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### Solving Inequalities in Phases and Shadings

```
Date: 01/19/98 at 12:38:53
From: rebeka
Subject: Solving Inequalities

Hello,

I am in the 9th grade taking Algebra I. I have a tutor but even she
doesn't understand Solving Inequalities. Here is an example of one
of these problems. x+1> -2 and 3x<6 How do you do this?
```

```
Date: 01/23/98 at 15:51:38
From: Doctor Ezick
Subject: Re: Solving Inequalities

Good Question!

There are two phases to solving a system of inequalities with one
variable.

First, you need to solve EACH equation as an inequality with x alone
on one side. For example, if you had -2x - 3 > 5, the you would first
add 3 to each side to get -2x > 8, and then divide each side by -2 to
get: x < 4. Notice that when you multiply or divide by a negative
value, you need to switch the inequality sign. (Not doing this is the
one biggest mistake that people make on these problems.)

The second step, once you have both of the equations in this form,
is to graph them on a number line.

Take a number line and draw a circle on the value in the inequality
(solid for <= or >= and open for < and >) (here <= means less than or
equal to, and >= means greater than or equal to), and shade in one
direction, representing the values that satisfy the inequality. The
shading acts as a way to denote solutions to the inequality. Just pick
one value on the number line which is not the value where the circle
is, and test that. If it satisfies the inequality, then shade in that
direction; otherwise shade the other way.

A good trick here is that if x is on the left, the inequality acts
like an arrow and shows you how to shade. For example, if x < 4, then
draw an open circle on 4 and shade to the left. After you plot BOTH
equations, the solution is the area of the number line that is shaded
for BOTH equations. This corresponds to the value that satisfies both
inequalities.

When you shade for both equations, one of three things can happen.

1. There might be no solution. If the two equations reduce to, for
example, x < 3 and x > 5, then clearly there is no solution since
there is no value that is both less than 3 and more than 5.

2. One equation might be redundant. This happens if both equations
result in shading the same way. For example, if x < 3 and x < 6 then
the solution is just x < 3, since anything that is less than 3 is also
less than 6.

3. The solutions might form a range. For example, x < 6 and x >= 3.
Here the solution would be 3 <= x < 6. You need to be careful here to
preserve the = in <= and >= - remember that solid dots are part of the
solution, but open ones are not.

Now to do the problem that you posed at the beginning.

1. x + 1 > 2  ====>  x > 1 :  by subtracting 1 from both sides.
2. 3x < 6     ====>  x < 2 :  dividing both sides by 3.

Now shade both on a number line.

(x > 1)    O----------------->
<-----------------O   (x < 2)
|---|
O---O  Solution:  (1 < x < 2)
<---+---+---+---+---+---+---+---+--->
-2  -1  0   1   2   3   4   5

As you can see, this is an instance of the third case above.

Hope this helps,

-Doctor Ezick,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra

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