How many solutions are there to solve a system of linear equations using substitution and get 5 = 3 ?
143708
Don't post every question on your homework assignment.
621340
The first one is (x-2)(x+9)(x-7).\nHowever, for the second part, everyone else has forgotten that imaginary zeroes come in CONJUGATE PAIRS.\nSo there are actually 5 zeroes to part 2, 4i, -5, -2i, 2i, -4i.\n\nSo the solution would be (x-4i)(x+4i)(x+5)(x-2i)(x+2i)\n(x^2+16)(x+5)(x^2+4)\nx^5 + 5x^4 + 20x^3 +100x^2 + 64x + 320\nViola...
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.
934633
By substitution, the first equation when substituting in y=2+x becomes:\n4x - 4(2+x) = -8 ==> -8 = -8 which is always true.  This means that what you have are two equations with an infinite number of solutions, as the two lines are the same line.  So any pair of the form (x,x+2) such as (0,2) or (1000,1002) will work.
731670
(-4x3 + 3x2 - 5) + (x2 - 5x + 6) =\n-4x3 + 3x2 - 5 + x2 - 5x + 6 =\n-4x3 + 4x2 - 5x + 1
895237
you do not get specific values, there are infinite number of solutions for each equations\n\n2x + y = 9, y= -2x + 9\n\n-x - y = - 8, y=-x+8\n\n- 3x + 4y = 2, y = 3x/4 - 1/2\n\n- 2x - y = - 3, y= -2x +3\n`
892382
To solve the first one, divide each part of the first equation by two, so 2x-3y=-13; since -2x+3y = 13, adding the two equations gives 0x+0y=0.  No help.  Try substitution.\nFrom the first equation, 4x = 6y-26; x = 1.5y - 6.5\nsubstitute into the second equation -2(1.5y-6.5) + 3y = 13; -3y +13 + 3y = 13; 13=13.  Still no help.\nBoth equations are equivalent; the line that represents them is the solution.\n\nProblem two:\nSubtract equation 2 from equation 1.\n3x-3x+2y-3 = 6 - y\n2y-3 = 6-y\n3y = 9\ny = 3\n3x+6=6\nx=0\nchecking\n3x+3=y=3\n3x=0\nx=0
704941
The second equation can be rewritten as:\n3x + 2y = -1\nMultiply both sides by 2\n6x + 4y = -2\n6x + 5y = -7 (this is the first equation)\nSubtract the first equation from the second one.\ny = -5\nNow, plug the y = -5 into the second equation:\n3x +2(-5) = -1\n3x - 10 = -1\n3x = 9\nx = 3\nSo your solution is x = 3 and y = -5
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
518063
Solving a system by addition involves simply adding the two equations together. First, however, you have to do a little prep work. Either the x's or the y's must be of the opposite sign. I'll show you how to do the y's first, because it is easier.\n\n5x-3y=13\n4x-3y=11\n\nThe y's need to be opposite. What is the simplest way to do so? Make one of the -3y's into a positive 3y. Thus, multiple one of the equations by -1.\n\n-1(5x-3y=13)\nbecomes\n-5x+3y=-13\n\nNow you have\n-5x+3y=-13\n4x-3y=11\n\n+3y + -3y = 0\n\nYou have just canceled out the y's.\n\nNow you have:\n-5x=-13\n4x=11\n\nAdd these together and you get:\n-x=-2\nThus, x=2\n\nNow you know x. Take either of the original equations, and substitute in your new x.\n\n5(2)-3y=13\n10-3y=13\n-3y=3\ny=-1\n\nThat's it! Hope I helped.
789310
Notice the first equation only involves x.\nx - 5 = 6\nAdd 5 to both sides\nx = 11\nNow replace x with 11 in the second equation\n3(11) - 2y = 5\n33 - 2y = 5\n-2y = 5 - 33\n-2y = -28\ny = -28/-2\ny = 14\nAnswer: (11,14)
720922
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)
824262
well... imagine 2 functions\n\n2x + 3y = 5\n-2x + 2y = 10\n\nadd the two togather\n\n5y = 15\ny=3\n\nplug y into one of the first equations\n\n2x + 3y = 5\n2x + 3(3) = 5\n2x + 9 = 5\n2x = -4\nx = -2 and y = 3\n\nyou can check it with the other equation too\n\n-2x + 2y = 10\n-2(-2) + 2(3) = 10\n4 + 6 = 10\n10 = 10\n\nand the system is solved... \n\nx = -2 and y = 3
937047
Well, there may be more solutions here than usual, but the first solution I see is:\n(3 L,5 L,8 L)\n(0/3,0/5,8/8)\n(0/3,5/5,3/8)\n(3/3,2/5,3/8)\n(0/3,2/5,6/8)\n(2/3,0/5,6/8)\n(2/3,5/5,1/8)\n(3/3,4/5,1/8)\n(0/3,4/5,4/8)
1103313
5  3  4     *   x                        19\n3 -1 1          y            =           1\n                   z\nthere are infinite solutions:\n\nx = (22-7z)/14\ny = (52-11z)/14
120172
y-2 = -2(x-3)\ny-2 = -2x + 6 \ny+ 2x = 6+2 \ny+2x  = 8  ( linear equation)\n\nFINDING SOLUTIONS FOR THIS LINEAR EQUATION :\n\ny+2x = 8\n(a)  Assume, x=2 \n     subtitute the assumed value of x in the equation\n     y+ 2 (2) = 8\n     y+4 = 8\n     y = 8-4\n     y = 4\n     HENCE,  when -->   x= 2 then y= 4\n\n   Similarly you can find more solutions for any\n   given linear equation by assuming the value of any one unknown ( x or y ) and then finding the value of the other by substitution in the same given equation.
881909
No, I don't think you are right.  I assume this is a system of simultaneous equations and you're trying to find a solution set.\n\n4x - 12y = 5\n-x + 3y = -1\n\nFirst, multiply the second equation by 4 on both sides, giving\n\n4x - 12y = 5\n-4x + 12y = -4\n\nNext, add the two equations, giving\n\n4x-4x +12y -12y = 5 -4\n0x + 0y = 1\n\nThere are no solutions (x,y) for which this is true, so the solution set is the empty set.\n\nTaking the solution set you've determined, (5,0), putting that into the first equation gives\n4*5 -12 * 0 = 5\n20 -0 = 5 or 20 = 5, which is not true, so (5,0) cannot possibly be a solution.\n\nYou've made several errors, as follows:\n\n4x-12y=5  OK\n-x+3y= -1 OK\n\n-x=3y+-1 NO!, should be -x = -3y + -1 or x = 3y +1\n4(3y+-1)-12y=5 OK\n12y+-4-12=5 NO!, should be 12y -4 -12y = 5 or -4 = 5 \n0y=1 I have no idea how you got this\ny=0 NO!, if 0y =1, y is not 0, there is no y such that 0y =1\n-x=12y+5  I have no idea how you got this\n-x=12(0)+5 I understand how you got this, but y does not equal 0\n-x=0+5 OK\n-x=-5 previously you had -x = 0 + 5 , which was wrong but if it was right, you should get x = -5, not -x = -5\n\nOMG