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Systems of Linear Equalities
By L.D.
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Problem 1
y > - 2x + 1
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Problem 1
y > - 2x + 1Now lots of people may not like graphing
inequalities like this, so an alternate way to
see this is y = -2x + 1, basically we are just
graphing an inequality in the y = mx + bmethod, m is the slope and b is the y
intercept. So lets graph it using the alternate
way first on the other page. The slope is red
and the y intercept is blue.
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Easy Right
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To make this an
inequality wemust look at our
original problem
y > - 2x + 1
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dMini Lesson
So, to tell the difference between > and >
when doing and inequality what you do is
you think that since > has more ink underit, it gets a full line like ______ and since
> has less ink under it, it can only afford a
dotted line _ _ _ _.
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After that short
lesson, we nowknow that the
line shouldnt be
solid since the
problem is
y > - 2x + 1. And
that inequality
sign doesnt have
enough ink.
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After that short
lesson, we nowknow that the
line shouldnt be
solid since the
problem is
y > - 2x + 1. And
that inequality
sign doesnt have
enough ink.
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Now we need to
bring in the
inequality part.
Since we dont know
what y is iny > - 2x + 1 we need
to fill up the empty
space on one side.
We choose the sideto shade based on if
the y is more or less
than the problem, if
it is more then we
shade the top, if it is
less, than we shade
the bottom. In this
case we shade
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THE TOP, THATS
ALL YOU NEEDSINCE THIS IS THE
FINAL ANSWER
(for problem 1)
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Problem 2
x 4y > -8
y 2x < -1
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Problem 2
x 4y > -8
y 2x < -1
So the first thing we
need to do is reorder
these so they fit the
y = mx + b format.
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Problem 2
x 4y > -8
y 2x < -1
(x 4y > -8) 4
x/4 y > -2
+y +yx/4 > -2 + y
+2 +2
x/4 + 2 > y
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Problem 2
x/4 + 2 > y
y 2x < -1
y 2x < -1
+2x +2x
y < -1 2x or -1 2x > y
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Problem 2
x/4 + 2 > y
-1 2x > y
Next we need to graph
them, just graph them,
dont shade yet. I will
use green for the topand purple for the
bottom.
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Now try shading
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The answer to
our problem is
the area theyboth shade over.
To check the
problem you can
choose a pointthey both shade
over, I choose (-
3,0), and then
you cansubstitute it into
both problems
individually.
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Problem 2
(-3)/4 + 2 > (0)
1 > 0
-1 2(-3) > (0)
-1 + 6 > 0
5 > 0
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Problem 2
(-3)/4 + 2 > (0)
1 > 0
-1 2(-3) > (0)
-1 + 6 > 0
5 > 0
Since both formulas
turn out correct, we cannow assume my
graphing was correct.
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Problem 3
x > 0
y > 0
y > 2x + 3
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Problem 3
If you are currently wondering how to deal with
the 0s, wonder no longer. Treat the zero like if
you had seen something like 5 > y or 6 < x. For
the former you would draw a horizontal lineacross the 5 on the y axis and for the latter you
would draw a vertical line across the 6 on the x
axis. Its the same with the 0s, you draw ahorizontal or vertical line for them. So now try to
graph the lines with shading.
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x > 0 (red dots)
y > 0 (green dots)y > 2x + 3 (blue dots)
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Problem 4
y > -5
y < 3
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Problem 4
This is the last one, so give it a try graphing-
wise. Remember to shade and check!
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