This isn't homework. This is a problem that someone has asked me. I was wondering, what is the answer x <= -8 or x > 3. I understand the x > 3, but now the x <= -8. Here is my work\n\n(x - 2)/(x + 3) <= 2\nx - 2 <= 2(x + 3)\nx - 2 <= 2x + 6\n-x <= 8\nx >= -8\n\nBut instead it states x <= -8. Where did i go wrong. I understand that if it wasn't x <= -8, then it wouldn't fit the problem, but how did they get their answer to become that.
138408
You actually need to solve this a different way.\nYou can't multiply both sides by x+3 because this could be positive or it could be negative...you don't know.\nInstead, bring everything to one side by subtracting 2, then get a common denominator.\n(x - 2)/(x + 3) <= 2\n(x - 2)/(x + 3) - 2 <= 0\n(x - 2)/(x + 3) -2(x+3)/(x+3) <= 0\n[(x - 2) - 2(x+3)]/(x + 3) <= 0\n[(x - 2) - 2x - 6)]/(x + 3) <= 0\n(-x - 8)/(x + 3) <= 0\nYour important values are -8 and -3\nTest values less than -8, between -8 and -3, and greater than -3.\nYou will see that you get negative values for\nx <= -8 and x >= -3
559180
take the equation x^2 - 7x + 12 and do the following  (x ) (x) now...see that there is a - and a + in the equation....you need to do this ( x -  ) ( x - ) 'cause you put the minus in the first factor in the second factor you take the - and the + and multiply them....So you get (x- ) (x -) the other 2 terms for each equation comes from the 7x and the 12.  What you do is to find 2 numbers that multiplied gives you 12 and added give you 7....The numbers are 3 and 4 ...Then put them into your factors (x-3)(x-4).\n\nFor the other 2 equations you do the same process\n\n(x^2 - 8x + 7) = (x-  ) (x-  ). In this one, you have to find 2 numbers that multiplied give you 7 and added give you 8. They are 1 and 7. Then (x^2 - 8x + 7) = (x- 1 ) (x- 7 ).\n\nw^2 - 8w + 15 = (w - ) (w- ). As you saw in one and two, three is the same....so the factors are 3 and 5. Then, w^2 - 8w + 15 = (w -3 ) (w-5 )\n\nFor the solve...\n1. x^2 - 6x = 0\n2. x^2 - 2x - 3 = 0\n\nTHe 1 is like this...you have to complete your quadratic equation then you do the following x^2 - 6x = 0 --> x^2 - 6x + 9 =0+ 9...The 9 in the right side was added in order to keep the two sides equal...IN the left side you got a perfect quadratic equation, hence it will be (x-3)^2= 9....Now you pass the 9 of right to the left ...then it will be [(x-3)^2]-9=0\n\n\nThe other equation x^2 - 2x - 3 = 0....This one is quite simple, it is the same as the part 1...The difference is that when you find the factors, they will be 2 numbers that multiplied give you 3 but substracted gives you 2.....The #s are 3 and 1....Here is the thing....when you break the equation into factors, you will have the following (x- ) (x+ )...Since you have a -2 that means that it is the -3 and the +1...Hence your factors will be the following (x-3)(x+1)
253991
Case 1 : 2x+10<4\n   \n-> 2x+10-4<0\n-> 2x + 6 <0\n-> 2(x+3)<0\n-> x < -3\n\nCase 2 : 2x+10>-4\n\n-> 2x + 14>0\n-> 2(x+7)>0\n-> x > -7\n\nNow we combine the results of the 2 cases\n\ni.e, x<-3 & x>-7\n\nthis will come out to be -7 < x < -3
93059
x can't equal 0, so if you take what's in the denominator, (2x-6), and set it equal to zero, you'll get what shouldn't be in the denominator.\n\nso you complete the following:\n2x-6=0\n2x=6\nx=3\n\nso x CANT be three, because then the denominator would be zero and undefined.\n\nso the domain is all real numbers except for three.
