To solve the inequality [tex]\(\frac{5 - x}{2x + 1} \geq 3\)[/tex], we need to follow a series of steps to determine the values of [tex]\(x\)[/tex] that satisfy this condition.
1. Rewrite the inequality:
[tex]\[
\frac{5 - x}{2x + 1} \geq 3
\][/tex]
2. Move everything to one side of the inequality:
[tex]\[
\frac{5 - x}{2x + 1} - 3 \geq 0
\][/tex]
3. Combine the terms on the left-hand side by using a common denominator:
[tex]\[
\frac{5 - x - 3(2x + 1)}{2x + 1} \geq 0
\][/tex]
4. Simplify the numerator:
[tex]\[
5 - x - 6x - 3 = 5 - x - 6x - 3 = 2 - 7x
\][/tex]
Thus, the inequality becomes:
[tex]\[
\frac{2 - 7x}{2x + 1} \geq 0
\][/tex]
5. Determine the critical points by setting the numerator and denominator to zero:
- Numerator: [tex]\(2 - 7x = 0 \Rightarrow x = \frac{2}{7}\)[/tex]
- Denominator: [tex]\(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)[/tex]
6. Create a number line and test intervals around the critical points [tex]\(x = -\frac{1}{2}\)[/tex] and [tex]\(x = \frac{2}{7}\)[/tex]:
1. Test an [tex]\(x\)[/tex] value in [tex]\((- \infty, -\frac{1}{2})\)[/tex]:
- Let [tex]\(x = -1\)[/tex]: [tex]\(\frac{2 - 7(-1)}{2(-1) + 1} = \frac{2 + 7}{-2 + 1} = \frac{9}{-1} = -9 \)[/tex], which is not [tex]\(\geq 0\)[/tex].
2. Test an [tex]\(x\)[/tex] value in [tex]\((- \frac{1}{2}, \frac{2}{7})\)[/tex]:
- Let [tex]\(x = 0\)[/tex] : [tex]\(\frac{2 - 7(0)}{2(0) + 1} = \frac{2}{1} = 2\)[/tex], which is [tex]\(\geq 0\)[/tex].
3. Test an [tex]\(x\)[/tex] value in [tex]\((\frac{2}{7}, \infty)\)[/tex]:
- Let [tex]\(x = 1\)[/tex]: [tex]\(\frac{2 - 7(1)}{2(1) + 1} = \frac{2 - 7}{2 + 1} = \frac{-5}{3} = -\frac{5}{3}\)[/tex], which is not [tex]\(\geq 0\)[/tex].
7. Test the critical points themselves:
- At [tex]\(x = -\frac{1}{2}\)[/tex]: The denominator is zero, leading to an undefined expression/division by zero. Therefore, [tex]\(x = -\frac{1}{2}\)[/tex] cannot be included.
- At [tex]\(x = \frac{2}{7}\)[/tex]:
- Substitute [tex]\(x = \frac{2}{7}\)[/tex] into the simplified inequality:
[tex]\[
\frac{2 - 7 \left(\frac{2}{7}\right)}{2 \left(\frac{2}{7}\right) + 1} = \frac{2 - 2}{\frac{4}{7} + 1} = \frac{0}{\frac{11}{7}} = 0 \geq 0
\][/tex]
- Since the expression equals 0, [tex]\(x = \frac{2}{7}\)[/tex] satisfies the inequality.
8. Combine the results:
The values of [tex]\(x\)[/tex] satisfying the inequality [tex]\(\frac{2 - 7x}{2x + 1} \geq 0\)[/tex] are [tex]\(x\)[/tex] in the interval [tex]\( \left(-\frac{1}{2}, \frac{2}{7}\right] \)[/tex].
Thus, the solution to the inequality [tex]\(\frac{5 - x}{2x + 1} \geq 3\)[/tex] is:
[tex]\[
\boxed{\left(-\frac{1}{2}, \frac{2}{7}\right]}
\][/tex]
