Solve:  The square root of 3x+6+4 is less than or equal to 7.?
606423
Here's how to work it out:\n\n(Sorry, because I can't use special symbols, I'm using _/ for the square root symbol and <= for the less than or equal to symbol.)\n\n_/3x + 6 + 4 <= 7\n_/3x + 10 <= 7\n3x + 10 <= 7 x 7\n3x + 10 <= 49\n3x + 10 - 10 <= 49 - 10\n3x <= 39\n3x/3 <= 39/3\nx <= 13
926616
2x+3<6x-1\n2x-6x<-1-3 \n-4x<-4\nx>1
222653
x+6(x^2)= 12\nx+6x^2  = 12\n6x^2+x-12=0\n(2x+3)*(3x-4)=0\n2x+3=0 3x-4=0\n2x=-3  3x=4\nx=-3/2 x=4/3
672081
Answer: x = 1 or 4\n\nEXPLANATION:\nx^3 - 6x^2 + 9x - 4 = 0\nThis can be written as:\nx^3 - x^2 - 5x^2 + 5x + 4x - 4 = 0\nor x^2(x-1) - 5x(x-1) + 4(x-1) = 0\nor (x-1)(x^2 - 5x + 4) = 0\nor (x-1)(x^2 - 4x - x + 4) = 0\nor (x-1){x(x-4) - 1(x - 4)} = 0\nor {(x-1)^2}(x-4) = 0\nTherefore either x = 1 or x = 4
811856
Okay multiply the top equation by 7 and the bottom equation by negative 3 to obtain 2 new equations with the x variable to eliminate:\n\n7(3x + 7y = -43)\n-3(7x + 6y = -59)\n\nthis becomes \n\n21x + 49y = -301\n-21x - 18y = 177\n\nNow cancel out the 21x's and add down.  Doing that, the equation becomes\n\n31y = -124   this can easily be solved for y\n\ny = -4\n\nNow plug -4 back into one of the original equations\n\n3x + 7(-4) = -43\n\n3x - 28 = -43\n\nadd 28 to both sides to yield\n\n3x = -15   and solve for x\n\nx = -5\n\nNow check x = -5 and y = -4 into both and see that you have solved this system correctly.
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)
741209
-2x-3 > 4x+7 .... subtract 7 from both sides, you get...\n-2x-10 > 4x   ..... then add 2x to both sides, and you get...\n-10 > 6x       ...... multiply both sides by 1/6....\n-10/6 > x    ..... flip it around and reduce....\n-5/3 > x  OR  x < -5/3    \n\nx is less than or equal to -5/3\n\nYou can double check your answer.  Pick a number less than x, such as -2 (which is less that -5/3).  Plug it in.  You get...\n-2(-2) - 3 = 4-3 = 1\n4(-2) + 7 = -8+7 = -1\nand 1 is greater than -1, so it works!
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.
970218
Do you require mathematical signs, or are standard operations OK (like square root)?\n\n1: (1 + 1 + 1)!\n2: 2 + 2 + 2\n3: 3 * 3 - 3\n4: sqrt(4) + sqrt(4) + sqrt(4)\n5: 5 + 5/5\n6: 6 + 6 - 6\n7: 7 - 7/7\n8: 8 - sqrt(sqrt(8 + 8))\n9: 9/sqrt(9) + sqrt(9)\n\nAnd for a bonus: (0! + 0! + 0!)! = 6
1091435
write this as 1+10/(3x^2+9x+7)\n\nthe minimum of 3x^2+9x+7 ( at x=-1.5 check it , do differentiation) 0.25\n so answer is : 1+10/.25 = 41
808574
4x^2-4x= -7\n4x^2-4x+7=0\nx=(4+/-√(4^2-4*4*7))/8\nx=(4+/-√(16-112))/8\nx=4+/-√(-96))/8\nx=(1/2)(1+/-√6 i)\nx=1/2+(1/2)√6 i\nx=1/2-(1/2)√6 i
880864
3(4x+6)+7\n\nexpand first (ie multiply what is in the bracket by what is out side the bracket)\n\n= (3 times 4x + 3 times 6) +7\n=12 x + 18 + 7\n\nthen collect like terms (i.e. similar powers of x)\n\n= 12x +25
581691
if by this you mean\n\n((x^2)/2) - ((7x)/4) = (-3/2)\n\nMultiply everything by 4\n\n2x^2 - 7x = -6\n2x^2 - 7x + 6 = 0\n(x - 2)(2x - 3) = 0\nx = 2 or (3/2)\n\nANS : x = 2 or (3/2)
1070204
suppose given equation √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + .... = x\nnow  x^2 = 6+√(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 +.....\n      x^2 = 6 + x          \n                    as the other part except 6 can be written as x again\n     x^2 - x -6 =0\n    x^2 -3x+2x-6 =0\n    x(x-3)+2(x-3)=0\n   (x-3)(x+2) = 0\n    so x can be either 3 or -2
372975
3(a+2)^2-(a+2)-4=0\n\nTake (a+2)=x\n\n.'. 3x^2-x-4=0\n\nroots of the equation:\n\n(  -(-1)+/-sqrt[1-4.3.(-4)]  )/2*3        \n\nsince  roots of the equation ax^2+bx+c=0 is \n  (  -b +/- sqrt[b^2-4.a.c]  )/2*a\n\nroots=(1 +/- 7)/6\nx=4/3,-1  \n\n.'. (a+2)=4/3     =>     a= -2/3\n      (a+2)= -1      =>     a= -3
819535
sqrt(x) - sqrt(3x + 3) = 1\nsqrt(x) = sqrt(3x + 3) + 1\n\nsquare both sides\n\nx = (sqrt(3x + 3) + 1)^2\nx = (sqrt(3x + 3) + 1)(sqrt(3x + 3) + 1)\nx = (3x + 3) + sqrt(3x + 3) + sqrt(3x + 3) + 1\nx = 3x + 3 + 2sqrt(3x + 3) + 1\nx = 3x + 4 + 2sqrt(3x + 3)\n\n-2x - 4 = 2sqrt(3x + 3)\n-x - 2 = sqrt(3x + 3)\n\nsquare both sides\n\n(-x - 2)^2 = 3x + 3\n(-x - 2)(-x - 2) = 3x + 3\nx^2 + 2x + 2x + 4 = 3x + 3\nx^2 + 4x + 4 = 3x + 3\nx^2 + x + 1 = 0\n\nx = (-b ± sqrt(b^2 - 4ac))/(2a)\n\nx = (-1 ± sqrt(1^2 - 4(1)(1)))/(2(1))\nx = (-1 ± sqrt(1 - 4))/2\nx = (-1 ± sqrt(-3))/2\n\nx = (1/2)(-1 ± isqrt(3))\n\nThere are no real solutions.
344350
=f^3(6f^2+6f+4-9f) \n=f^3(6f^2-3f+4)\n\nsolve 6f^2-3f+4 using f=[3 +/- sqrt(9-4(6)(4))]/12\n\nNote that the solution (other than f=0) is imaginary since the term under the sqrt sign is negative