Let's solve the inequality [tex]\(\frac{1}{2} x + 3 \geq 5 x + 2\)[/tex].
1. Write down the inequality:
[tex]\[
\frac{1}{2} x + 3 \geq 5 x + 2
\][/tex]
2. Move all terms involving [tex]\(x\)[/tex] to one side of the inequality and constant terms to the other side:
[tex]\[
\frac{1}{2} x + 3 - 5 x \geq 2
\][/tex]
3. Combine like terms:
[tex]\[
\frac{1}{2} x - 5 x + 3 \geq 2
\][/tex]
Convert [tex]\(\frac{1}{2} x\)[/tex] to a fraction with a common denominator for easier combination:
[tex]\[
\frac{1}{2} x - \frac{10}{2} x + 3 \geq 2
\][/tex]
[tex]\[
\left(\frac{1 - 10}{2}\right) x + 3 \geq 2
\][/tex]
[tex]\[
\frac{-9}{2} x + 3 \geq 2
\][/tex]
4. Isolate the term with [tex]\(x\)[/tex] by subtracting 3 from both sides:
[tex]\[
\frac{-9}{2} x + 3 - 3 \geq 2 - 3
\][/tex]
[tex]\[
\frac{-9}{2} x \geq -1
\][/tex]
5. Solve for [tex]\(x\)[/tex] by dividing both sides by [tex]\(\frac{-9}{2}\)[/tex] (note that dividing by a negative number reverses the inequality sign):
[tex]\[
x \leq \frac{-1}{-\frac{9}{2}}
\][/tex]
[tex]\[
x \leq \frac{-1 \cdot 2}{-9}
\][/tex]
[tex]\[
x \leq \frac{2}{9}
\][/tex]
So the solution to the inequality [tex]\(\frac{1}{2} x + 3 \geq 5 x + 2\)[/tex] is:
[tex]\[
x \leq \frac{2}{9}
\][/tex]
This means that any value of [tex]\( x \)[/tex] that is less than or equal to [tex]\(\frac{2}{9}\)[/tex] will satisfy the given inequality.
