After you graph a system of linear inequalities, how can you use algebra to check whether the correct region-?
959832
Pick a point in the shaded region, plug that coordinate into the equations. If the result you produce is true, then you have done your job well!
916482
Sorry, but I can't graph and shade here, but if you've already done that, then you'll see that inequality 1 meets with inequalities 2 & 4 and that inequality 3 also meets with inequalities 2 & 4.\n\nSo now you can set up 4 simultaneous equations to find the corner points.\n\nLet Sim 1 be : -x + 3y = 20 and 13x + 3y = -50 (1 & 2)\nLet Sim 2 be : -x + 3y = 20 and 5x + 2y = 36 (1 & 4)\nLet Sim 3 be : 13x + 3y = -50 and -3x + 5y = -34 (2 & 3)\nLet Sim 4 be : -3x + 5y = -34 and 5x + 2y = 36 (3 & 4)\n\nSim 1 gives the point (x, y) = (-5, 5)\nSim 2 gives the point (x, y) = (4, 8)\nSim 3 gives the point (x, y) = (-2, -8)\nSim 4 gives the point (x, y) = (8, -2)
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
935919
Please calm down. Now do you want the line between (2,0) and (-2,-2) The way I do it is to get the point vector solution of the line and eliminate t. The point is (2,0). The vector is <-2 - 2, -2 -0>. Just subtract the first coordinate from the second. So the vector is <-4, -2> \nNow set up the point vector equation describing the postion of the graph at time t.\nJust point + vector*t in component form.\nThe x-components: x = 2 - 4t \nThe ys: y = 0 - 2t\n\nThere, that is it. But some nasty professors don't like this form, so we'll eliminate t. We plug y = -2t into first equation and get x = 2 + 2y.\n2y = x - 2. y = x/2 -1.
979117
From a table -- see which y-value corresponds to x=0\nFrom a graph -- see where the graph crosses the y-axis\nFrom an equation -- get the equation in y=mx+b form; the y-interecept is b
759182
x+2>9\n .-2 .-2\n____\n|x>7|\n```````\nx-18<35\n +18 +18\n_____\n|x<53|\n`````````\n 30+x<16\n-30 . . -30\n_______\n|x < -14|\n```````````\nx-9.5<9.5\n +9.5 +9.5\n_____\n|x<19|\n`````````\nAnd when you graph the inequalities, just draw a number line and draw a dot(open) on the mark that shows your number. Then, if the variable is greater than your number, draw an arrow to the right, and if the variable is less than your number, draw an arrow to the left.\nLike, x<3 is:\n . -1 . . 0 . . 1 . . .2 . . .3 . . 4 . . .5. . .6 . . .7\n< =l===l===l===l===o---- l-------l-------l-------l-->\nThe "=====" is the graphing part, and the "----" is just the original number line. Make the "===" darker so you know where you drew the line. this only work is the inequality is "greater than" or "less than". If the inequality is "greater than or equal to" or "less than or equal to", than the dot should be closed, or solid.
500255
This is called an inequality. First, find the zeros (in this case, -4 and 9). Next, make a number line:\n<------(-4)----------(9)----->\nNow, determine where inside and outside the zeros the inequality will be true or false. Finally, express your answer in terms of the inequality.\nExample: x+12 > 0\nAnswer: x > -12
859533
When the whole equation is greater than zero, graph it like normal. When it's less than zero, graph it's negative. For example:\n\ny = /x + 1/\n\nWithout the absolute value signs, this would be positive when x > -1. So, for x >= -1, graph it like normal. For x < -1, graph -x - 1. You should get a v-shape, with only positive values.\n\nIf you have values added on outside the absolute value, then you just raise and lower the graph accordingly.\n\nAnd if you're still confused... try setting the expression equal to zero, and find x. Then, since it's linear, just plug in a value both less than and greater than this x to get two more points. Graph the three points, and connect with lines to get a v-shape.\n\nHope this helps!
