A1Side
Solving Basic Linear Inequalities
(single variable)

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statement
If you can solve a linear equation, you can solve a linear inequality. The process is the same, with one exception ...

... when you multiply (or divide) an inequality by a negative value,
you must change the direction of the inequality.

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ex1

Solve and graph the solution set of:   x - 3 > -2
Solve as you would a linear equation
(add 3 to both sides of the statement).
The solution set for this problem will be all values that satisfy x > 1.

Graph using an open circle for x =1 (since x can not equal 1), and an arrow pointing to the right.

x - 3 > -2
x > 1


Note: When you finish solving an inequality, you may want to CHECK a couple of values to see if the number line graph is correct. In Example 1, choose 5 from the number line and see if it solves the original inequality. It does (5 - 3 > -2),
It is important to check back to the "original" inequality, just in case you make an algebraic error.
You can also check values to the left of 1 on the number line to see that they do NOT solve the original inequality.

ex1

Solve and graph the solution set of:   x + 5 < 8
Solve as you would a linear equation
(subtract 5 from both sides of the statement).
The solution set for this problem will be all values that satisfy x < 3.

Graph using a closed circle for x = 3 (since x can not equal 3)., and an arrow pointing to the left.

x + 5 < 8
x < 3

ex1

Solve and graph the solution set of:   4x < 24
Again, proceed as you would when solving a linear equation:
Divide both sides by 4.

Note: The direction of the inequality stays the same since we did NOT divide by a negative value.

Graph using an open circle for 6 (since x can not equal 6) and an arrow to the left (since our symbol is less than).

iqmath7

circleg3

ex2 beware

Solve and graph the solution set of:   -5x greaterequala 25
Divide both sides by -5.

Note: The direction of the inequality was reversed since we divided by a negative value.

Graph using a closed circle for -5 (since x can equal -5) and an arrow to the left (since our final symbol is less than or equal to).

iqmath8

circlegraph8a

ex3

Solve and graph the solution set of:   3x + 4 > 13
Proceed as you would when solving a linear equation:
Subtract 4 from both sides.
Then, divide both sides by 3.

Note: The direction of the inequality stays the same since we did NOT divide by a negative value.

Graph using an open circle for 3 (since x can not equal 3) and an arrow to the right (since our symbol is greater than).

iqmath1

circleg3

ex3 beware

Solve and graph the solution set of:   9 - 2x lessequal 3
Subtract 9 from both sides.
Then, divide both sides by -2.

Note: The direction of the inequality was reversed since we divided by a negative value.

Graph using a closed circle for 3 (since x can equal 3) and an arrow to the right (since our symbol is greater than or equal to).

iqmath4

circleg4

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For the following problems, alternate forms of expressing the solutions are included.
For more information on "alternate notations", see Notations for Solutions.

ex3

Solve and graph the solution set of:   qmath3a
Proceed as you would when solving a linear equation with a fraction:
Add 6 to both sides.
Multiply both sides by threehalfs.


Note:
The direction of the inequality stays the same since we did NOT multiply by a negative value.


Graph using a closed circle for 6 (since x can equal 6) and an arrow to the left (since our symbol is less than or equal to).

 

iqmath5a

circlegraph5

Other notations:
ex1notation

ex2

Solve and graph the solution set of:   5(x - 3) > 10
Distribute (remove the parentheses).

Add 15 to both sides.

Divide both sides by 5.

Note: The direction of the inequality stays the same since we did NOT multiply by a negative value.

Graph using a open circle for 5 (since x can NOT equal 5) and an arrow to the right (since our symbol is greater than).

1qmath9

circlegraph11

Other notations:
ex1notation



ex3

Solve and graph the solution set of:    4(x - 1) > 3(x - 2)
Distribute across both sets of parentheses.

Subtract 3x from both sides.
The solution is easier if you move the smaller x value.

Add 4 to both sides.

Note: The direction of the inequality stays the same since we did NOT multiply or divide by a negative value.

Graph using an open circle for -2 (since x can not equal -2) and an arrow to the right (since our symbol is greater than).

iqmath6

circlegraph6

Other notations:
ex1notation


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