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Solving Inequalities in Phases and Shadings

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


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 

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 

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
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Associated Topics:
High School Basic Algebra

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