119811
well lets take the first example- ok so the middle number is 11. and the number with no 'x' in it is 30. what makes 30 = 6 x 5 or 3 x 10. so try adding and subtracting those numbers in your mind: 6+5=11 ... 3+10=13 ...6-5=1...5-6=-1...10-3=7...3-10=-7. so which one of those give 11  ->> 6 + 5. \nso all u do is x^2+ 6x + 5x + 30 [split 11 as 6 + 5]\nur next step ; take 'x' common for the first to terms and 5 common for the 2nd two terms, u get\nx(x+6)+ 5(x+6)\ntake the bracket term multiplied by the un-bracket terms\n(x+5)(x+6)\n\nvoila...problem solved
409208
ill give it a shot - im not sure , anyways i got 2 answers..\n\n(1st)\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6x = 2-4x\n\nx+4x+6x = 2+1\n\n11x = 3\n\nx = 3/11\n\n\n(2nd possible answer)\n\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6/x = 2-4/x\n\n-5 = -2/x\n\n-5x = -2\n\nx = -2+5\n\nx = 3
649433
x^2-x-6 = (x-3)(x+2)\nFor x=-2 or x=3, this equals 0.  So look at three cases:\n\nCase 1: x<-2\nHere x-3<0 and x+2<0.  So (x-3)(x+2)>0 (Product of two negatives is positive).\n\nCase 2: -2<x<3\nHere x-3<0 and x+2>0.  So (x-3)(x+2)<0 (Product of a negative and a positive is negative)\n\nCase 3: x>3\nHere x-3>0 and x+2>0.  So (x-3)(x+2)<0 (Product of two positives is positive)\n\nSo the only case where it's negative is case 2, and your answer is b) {x|-2<x<3}
79133
I don't think (x-1)(x+3) are factors for x^2 + x - 3 = 0\nExpanding that will give x^2+2x-3=0\n\nQuadratic formula?\n\nFactors for x=(-1+sqrt13)/2, (-1-sqrt13)/2\nHence x= 1.30, -2.30\n\nTo solve for inequality of x^2+x-3<0\n\nTest x=2, (2)^2+2-3>0 [This is to test x>1.3)]\nTest x=0, (0)^2+0-3<0 [This is to test -2.3 <x<1.3]\nTest x=-3, (-3)^2-3-3>0 [This is to test x<-2.3]\n   \n\nHence solutions for x^2+x-3<0 are \n\n  -2.30 < x < 1.30 \n\nOR better still (-1-sqrt 13)/2 < x < (-1+sqrt 13)/2\n\nmeans that when x values are between this two numbers, x^2+x-3 is less than zero.
447488
Now my Algebra is a bit rusty but I think there is only one answe and it is {all reals}.   Here is my work...\n\n2x=x+x (subtract x from the right side)\n\n2x-x=x  (subtract x again)\n\n2x-2x=0\n\n0=0\n\nLike I said my Algebra is rusty, but I think I did that correctly.  The major thing I remember is to set one side equal to 0 then pick a variable to isolate.  We can do this another way too, I just noticed.\n\nx+x=2x therefore\n\n2x=2x \n\nThe answer is a set that contains all real numbers.
325201
If the two equations are related, and if the second equation is correct, then the first equation is wrong.\nIf 3x=2(x-4), then 3x-2(x-4)=0\nOn the other hand, if the two equations are not related, then x could be different in each, and therefore both could be correct.
587078
This is the answer i got when i worked out the problem\n1. You have to distribute the (x + 2) & (x + 1)....you do the by using the FOIL method....Firsts Outsides Insides Last and you get: x + 2 = x^2 + 3x + 2\n\n2. Get rid of the 2 on the left side by subtracting it on the right side and you end up with : x = x^2 + 3x\n\n3. Now you can move the x on the left to the right side too so you have: 0 = x^2 + 2x\n\n4. Pull out and x leaving you with 0= x (x+2)\n\n5. x = 0 and x = -2\n\nMy algebra is rusty but I think this is how it's done.
410726
nope\n\n2+3(x-4) < 3(2x-5)\n\n2+3x-12 < 6x-15\n\n3x-10 < 6x-15\n\n-3x < -5\n\nx < 5/3
636318
when you switch -x to x, you are multipling by -1.  This needs to be done to both sides of the equation.  One of the rules of algebra is that when ever you multiple or divide both sides of the equation by a negative number, you must flip the > to a <, or the < to a >.  So...\n\n-x > -2/3\n-x * -1 < -2/3 * -1\nx < 2/3\n\ni hope that helps!
197403
My math is rusty... but let me try.\n\nX can not be -1 here's why.\n\nYour fourth step was wrong: x=(x-1)/-(x-1)\n\nHere's the right one from step 3:\nx(x-1)=(1-x)\n-x(x-1) = -(x-1)  {multiply the two sides by -1}\nx =(x-1)/(x-1)... \nx=1... however x can not be 1, because it will yield 0/0 which is undefined.\n\nthe problem is... \nsay x = 1\n\njust because x = 1 is not the same as x^2 = 1. remember (-1)^2 = 1 too. So you are introducing an extra value.
389705
You can see thats wrong, because the two constant terms multiply to -14 instead of 3.\nIts going to either be (8x+?)(x+?) or (4x+?)(2x+?).\nLets try the second one:\n(4x+a)(2x+b). You need a and b to multiply to 3, while 4*b + 2*a = -14. Trying a couple of different possibilities shows that b = -3 and a = -1 works.\nSo (4x-1)(2x-3).
475165
x^2 - 8x + 16\n\n\nthe person above me said it is -x^2 but that is incorrect because -x * -x = x^2
584694
i think you made an arithmetic error in the first step.\n\nstart with: |5x - 3| + 8 = 22\nthen subtract 8 from both sides, and you will get\n|5x-3| +8 - 8 = 22 - 8\n|5x-3|  = 14\nand not |5x - 3| = 24\n\nfrom there:\n|5x-3|  = 14\n5x-3 = 14 or 5x-3= -14\n5x-3+3=14+3  or 5x-3+3= -14+3\n5x = 17 or 5x = -11\nx = 17/5 or -11/5