use the rule a*b = (3a+2b) (3a-2b)  find 4*1?
98910
bloodless, you messed up the math:\n= (12 + 2) (12 - 2)\n= (14)(10)\n= 140
966048
(3ab^2)(3ab)^2\n(3ab^2)(9a^2b^2)\n27a^3b^4
980443
-4w -4 < or = 5/6w -1\n-4 < or = 4w +5/6w -1\n-3 < or = 4 5/6 w\n-3 <or = 29/6w\n-3(6/29 < or = w\nw = or > - 18/29
815598
a+b+ c+ab+cb+ac = abc + 1\nor a+b+c - abc = 1- ab-cb-ac\nor (a+b+c - abc)/(1-ab-bc-ac) = 1\n\nputting a = TAN A , b= tan B, c = tan C\n\nwe get LHS = tan(A+B+C) = 1\nor A+B+C = tan ^-1 = pi/4+npi\n\n chose any a and b then c can be chosen as\n\ntan^1(pi/4 - tan A - tan B)
1109869
multiply them together to find out 7/16 x 4/3 x 1/2 = ????
62092
if you mean 3(2ab-x) ... then 6ab-3x
1144353
4a+4b+4c=2ab+2bc+2ac \n4(a+b+c)=2(ab+bc+ac)\na=2\nb=2\nc=2\n\nor \na=1\nb=2\nc=4
747340
-2x-x > -6 + 4\n-3x > -2 \nx < 2/3
775644
Duch, there's some confusion over what the problem is.  It could be a/(3+4b) - a/(3+b)  OR  a/3 +4b -a/3 -b\n\nIf the former, then it boils down to -(11ab)/(9 + 15b + 4b²)\n\nIf the later, 3b
560506
that works... so does this:\n\n-2 x -1 x 1 x 2 x 3 = 12\n\n-2*-1=2\n2*1=2\n2*2=4\n4*3=12
333726
Perform operations in this order:\ni)   brackets\nii)  powers\niii) divide\niv)  multiply\nv)   add\nvi)  subtract\n\nSo work it out like this:\n\n2 + 2 (4 - 6 * 7) + 4\n= 2 + 2 (4 - 42) + 4\n          By doing multiplication 6 * 7 in brackets first.\n= 2 + 2 (-38) + 4\n          By doing subtraction 4 - 42 in brackets.\n= 2 + -76 + 4\n          By doing multiplication 2 * -38\n= 2 - 76 + 4\n          Since + and - sign in front of 76 combine into a -\n= -74 + 4\n          By doing subtraction 2 - 76\n= -70\n          By doing addition -74 + 4
231477
1/3 x 1/4 = (1x1)/(3x4) = 1/12
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
854360
S=(c-b)/(b-a)\nSb-Sa-c+b=0\nc=Sb-Sa+b\nc=S(b-a)+b\nHowever you like it.
50939
Use a division sign in parentheses. That way three fourths will look like (3/4).
1027973
the 3rd one is correct, but I get -(3+b)/b+2 and 2 for the first 2
770101
3ax+4by-2ay-6bx\n=3ax-6bx-2ay+4by\n=3x(a-2b)-2y(a-2b)\n=(3x-2y)(a-2b)\n2ax-2by+ay-4bx\n=2ax-4bx+ay-2by\n=2x(a-2b)+y(a-2b)\n=(2x+y)(a-2b)\nso 3ax+4by-2ay-6bx over2ax-2by+ay-4bx\n=3ax+4by-2ay-6bx /2ax-2by+ay-4bx\n=(3x-2y)(a-2b)/(2x+y)(a-2b)\n= (3x-2y)/(2x+y)