832106
1) 1/-4<m/-4\nmultiply both sides by negative 4 and flip the inequality sign\n4>m.\n\n2) -1/8< or equal to n/-32\nmultiply both sides by negative 32 and flip the inequality sign\n4 greater than or equal to n\n\n3) x/-12>1/4\nmultiply both sides by negative 12 and flip the inequality sign\nx<-3\n\n4) -1/2x< or equal to 2\nmultiply both sides by negative 2 and flip the inequality sign\nx> or equal to -4
816270
is this for a algebra class? calculs class? matrix methods?\n\nTry this.\n\nWe want to simplify the complicated list of 3 constraints to two simple constraints: a range for y and a range for x.\n\nThe lower bound is simple: constraint III says all x and y are greater than or equal to zero.\n\nFind a range for y. Algebra can be done on these eqations the same as with an equal sign as long as the alligator is in the correct direction (it is). So eliminate x from the equations by using, "Constraint II minus Constraint I"; or;\n\n  x+3y<_6\n-(x+ y<_2)\n\n=2y<_4....... or, y<_2. Therefore, 0<_y<_2.\n\nSimliarly for range x, eliminate y by "Three times Constraint I plus Constraint II.", or:\n\n3*(x-y<_2) =  3x- 3y<_6\n                    +( x+3y<_6)\n                    = 4x<_12.... or, x<_3. Therefore, 0<_x<_3.\n\nNow we simply have an equation of three variables with two constraints.\n\nz=3x+2y;  0<_y<_2\n                 0<_x<_3\n\nI'm not sure what you want from here, but I think I understand.\n\nThe graph will be a plane in the x,y,z coordinate system.\n\nThe maximum z is achieved by using maximum values of x and y.\n\nThe corners will be defined by the extreme values of x and y with the correseponding z.\n\nFor instance, with (x,y,z)...\n\n(0,2,z1)\n(3,0,z2)\n(0,0,z3)\n(3,2,z4)\n\nThis will create a "polygon...?" in the x,y,z plane with the corners at the locations shown above.\n\nThis might be able to be done more simply by using matrices but I can't see an obvious way.
891120
One easy way is to plug in a number for x.  From there, find numbers for y that work with that x.\n\nStart with x = 0 and find a y that works for both inequalities.\nx - 6y < 6\n-x + y < 4\n\nx - 6y < 6\nPlugging in 0 for x, we get\n-6y < 6\ny > -1\n\n-x + y < 4\nPlugging in 0 for x, we get\ny < 4\n\nWhen x is 0, y has to be greater than -1 and less than 4.\nThat gives us four solution points already.\n(0,0), (0,1), (0,2), (0,3)
973040
Ok, since you didnt give any equation, i would assume that the graphs are linear in nature\n\nThen for find the points where the graph intersects the Axes,\n\n  Substitute x=0 in the equation to find the point where the point where the graph intersects the Y axis\n\n    Substitute y=0 in the equation to find the point where the point where the graph intersects the X axis\n\nEg.   2x + y = 4\n\nSubstituting y=0,\n\n                          2x=4\n                           x=2\n\nHence the the graph meets the x axis at (x,y) that is (2,0)\n\nSimilarly, substituting for x=0, we get..\n\n                         y=4\n\nTherefore the point where hte graph meets the x axis is (x,y), in this case (0.4)
826809
It is probably best to determine the range after you have the graph.  You will need to look at the local maximums, local minimums, and the horizontal asymptotes to determine the range.  (In other words, it is not a fast problem).
538748
Yes, but you need to account for two cases. The full solution:\n\n(x-1) / (x-10) < 0\n\nx-1 < 0 [if x > 10] OR\nx-1 > 0 [if x < 10] (by "flip the sign" rule; x cannot be 10)\n\nFirst inequality can never be satisfied.\nSecond inequality means 10 > x > 1.\n\nYou do not use foil for that.\n\nClassical (and Kemmy) probably interpreted it as a number/phase line, which can be faster in a simple case like this. For more difficult inequalities it's best to write out steps.
879256
Zeroes: -3\ny-int: 2\nVA: x=1/2\nHA: y=-1/3\nThe graph looks like this: one wing opens on the right down and the other opens on the right up.
740142
You're going to want to show, on a number line, what values of x satisfy the inequality.  Because you know that (-8) and (2) are both zeros of the function, you only have to test three regions:\n1) x is below -8.\n2) x is above -8 and below 2.\n3) x is above 2.\nFor each region, if the test point gives the function a value greater than 0, shade the region, otherwise, do not shade it.\n\n(You will end up shading one region.)
411757
since y=+-(x)^(3/2) Range [0,oo[\nnP=(n,n*n^1/2) and x_n/y_n= [(n^1/2^)/n]=n^-1/2