how do you solve: e raised to the 3x+1 minus e raised to the negative 2x+6?
487411
e^3x+1=e^-2x-6\nor 3x+1=-2x-6\nor 3x+2x=-6+1\nor e=-5/5=-1
948937
(-3,-3);(-2,0);(-1,1);\n(0,0);(1,-3)
631123
x(5x-3)=1\nx(5x) + x (-3) = 1\n5x^2 -3x = 1\n5x^2 -3x -1 = 0
774101
6+4x=3x+2\n4x-3x=2-6\n1x=-4\nx= -4
805693
#1\n4x+1+3=0\n4x+4=0\n4x=-4\nx=-1\n\n#2\n2y+7+4=y\n2y+11=y\ny=-11
1068964
simplify expression: -2(y+3)+6(y+1)\n\nsimply distribute then add.
261181
assuming that 2x+5=0, then I deny that solution. The Pair(3,-5) workes for 6y-8=-54 only.
117592
Multiply the 1st equation by 3\n\n3 [ 2x + y = -4]\nthat gives.. 6x + 3y = -12\n\nsubtract the 2nd equation from this\n\n6x + 3y - (-5x + 3y ) = -12 - (-1)\n\n6x + 3y + 5x - 3y = -12 + 1\n\n11x = -11\n\nx = -1\n\nput this into 1st equation\n\n2(-1) + y = -4\n\n-2 + y = -4\n\ny = -4 + 2\n\ny = -2\n\n\nx = -1 & y = -2\n\nyou can check this with 2nd equation\n\n-5(-1) + 3(-2) = -1\n+5 - 6 = -1 . hence the solution x = -1 & y = -2 is CORRECT
919399
6(x+4)-2x=-8\n6x+24-2x=-8\n6x-2x+24+8=0\n4x+32=0\n4x=-32\nx=-32/4\nx=-8
554172
Um...\n\n2(-X) + 3(-X) + X + 8\n\n-2X - 3X + X + 8\n\n-5X + X + 8\n\n-4X + 8\n\n8 - 4X
232213
I'll do the fourth problem.\n\n-3x< -6  or x+5< -2\n\nfirst deal with one half- \n-3x< -6 (now divide by -3)\nx>2 (when you divide by a negative the sign is reversed)\n\nthen the second half- \nx+5<-2 (get x alone, so subtract five from each side)\nx< -7\n\nso you have:\n2 < x  or  x < -7\n\nto graph this there is a line with an arrow to the left of the first slash and another line to the right of the second slash\n\n<-----/-------0--/-->\n____-7______2
1060834
(5x+2)/3 -(2x+4)/6 = -4/3\n\nMultiply everything by 6:\n\n10x + 4 - (2x+4) = -8\n10x + 4 - 2x - 4 = -8\n8x = -8\nx = -1
984158
The general approach is to move the x's to one side of the equation and the numbers to the other side, then solve for x. In your example:\n\n5x - 6 = 4x -+ 8\nFirst step: move 4x to left side. Since you can subtract the same amount from both sides of an equation the detailed step looks like this:\n5x - 6 - 4x = 4x + 8 - 4x\nor, in short form,\n5x - 4x - 6 = 8\n\nThen, move the 6 to the other side by adding 6 to both sides:\n5x - 4x - 6 + 6 = 8 + 6\nor, in short form:\n5x - 4x = 8 + 6\n\nThen, reduce both sides of the equation:\nx = 14\n\nHope this helps!
935326
You have got to use expansion series!\n\nX->0+ [e-(1+x)^(1/x)]/x =e/2 \nlet x=0+h\n=>as x->0, h->o\n\nh->0 [e-(1+h)^(1/h)]/h\n\nlet y=(1+h)^(1/h)\n=>log y=(1/h)log(1+h)\n=>y=e^[(1/h)log(1+h)]\n\nexpansion of log(1+x)=x - (x^2)/2..................\n=>e^[(1/h)log(1+h)]= e^[(1/h)[h - (h^2)/2!............]\n\n=>e^[(1/h)[-h - (h^2)/2!............]=e^[1-h/2!........]\n\ntherefore,\n\nlet y =lim h->0 [e-(1+h)^(1/h)]/h=lim h->0 {[e- e^(1-h/2!...)]/h}\n\nexpansion of e^x=1+x+(x^2)/2!.................\n\n=>lim h->0 {[e- e^(1-h/2...)]/h}=lim h->0 {-e[e^(-h/2.....) -1]}/h\n\nexpansion of e^x=1+x+(x^2)/2!.................\n=>lim h->0{-e[1+(-h/2-(h^2)/3.....)+ ........-1]/h}\n\n=>lim h->0{-e[(-h/2-(h^2)/2...)+ [(-h/2-(h^2)/2...)^2]/2!+.....]}/h\n\n=>lim h->0{-e(-1/2 -..(terms containing highter powers of h))}\n\n=>e/2\n\n\nP.S.  By the way L' Hospitals rule would fail here as limit is not of form 0/0 or ~/~. It is of form a/0 where a is constant.
1155015
Original:\n13 + (-12) - (-5)\n\nSubtracting the negative (-5), reverse the signs and add\n13 + (-12) + 5\n\nNow just go left to right:\n13 - 12 + 5\n(13 - 12) + 5\n1 + 5\n6\n\nThe answer is 6
520921
The (x-3)s cancel out because they will produce the same answer for all values of x.\n\n(x-5)/(5-x) produces an answer of -1 because they'll always be the opposite (negative) answer for all values of x.\n\nThe answer is -1
350975
-2 * x -2 * 6 > x +4\n-2x - 12 > x + 4\n-2x - 12 + 12 > x + 4 + 12\n-x - 2x > x - x + 16\n-3x > 16\n-3/-3x > 16/-3 flip the greater than less than for the negative\n\nx< -16/3\nx<-5.3333