SOLVE AND GRAPH LINEAR INEQUALITIES IN ONE VARIABLE
Linear inequalities will have a <, ≤, >, or ≥ symbol in the equation.
To solve linear inequalities is similar to solving linear equations. Your solution will be the
values for the variable in which the inequality will be a true statement.
KEEP IN MIND:
1) The object in solving an inequality is to isolate the variable to one side with a coefficient of
positive one.
2) If you are bringing a term to the opposite side of the inequality symbol, then do the
opposite operation.
You will follow the same steps as you would when solving an equation.
The exception is the following:
When applying the multiplication/division property, if you multiply or divide both sides of
the inequality by a negative number, then reverse the direction of the inequality symbol.
1)
2)
2)
3)
4)
5)
6)
Simplify any grouping symbols.
If you would like to eliminate fractions, then multiply each term by the LCD.
Combine like terms.
Bring variables to one side of inequality symbol.
Apply the addition property.
Apply the multiplication/division property. Reverse inequality, if necessary.
Check your solution in the original equation.
EXAMPLES:
1) Solve 3(x - 5) < 4 - (2 - 2x).
3x - 15 < 4 - 2 + 2x
3x - 15 < 2 + 2x
x - 15 < 2
x < 17 is the solution.
NOTE:
All real numbers less than 17 is the solution. There is no one value. In most inequalities, the
solution may be infinite.
2) Solve 5x - 12 ≥ 7x + 4.
-2x - 12 ≥ 4
-2x ≥ 16
x ≤ -8 is the solution.
dividing both sides by -2 changed the
the direction of the inequality.
The solution, x ≤ -8, is given as an algebraic expression. However, for inequalities, the solution
is also beneficial as a graph.
Graph of Linear Inequalities in One Variable
The solution to the linear inequality will be graphed on a number line. There are two pieces to
graphing the linear inequality.
I. Boundary Point
(for example, -8 would be considered the boundary point in the expression, x ≤ -8)
if the inequality is < or >, then there will be an open circle at the boundary point.
if the inequality is ≤ or ≥, there will be a closed/shaded circle at the boundary point.
If the boundary point has an open circle, then that indicates the boundary point value is not part
of the solution.
If the boundary point has a closed/shaded circle, then that indicates the boundary point value is
part of the solution.
II. Shading
The shaded part of the graph are the solutions to the original inequality. If you choose a real
number within the shaded part and substitute into the original inequality, then you will get a true
statement.
if the inequality is x < or x ≤, then you shade to the left of the boundary point.
if the inequality is x > or x ≥, then you shade to the right of the boundary point.
NOTE:
Read the variable to know the direction in which to shade.
For example, x ≤ -8 is read as x less than or equal to -8 and would be shaded to the left and 4 ≥ x
is read as x less than or equal to 4 and would be shaded to the left.
EXAMPLES:
The red shaded part of the number line will represent the solution in the graph.
1) Solve and graph 5x - 22 < 8.
5x < 30
x < 6 is the algebraic solution.
Graph:
<--------|----------------------|-----------------------()------------->
-6
0
6
2) Solve and graph 2x + 1 ≤ 5x - 2 .
3
3
3
2x + 1 ≤ 5x - 6
(multiply by LCD)
1 ≤ 3x - 6
7 ≤ 3x
7/3 ≤ x is the algebraic solution.
Graph:
<--------|----------------------|-------------|----(|)--------|------------>
-6
0
2 7/3
3
NOTE:
1) 7/3 = 2 1/3 which is between 2 and 3. When plotting rational numbers, you should
indicate between which two integers it falls.
2) The shading is to the right because the solution when read is x is greater than or equal to
7/3.
3) Solve and graph
(x + 3)(x - 2) - x ≤ (x - 4)(x - 1).
x2 + x - 6 - x ≤ x2 - 5x + 4
(multiply binomials)
2
2
x - 6 ≤ x - 5x + 4
- 6 ≤ - 5x + 4
(subtract x2 on both sides)
-10 ≤ - 5x
2 ≥ x is the algebraic solution.
(divide both sides by
neg # then reverse inequality)
Graph:
<----------------|----------------|--------------()-------------->
-2
0
2