if you use (2) to get (3), you must substitute in (1). This is to make sure that both equations are used in finding the values of x and y.\n\nSubstituting in (1) we have:\n3x+2(9 -3x) =12\n3x + 18 - 6x = 12\n3x - 6x = 12 - 18\n-3x = -6\n-3x/-3 = -6/-3\nx = 2\n\nSubstitute the value of x in (3) as this is the only equation that has not been used.\n\ny = 9 -3x \ny = 9 - 3(2)\ny = 9 - 6\ny = 3\n\nTherefore x = 2, y = 3\n\nCheck ...\nTo check that the answers are correct we substitute the values of x and y in equations (1) and (2)\n\n3x+2y=12 ... (1)\n3x+y=9 ... (2)\n\n3(2) + 2(3) = 12 ... (1)\n6 + 6 = 12\n12 = 12!\n\n3(2) + 3 = 9 ... (2)\n6 + 3 = 9\n9 = 9!
683580
y^2 + 7y + 3y + 21\ny^2 + (7 + 3)y + 21\ny^2 + 10y + 21\n(y + 7)*(y + 3)
334300
A linear equation can have three solutions...\n-All numbers that fit the equations\n-No solution\n-Or just one set\n\n\nI just had a test on that today.