SOLVE AND GRAPH LINEAR COMPOUND INEQUALITIES
SOLVE AND GRAPH LINEAR COMPOUND INEQUALITIES
Compound inequality are formed by two or more inequalities that are joined by the words "and" or "or".
The compound "and":
The "and" compound inequality solution are values that satisfy both inequalities.
Example:
-2 < x < 5 translates into -2 < x and x < 5
3 is a solution because -2 < 3 and 3 < 5
8 is not a solution because it does not satisfy both: -2 < 8, but 8 not < 5
To solve compound "and" inequality, you solve each inequality and connect with the word "and".
To graph a compound "and" inequality:
1) graph each inequality on the same number line.
2) the final graph solution will be the portion where both graphs overlap/intersect. When giving the final graph, show clearly only the portion that overlapped/intersected.
Recall: The shading in the final graph indicates the real numbers that are the solutions of the inquality
EXAMPLES of Compound "and":
1) Solve and graph: 3x - 4 > 8x + 6 and 1/2 x > -3
solve: -5x - 4 > 6 x > -6
-5x > 10
x < -2
The algebraic solution is x < -2 and x > -6.
Graph:
x > -6 -6
x < -2 -2 0
The final graph solution is
-6 -2 0
2) Solve and graph: -6 ≤ -2(1 - x) < 6
-6 ≤ -2 + 2x < 6
-4 ≤ 2x < 8 (add 2 to all three sides)
-2 ≤ x < 4 (divide all three sides by 2)
-2 ≤ x < 4 is the algebraic solution.
The original inequality can also be written as -6 ≤ -2(1 - x) and -2(1 - x) < 6 . You would solve as done in example 1. The algebraic solution is -2 ≤ x and x < 4.
Graph:
-2 ≤ x -2
0 4
x < 4
The final graph solution is
-2 0 4
3) Solve and graph: 5x - 12 < 13 and 3x + 6 < 12
5x 5
-5 is a solution because it satisfies x < 2
3 is not a solution because it does not satisfy either inequality
To solve compound "or" inequality, you solve each part and then combine with the word "or".
To graph a compound "or" inequality:
1) graph each inequality on the same number line.
2) the final graph solution will be the union or combination of both graphs.
Recall: For the final graph solution, the shading indicates the real numbers that are the solutions to have a true statement.
EXAMPLES of Compound "OR":
1) Solve and graph: 4(x - 5) ≤ 2x - 20 or 7x + 1 ≥ 15
4x - 20 ≤ -20 7x ≥ 14
4x ≤ 0 x ≥ 2
x ≤ 0
The algebraic solution is x ≤ 0 or x ≥ 2 .
Graph:
0 2
2) Solve and graph: 2 x - 1 > 13 or 14 - 12x < 6 - 8x
3 5 15
10x + 3 > 13 14 - 4x < 6
10x > 10 -4x < -8
x > 1 x > 2
The algebraic solution is x > 1 or x > 2.
Graph:
0 1 2
3) Solve and graph: 6x + 25 ≥ 55 or 12x ................
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