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3x+y<=5 it says graph the inequality
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Start with a point like (1,2). This is on the included border of trueness.
Now it is linear, because it is in the form of a line.
x and y are both positive on one side of the equation, so they have to be less than 5.
So all points leftward and downward from (1,2) are true.
Also you can draw the borderline, which you know has a slope of either -1/3 or -3.
Just figure it out by trying to translate (1,2) by (-1,-3) or (-3,-1) and see which one is true.
(1-1,2-3) is (0,-1), and 3(0)-1 <= 5 is TRUE, but not on the borderline of equality.
Now (1-3,2-1) is (-2, 1), and 3(-2) + 1 <=5 is TRUE, but not on the borderline of equality.
So I made a mistake, you find it...
...Oh I see my mistake, oh dumb...
A negative slope has one axis translation negative and the other positive, my mistake!!
Let's try again. By the way the slope is -3, just subtract 3x from both sides and you
can see the y = mx + b form where y is less or equal to the other side.
Anyway, take the (1,2) and translate with (+3,-1).
(1 + 3, 2 - 1) is (4,1) and this is wrong since 3(4) + 1 <= 5 is not true.
So try the other reciprocal slope (which is a mirror image on 45 degrees), or
do the translation (1,2) by adding (+1, -3) and get (1 + 1, 2 - 3) is (2, -1) and
plug (2,-1) in and get 3(2) - 1 <= 5, which is TRUE, yeah!!!
So now you have another point on the true-included border, which slopes right 1 and down 3.
So sloping right 1 and down 3 is very steep, actually
Last edited by John E. Franklin (2007-02-10 07:39:37)
igloo myrtilles fourmis
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yep you did get me throughly confused and then unconfused me sure am glad you caught that as I would have probably stayed confused.. Thanks
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