SOLVING INEQUALITIES

INEQUALITIES

BASIC RULE OF INEQUALITIES

Division or multiplication by a negative,

FLIPS THE INEQUALITY.

So if 3x > 18 then x < 6.

1) LINEAR INEQUALITIES:

Solve like linear equations but follow the basic rule.

ex: 3x 2 < 5x + 10 becomes 2x < 12 becomes x > 6

2) QUADRATIC INEQUALITIES:

Locate the zeros by factoring or with the quadratic formula,
then use a number line or a graph to find the intervals that satisfy the inequality.

ex: x2 > 3x +10 becomes x2 + 3x 10 > 0 becomes (x + 5)(x 2) > 0
the zeros are
5 and +2

It's a parabola opening upwards, changing from > 0 to < 0 at 5 and +2.

since we want x2 + 3x 10 > 0, we need the x's that make y positive.
This means the curve must be above the x-axis,
so the solution is
x < 5 and x > 2.

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3) FRACTIONAL INEQUALITIES (constant denominators):

Multiply though by the lcm of the denominators and solve like a linear inequality. Most students call this cross-multiplication because that's what it looks like. Really, we're multiplying the entire equation through by the lcm of the denominators.

example: the lcm here is 6.

4) FRACTIONAL INEQUALITIES (variables in denominators)

Relate to zero !! Find the common denominator and combine to a single fraction. Then use a number line like in quadratic inequalities. Mark the zeros then check the function's sign in each of the intervals defined by those zeros to see if it is positive or negative.

example: Solve for x:

We mark these values on a number line to define the intervals where the
y-values of the function are not equal to zero,
where they are either positive (> 0) or negative (< 0).

We have four intervals to check:

x < 3, 3 < x < 1/5, 1/5 < x < 5, x > 5.
We test them to see where the fraction is positive since we want it > 0.

We set x = 4, x = 0, x = 2, and x = 6.

We find that the fraction is positive ( > 0) when 3 < x < 1/5 and when x > 5.

5) ABSOLUTE VALUE INEQUALITIES:

a) less than absolute value inequality:

| 2x 3 | < 5 means that 2x 3 is sandwiched between 5 and + 5, since
that interval includes all the numbers with an absolute value smaller than 5.

Therefore, we solve an absolute value, "less than" relation by sandwiching the expression between the negative and positive values of the constant.

If | thing | < a then a < thing < a.

Let's finish the example above:

| 2x 3 | < 5 becomes 5 < 2x 3 < 5 becomes 2 < 2x < 8

so – 1 < x < 4,

the solution is an interval denoted (1, 4) or ] 1, 4 [

b) greater than absolute value inequality:

We set the expression either less than the negative value of the constant (left tail)
or greater than the positive value of the constant (right tail).

Example: | 2x 3 | > 5.

Here, 2x 3 would have to be less than 5 or bigger than 5 as shown.

So| 2x 3 | > 5 we break into two separate inequalities like this:

| 2x 3 | > 5 becomes 2x 3 < 5 or 2x 3 > 5

Note: the solution is two separate intervals, x's less than 1, or x's greater than 4.

If | thing | > a then thing < a or thing > a

Less than: sandwich thing between a and a. The solution is a single interval.

Greater than: put thing in the tails, solve 2 inequalities. The solution is 2 separate intervals.

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Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s), then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again.

PRACTICE EXERCISE

Solve these inequalities:

1) | 2x 3 | < 7 2) | 3x + 6 | > 9 3) | 5 2x | < 3 4) | 4 x | > 5
5) x² 3x > 18 6) 7)
8) 9) 10)

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SOLUTIONS

1) 7 < 2x 3 < 7

so 2 < x < 5

2) 3x + 6 < 9 or 3x + 6 > 9

x < 5, or x > 1

3) 3 < 5 2x < 3
8 < 2x < 2

1 < x < 4

4) 4 x < 5 or 4 x > 5
x < 9 or x > 1

x > 9, or x < 1

5) x 2 3x 18 > 0
(
x + 3)(x 6) > 0

x < 3, or x > 6

6) zeros at 5/2, 2, and 4
we need > 0

x < 5/2, or 2 < x < 4

7) multiply by 12,
8
x + 24 < 3x 12
we need < 0

x < (36/5)

8) zeros at 5, 2, and 4
we need < 0

5 < x < 2, or x > 4

9) note 2x + 6 = 2(x + 3)
we need > 0

x > 3

  10) zeros at 5, 0, 4, and 7

x < 5, or 0 < x < 4, or x > 7

 

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