Solve: x + 3 > 2. Write the solution set in interval notation and graph it. We will use a property of inequality to isolate the variable on one side. To solve the original inequality, we want to find a simpler equivalent inequality of the form x > number or x < number, whose solution is obvious. We can use the subtraction property of inequality to isolate x on the left side of the inequality. We can undo the addition of 3 by subtracting 3 from both sides. x + 3 > 2 This is the inequality to solve. x + 3 - 3 > 2 – 3 Subtract 3 from both sides. x > -1 All real numbers greater than -1 are solutions of x + 3 > 2. The solution set can be written in set-builder notation as {x | x > -1} and in interval notation as (-1, ∞). The graph of the solution set is shown below. -3 -2 -1 1 2 Since there are infinitely many solutions, we cannot check all of them. As an informal check, we can pick two numbers in the graph, say 0 and 30, substitute each for x in the original inequality, and see whether true statements result. Check: x + 3 > 2 x + 3 > 2 0 + 3 3 4 x 30 + -5x 3 2 Substitute 0 for x. Substitute 30 for x. -1 x True. |-5 x > 2 True.
Solve: x + 3 > 2. Write the solution set in interval notation and graph it. We will use a property of inequality to isolate the variable on one side. To solve the original inequality, we want to find a simpler equivalent inequality of the form x > number or x < number, whose solution is obvious. We can use the subtraction property of inequality to isolate x on the left side of the inequality. We can undo the addition of 3 by subtracting 3 from both sides. x + 3 > 2 This is the inequality to solve. x + 3 - 3 > 2 – 3 Subtract 3 from both sides. x > -1 All real numbers greater than -1 are solutions of x + 3 > 2. The solution set can be written in set-builder notation as {x | x > -1} and in interval notation as (-1, ∞). The graph of the solution set is shown below. -3 -2 -1 1 2 Since there are infinitely many solutions, we cannot check all of them. As an informal check, we can pick two numbers in the graph, say 0 and 30, substitute each for x in the original inequality, and see whether true statements result. Check: x + 3 > 2 x + 3 > 2 0 + 3 3 4 x 30 + -5x 3 2 Substitute 0 for x. Substitute 30 for x. -1 x True. |-5 x > 2 True.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Transcribed Image Text:Solve: x + 3 > 2. Write the solution set in interval notation and graph it.
We will use a property of inequality to isolate the variable on one side.
To solve the original inequality, we want to find a simpler equivalent inequality of
the form x > number or x < number, whose solution is obvious.
We can use the subtraction property of inequality to isolate x on the left side of the inequality. We can undo the addition of 3 by subtracting 3 from both sides.
x + 3 > 2
This is the inequality to solve.
х+ 3 —
3 > 2
- 3
Subtract 3 from both sides.
х> -1
All real numbers greater than -1 are solutions of x + 3 > 2. The solution set can be written in set-builder notation as {x | x > -1} and in interval notation as (-1, o). The
graph of the solution set is shown below.
-3
-2 -1 0 1
2
3
Since there are infinitely many solutions, we cannot check all of them.
As an informal check, we can pick two numbers in the graph, say 0 and 30, substitute each for x in the original inequality, and see whether true statements result.
Check:
X + 3
2
x + 3
0 + 3 4 X
Substitute 0 for x.
30 + |-5X 2
Substitute 30 for x.
-1X
> 2
True.
-5X
>
True.
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