The solution for the quation 6( 8s-3) =48s -18 is given as 0.  Is this correct? Why?
404498
yes ... it most definitly is correct!\n\n6(8s-3)=48s-18\n6(8(0)-3)=48(0)-18 (anything times 0 is 0)\nso you get 6(0-3)=0-18\n6(-3)=-18\n-18=-18\n\nbecause they equal the same at the bottom when you sub in the number, it is correct
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
1133153
7 - 2[4-6+2x] - 8+ x=6\n\n7 - 2[-2+2x] - 8 +x =6\n7 + 4 - 4x - 8 + x =6\n3 - 3x = 6\nx = -1\n\nDon't listen to the math goober - plug the value in and check it.
534103
4(2X+6)=-8\n8X+24=-8\n8X=-36\nX=-4\n\nThere you are =)
137681
Ok...\nI see what you are doing...\nYou merely substitute the number in for x...\nso the problem becomes 6-2*(-1)\n6+2\n8
1160074
18r^4s^5t^6\n---------------- =\n-3r^2st^3\n\n-6r^(4-2)s^(5-1)t^(6-3)= -6r^2s^4t^3\n\nhope it's not too late.\n\nGood luck!
160067
When you substitute with y=x^(1/6), you're assuming :\nI) x>0 (which should be put as a starting condition)\nII) y>0 . This is to be considered after solving the 2nd degree equation, as you get a fake solution y= -3 that you must reject as non-acceptable.\n\nThe only acceptable solution is x= 1/64
809908
2(y-3) = y-6\n\nThis is what you are supposed to calculate : a value for the y, such that the equation holds.\n\nfor instance try y = 1\n\n2(1-3) = 1-6\n2 - 6 = 1- 6\nand you see that this is not so, thus y=1 is not a solution.\n\nthe standard way to solve is : move all y''s to the left of the=, and all numbers to the right of the =\n\n2(y-3) = y - 6\n2y - 6 = y - 6\n2y -y - 6 = -6 ( move left y to the right, sign changes )\n2y - y = -6 + 6 moved left -6 to right, change sign\n\ny = 0. ( adding the 2y - y together, and the -6 + 6\n\nso the solution is y = o
1013241
no.\n\n36/6 = (6*6)/6 = 6/1\n\n6/0 is undefined
610284
The equation has 6c3= 20 possible solution but since your looking for positive integral solutions its 20/2=10...
550482
-2a(b+2)-6b=\nor\n-2(ab+2a+3b)
909799
gopal's solution:\n\nf(x)=x^3-6x^2+9x+2\nf'(x)=3x^2-12x+9=0 for extremas\n3(x^2-4x+3)=0\n3(x-3)(x-1)=0\nx=3 or 1\nf''(x)=6x-12\nat x=1 6(1)-12=-ve so max\nat x=3 6(3)-12=+ve so min\nso the absolute maximum is at x=1 and\n=6\nabsolute min is at x=3 and is=2  <------ However still need to check endpoints of the interval [-1,4] as these turnng points are actually LOCAL maxima and minima!\n\nSo f(-1) = -1 - 6 - 9 + 2 = -14 < 2\nSo ABSOLUTE minimum = -14 at x = -1\n\nAlso f(4) = 64 - 96 + 36 + 2\n= 6\nSo ABSOLUTE maximum = 6 and occurs at both x = 1 AND at x = 4
445500
(18)(-3)=(-11)r\n-54=(-11)r\n54/11=r
431729
6(8s-3)=48s-18\n\n\nactually, the correct answer is all real numbers. any number you substitute for "s" will work
235841
A(8) = 1/2 (-3)* 8 -1 = -12-1=-13
957282
|3(x + 1) + 6| =18\nie |3x + 9| = 18\nwhence |x + 3| = 6\nThere are 2 different methods to solve such an equation\n\nOne is to use the definition that\n|N| = N if N ≥ 0\nand = -N if N < 0\n\nThe other is to use the propety that |N|² = N²\n\nMethod 1\nSo x + 3 = ±6\nWhence x = -9 or 3\n\nMethod 2\n|x + 3|² = 6²\nie x² + 6x + 9 = 36\nx² + 6x - 27 = 0\n(x + 9)(x - 3) = 0\nx = -9 or 3
905134
2x - 3y - 6 = 0\n\n2x - 3y - 6 + 6 = 0 + 6\n\n2x - 3y = 6\n\n2x - 3y + 3y = 6 + 3y\n\n2x = 6 +3y\n\n2x/2 = 6= 3y/ 2\n\nx =  6 + 3y/2\n\n- - - - - - - - - -s-