solve a+b/(a-2b)-1/2b-a?
354066
you need to give more information, solve for what? this isn't an equation, there is no equals sign so nothing to solve. Its just a statement
156464
u can't devide by (a-b) as (a-b)=0
652728
a=b=i^(1/3)\ni=-1\nab+a^2b+ab^2 = [i^(1/3)*i^(1/3)] + [i^(2/3)*i^(1/3)] + [i^(1/3)*i^(2/3)]\n= [i^(2/3)] + i + i\n= -2+(-1)^(2/3)
626693
ax - b = -c\nax = b - c\nx = (b - c) / a.
641733
[-b +/-  sqrt(b^2-4ac)]/(2a)
388695
"the sum of two numbers is 24" a+b=24\n\n"twice the first" -- 2a -- "plus the second" -- b -- "is 26" so 2a + b = 26\n\nNow you have two variables and 2 equations, so you can solve the system of equations\n\na + b = 24\n2a + b = 26  Subtract the first equation from the second one.\n\n2a - a + b - b = 26 - 24\na = 2\n\nNow substitute this into one of the equations and solve for b.\n\n2 + b = 24\nb = 22
61956
try to follow\n\nx/2-10=a\na/2-4=b\nb/2-2=6\n\nsolve and substitue\nb=16\na=40\nx=100 --the original amount\n\nhope that helped
937737
3a2(a-b) -6a(a-b) +21(a-b)\n(a-b)(3a^2-6a+21)\n3(a-b)(a^2-2a+7)
272590
1/a + 1/b = c  multiply both sides by ab\n\nb + a = abc  get the terms with b on one side\n\na = abc - b factor out a b\n\nb(ac - 1) = a  divide by ac - 1\n\nb = a/(ac - 1)
538044
hmmm...very tough one!\n\nill sovle by the formula!\n       _________\n-b +- V (b^2)-4ab\n------------------\n       2a\n       _________\n-10 +-V 100+512\n------------------\n       2\n\n       _________\n-10 +-V 612\n------------------\n       2\n\n-10 +- 24.7\n------------\n     2\n\n14.7/2\n\n34.7/2\n\n(7.35, 17.35)
912865
bring the a's to one side... u then get 9a = 27 ... divide and answer is a = 3
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
825318
(a^2 -2ab + b^2) = (a-b)(a-b) = (a-b)^2\nYou cannot cancel anything, so \n(a^2 -2ab + b^2)/(a+b) does not simplify.\n\nYou may wish to write it as (a-b)^2 /(a+b)
339354
(2x + b) * (a + 3c)\n\n\nsteps:\n2ax + 6xc + ba + 3bc \n1) group first two, and last two\n(2ax + 6xc) + (ba + 3bc)\n\n2) factor out 2x from the first, and then b from the second group\n2x*(a + 3c) +  b*(a + 3c)\n\n3) since (a + 3c) is common, combine 2x and b\n(2x + b)*(a + 3c)
966048
(3ab^2)(3ab)^2\n(3ab^2)(9a^2b^2)\n27a^3b^4
534750
1/2 + 1/2b + 2b^2 + 6bc + c^2\n\n(1/4b)(1/4b) + 4bc + (bc)(bc)\n1/2 + 1/4b +1/4b + b^2 +4bc + b^2 + c^2 + 2bc\n1/2 + 1/2 + 2b^2 + 6bc + c^2\n\nI think this is right... email me if I'm wrong?
982041
Well dear\n\n-If the first root is " α " and the second roots is " α + 3"  so have  \n▪x ² - (α + β) + α β = 0 \n▪( x - α )(x - β ) = 0 \n-b/a = α+ β  \n&\nc/a = α β\nso we have ; \nc/a = m/4  \n-b/a = - ( -16) / 4 = 4 \nα + α + 3 = 4 , 2α + 3 = 4 \nα = 1/2  ,  β = 3 1/2 \n2 /a = α β  =>  m/4 =  1/2 * 7/2  , m = 7\n\nGood Luck dear