I solved this:  -2x - 5 < -2 or x - 3 < -10\n\nAnswer was this: x > -3/2 or x < -7\n\nBut now the teacher wants the solution set for this inequality. I dont even remember learning solution sets...\n\nI did some brain cell killing and got this {x|x < -7 or x < -3/2} but I dont think that it is right, but I know I'm close. \n\nWhat do you think?
214994
The solution set is (-infinity, -7) U (-3/2, infinity)
415657
1. Square everything in the first set of inequalities.\n\nIf 4 < sqrt(x) < 9, then 4^2 < [sqrt(x)]^2 < 9^2\n\nor, 16 < x < 81.\n\nAs a test, let's choose x = 64. I picked it because it has an easy square root.\n\nIf x = 64 (and note that 16 < 64 < 81), then sqrt(x) = 8, which is greater than 4 and less than 9.\n\n2. If the total cost of x apples is b cents, then the cost of 1 apple is b/x cents. x * b/x = b, which is indeed the cost of x apples according to the problem.\n\nIf 1 apple costs b/x cents, then y apples cost yb/x cents.\n\nAs a test, suppose that 5 apples cost 20 cents. (That is, x = 5, b = 20).  In this case, 1 apple costs 4 cents.\n\nHow much are 6 apples (Let's let y = 6). 6 apples would cost 24 cents.\n\nNow, what is yb/x?\n\n6*20/5 = 120/5 = 24.\n\nIn this case, it checks out. (It will for all cases, but one case doesn't actually prove that it works. It really only justifies it. But it is right :D)
134105
Generally, you break an inequality such as this into two parts and graph each separately.  So you'd be graphing x=-1 and x=3.\n\nThe type of line depends on the sign: if it's a <= or >=, then use solid lines to indicate anything on the line satisfies the original inequality.  If it's just < or >, then use a dashed line.  (This is similar to open and closed circles on a number line.)  Then, you pick points in the different regions that you created with the lines and determine if they satisfy the inequality.  If so, the region gets shaded.\n\nThere is no y dependency, so you basically have two vertical lines for your graph for this problem.  Checking the regions, only the area between the lines satisfies the inequality, so that's the only region that gets shaded.
472298
You differentiated and found it has relative min at x=1/3?  but if you differentiate again you find second deriv is\n54x - 1/(x^2), and I think this goes from \nnegative when x<1/cuberoot(54) to positive when it's greater, so that means it has an inflection.  Also means it's concave down when \nx<1/cuberoot(54), so the answer must be 5, not 2.  Unless I made a mistake!\n\nPS  Dave is wrong about 1.  Domain is all positive real numbers, because lnx is undefined for 0 and neg.
94827
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
474061
29x - 515 < -14(8 - 3x) given\n29x - 515 < -112 + 42x distribute\n-515 < -112 + 13x subtract 29x from each side\n-403 < 13x add 112 to each side\n-31 < x divide by 13\nx > -31 just turn it around because it looks better\n\n(the answer above this one would be correct, but you need to reverse the direction of an inequality when dividing by a negative number.)
367814
To get x by itself in an inequality, divide the side with x by the constant (the nonvariable) next to it. So here, you would divide both sides of the equation by -2/5, which is the same thing as multiplying both sides by -5/2 (reciprical). This will cancel out the -2/5 next to the x, so 6 * -5/2 = -15. Also, when multiplying or dividing an inequality by a negative, you switch the "<" or ">" sign to the opposite sign. therefore, x< -15.
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
558458
For the first one, 4<-2<6 does become -2>x>-3 (you do have to flip the signs) and yes, it is still an "And" statement, just in reverse (or in other words, -2>x>-3 is the exact same thing as -3<x<-2). For the second one, you don't actually have to multiply by 10, but you should get the same answer either way. I got 0.84<x<1.76.
533983
If the sign in front of the 4 is a single negative sign then the answer is x>4, because you change the sign when you divide by a negative number in an inequality. If the inequality is actually -(-4)<-16, then your answer is right.
443691
1st one you are right because 2/3 < 4/5 so to the left\n2nd one you are wrong becuase it can be both. if x is positive -x is negative and if x is -ve -x is positive\n\n3rd one you are right
392097
Here's what you should think: 3 times a number that has to be zero or higher, then having another 3 added to it, cannot possibly equal 1 (it must be at least 3).  So there is no solution.\n\nYou did say it was an inequality, though.  If the = was really > or >=, the answer is all real numbers.  If it was < or <= the answer is the empty set.
343769
4x + 6 is not quite right\nwatch the order of operations and distributive properties\n\nrepost your answer and I'll tell you if you got it\n\n(you didn't think I'd just give up the answer, did you ?)
347375
Alright, don't listen to the above post...\n\nGrade 12 here, answering questions all day for mere fun.\n\n12)\ny - 3 = -1/3 (x + 4)\n\nEXPLAINED\n\nadd 3 to both sides\ny = -1/3 (x + 4) + 3\n\nexpand 1/3(x + 4) \n= x * 1/3 + 4 * -1/3\n= x/3 + 4/3\ny = -1/3x - 4/3 + 3\n\nTo add -4/3 and 3, use 3 = 9/3\ny = -1/3x - 4/3 + 9/3\ny = -1/3x + 5/3\n\nTrue, my bad. I originally added 4/3 instread of -4/3\n\n\n16)\n(-2, -1) (4, -4)\n\nm = Dy / Dx\nm = ((-1) - (-4)) / ((-2) - (4))\nm = 3 / -6\nm = -1/2\n\ny = mx + b\n(-1) = -1/2(-2) + b\n-1 = 1 + b\n-1 - 1 = b\n-2 = b\n\ny = -1/2x - 2\n\nCHECK\ny = -1/2(4) - 2\ny = -2 - 2\ny = -4\nTherefore it works.\n\n18)\n(0, 0) (2, 3)\n\nm = Dy / Dx\nm = ((3) - (0)) / ((2) - (0))\nm = 3/2\n\nb = 0\n\ny = mx + b\ny = (3/2)x + 0\ny = (3/2)x\n\nNote, the lower post copied my answers
70545
You prove that for any tiny little e>0 you pick, there exists a little d>0 such that \n\nif abs(1/(x^2)-1/x^2))<e, then abs(x-x)<d,\n\nand since x-x=0 (this is a trivial limit unless you've messed up the parentheses in the problem), just let d=e, or e/2, or any value between e and 0.
401953
Solve it just like a regular equality:\n\nx-6 ≤ 7\n\nAdd 6 to both sides, to get:\n\nx ≤ 13  ← this is your answer\n\nWhen working with inequalities, you can add and subtract from both sides just as you would in a regular equality.  You can also multiply and divide both sides, but the trick to keep in mind is that if you multiply or divide by a negative number, you need to change the sign!  For example, if you had:\n\n-2x ≤ 10\n\nDivide both sides by -2 AND REVERSE THE INEQUALITY to get:\n\nx ≥ -5\n\nHope that helped!\n\n~ ♥ ~\n\nEDITED TO ADD:  A "regular equality" would be where you have an equal sign rather than a < or > sign.  So x-6 = 7 is an equality.\n\nGood luck kid ... sounds like you will need it!
476897
It is false no solution\n\nLook at \n Example 5:  on this site\n http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut14_solve.htm\n \nWhere did our variable, y, go???  It disappeared on us.  Also note how we ended up with a FALSE statement, 0 is not equal to 11.  This does not mean that y = 11 or y = 0.  \nWhenever your variable drops out AND you end up with a false statement, then after all of your hard work, there is NO SOLUTION. \n\nSo, the answer is no solution.