Suppose you try to solve a system of linear equations & get the following result:0=0 How many solu does it hav
138490
Infinite.\nIn this case all variables and constants have canceled out, leaving a trivially true statement. The original equation, therefore, would be called an identity and one would not consider the graph (it would be the entire xy-plane). An example is 2x + 4y = 2(x + 2y). The two expressions on either side of the equal sign are always equal, no matter what values are used for x and y.
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.
197379
x is anything less than 1.\n4 x1=4 so 1 cannot be the answer (unless the < had a line underneath it)\n4x0=0 which is less than 4\n4x -2= -8
1079419
The answer is b, multiple solutions, if you work the problem so that it's in reduced row echelon form you will see that there are many possible solutions to the linear equations.
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
645941
4x + y >= 4\ny >= -4x +4\n\n-------\ny = 0\n0 >= -4x + 4\nx = 1\n-------\nx = 0 \ny >= 0 + 4\ny = 4\n-------\n\n\nTherefore, shade the region above the line.\n\nTwo points on the line are:\nx-intercept: (0, 4)\ny-intercept: (1, 0)
67585
Formulaically it reduces down to x*z = x*z, so no matter which numbers you plug in for any of the values it will be a valid equation.  \n\n(x*(y+z))-(x*y)=x*z      becomes\nx*y + x*z - x*y = x*z     rewrite that as \nx*y + x*z + -(x*y) = x*z    the positive and negative x*y terms cancel,leaving\n\nx*z = x*z\n\nSince that is a truism, an infinite number of solutions exist.
274355
Depends on the system you are trying to solve. check your limits of integration. from -infinity to 0 plus 0 to infinity.
405630
I'll take a guess and say zero, since if a*b = 0, then either a is zero or b is zero.
797702
Since there is no = sign in that equation, I assumne it should be equal to zero.  If that is the case then m= -1/8
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
148434
So, continuing with mr_w's excellent example, we have 4 equations in 4 unknowns. So you can set this up as a matrix-vector problem.  Define matrix M such that M * Number of games bought vector = Cost & total games vector.  Assigning the rows of M in order to the total cost, the PS3 and Xbox360 cost, the PS1 and Xbox360 cost, and the total number of games, \nM = \n70 150 300 450\n0 0 300 450\n70 0 0 450\n1 1 1 1\nNumber of games bought vector =\na\nb\nc\nd\nand Cost & total games vector =\n13910 12000 3050 54\nThe solution will find the values of a, b, c and d.\nWhen I ran the solution I ended up with noninteger values for each and the values checked with the problem statements.  I hope those values were mr_w's solution.
754976
It's true.\nThe answer is indeed y = 0.\nYou can find this by dividing each side by -4.\n-4y = 0\n-4y/-4 = 0/-4\ny = 0\nRemember that zero divided by any nonzero number is always zero.
234829
With factorisation you turn the equation x^2 + px + q = 0 into two smaller equations in brackets like this:\n\n(x+a)(x+b) = 0\n\nSo that a+b = p\nand a*b = q\n\nEG. Your first example you have p = 4 and q = 3\n\nSo you've got a+b = 4 so a could be 2 and b 2 or a 1 and b 3 etc\n\nand a*b = 3 so a could be 1 and b 3... Yes it matches.\n\n(x+3)(x+1) = 0\n\nThe solution is just x = -a and x = -b\n\n-> x = -1 or x = -3
610821
Easy...\n0 = 0\n\nadding -2 on both sides\n\n-2 = -2\n-> -3 + 1 = -6 + 4\n-> -2*3/2 + 1 = -4*3/2 + 4\n\nadding 9/4 on both sides..\n\n-> 9/4 -2*3/2 + 1 =9/4 -4*3/2 + 4\n->\n(3/2 - 1)^2 = (3/2 - 2)^2\n\ntaking +ve sqr root\n3/2 - 1 = 3/2 - 2\n\n-> 1= 2\nadding 1 on both sides\n2 = 3\n-> 1 + 1 =3
885957
You have probably figured out that no number will make this statement true. But don't forget about zero. \nYou will also find problems where even zero does not work. In those cases, "no solution" is the correct answer.
339198
2x0 = 0\nIt'll be the same as today... :P