Can you give a real-world example when the solution of a system of inequalities must be in the first quadrant?
374378
Sure, just let the feasible region describe the number of items you're going to sell, what they weigh, or how long it takes to make them.  None of those things can be negative.  With a little thought you should find dozens of examples.
840828
solve the equations of any two altitudes and make sure the solution set satisfies the equation of the third altitude
101802
I don't think there is a unique solution for it. You have more unknown variables than the equations.
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
1076564
What do you want to solve for? x? y? or do you want an ordered pair (x,y) that works (integers maybe?)
1051867
The range is:\n(-1)<x<=(1)\nx=1 is admissible
1047475
U r correct. the end points in a closed interval can be the value of c because c belongs to [-1,1].
909722
The only case that I could think of that A is not equal to B where AC=BC is when C=0.  Other than that A must equal B for the expression to be valid
30411
You first graph the inequality as if it were an equation. If it's great or less than you draw a dotted line, and if it's greater than or equal to or less than  or equal too you draw a solid line.  Then you check and see if the point (0,0) works for the inequality.  If it does, you shade the side of the line with (0,0) in it. If not, you shade the otherside.  You do the same thing with the other inequality, and then the area of numbers shaded by both inequalities is the region of the system.
1031756
It would help if you used some parentheses to indicate which terms are being divided into which terms!  I suspect that just using the order of precedence for operations will not result in the problem you are actually trying to solve.
838574
Part 2 is easy.  The function y = 1/(1+x) has an infinite discontinuity at x = -1.  Also, y(0) = 1, and for large x (positive and negative), the function approaches the x-axis as an asymptote.  For positive x, y(x) > 0, and for large negative x, y(x) < 0.  It's easy to draw the graph.\n\nIn this problem, we care about y(-3) = -1/2; y(-2) = -1; and y(1) = 1/2.  We want the parabola y = ax^2 + bx + c that passes through the points (-3, -1/2), (-2, -1), and (1, 1/2).  Since all three points must satisfy the quadratic equation, we can plug in values and write these three equations:\n\n(1)  For x = -3:  9a - 3b + c = -1/2\n(2)  For x = -2:  4a - 2b + c = -1\n(3)  For x = 1:  a + b + c = 1/2\n\nThe solution to this set of three linear equations in three unknowns is a = 1/4; b = 3/4; and c = -1/2.  (Details of the solution omitted.)\n\nThe quadratic is y = 1/4 x^2 + 3/4 x - 1/2.  This function passes through the three points we identified, and\n\n1/(1+x) < 1/4 x^2 + 3/4 x - 1/2 over the range specified in the problem.  (Actually, there's equality at three points.)\n\nThat inequality can also be written as\n\n4/(1+x) < x^2 + 3x - 2\n\nI'll answer the other part of the question as best I can in the morning.  Right now, it's time to sleep.
859022
x - 2y ≥ 4\n\nGraph x - 2y = 4 (ie y = ½x - 2 Slope = ½, y -intercept = -2)\n\nNow test the origin (0, 0)\n0 - 0 ≥ 4 is FALSE, so the origin is NOT in the region\n\nSo the requred region for this inequality is the half plane bounded by the line NOT containing the origin.\n\nLightly shade this region\n\nx ≤ 4\n\nGraph x = 4 ... a vertical line through 4 on the x-axis\nTest the origin (0, 0)\n0 ≤ 4 is TRUE so the origin is in the region\nSo the required region is the half plane bounded by the line x = 4 that contains the origin\nLightly shade this region differently to the first region\n\nThen the required condition of x - 2y ≥ 4 and x ≤ 4 is the overlap of the two above regions
363348
Are we solving for x?
908083
Tried to do it algebraically, that was a booger.\n\nI plugged that into mathematica and it gave me an error normally generated when the inverse of a function is needed to represent a solution and the solution variable occurs in more than one argument of that function.\n\nThat one might not have a solution?
110904
No. you may count yourself lucky to have these conditions because with these you are able to get an unique solution ψ.\nBut there are 4 main conditions:\n\nNorm: |ψ| := 1\n\nContinuity of ψ\n\nContinuity of ψ'    (only for finite potentials)\n\nLimit: ψ(x) -> 0 for x -> ∞
997478
system of inequalities: \n12x+20y<=800 i.e. 3x+5y<=200\nx+y<=50\nx>=10\ny>=15\n\nafter graphing it, you will find that\nx>=10\ny>=15\ny<=25
24754
5t+1 > 3t-13\n=> 5t - 3t > -13 -1\n=> 2t > -14\n=> t > -7\n\nAgain.. the following site will help you learn the concepts -\nhttp://www.quickmath.com/www02/pages/modules/inequalities/index.shtml