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Solve the following inequality for [tex]r[/tex]. Write your answer in simplest form.

[tex]-10r - 3 \geq 10r + 6[/tex]

Answer :

We start with the inequality
[tex]$$
-10r - 3 \geq 10r + 6.
$$[/tex]

Step 1: Add [tex]$10r$[/tex] to both sides.

Adding [tex]$10r$[/tex] eliminates the [tex]$r$[/tex] term on the left-hand side:
[tex]$$
-10r - 3 + 10r \geq 10r + 6 + 10r,
$$[/tex]
which simplifies to
[tex]$$
-3 \geq 20r + 6.
$$[/tex]

Step 2: Subtract [tex]$6$[/tex] from both sides.

Subtracting [tex]$6$[/tex] from both sides gives:
[tex]$$
-3 - 6 \geq 20r + 6 - 6,
$$[/tex]
resulting in:
[tex]$$
-9 \geq 20r.
$$[/tex]

Step 3: Divide both sides by [tex]$20$[/tex].

Since [tex]$20$[/tex] is a positive number, the direction of the inequality remains the same:
[tex]$$
\frac{-9}{20} \geq r.
$$[/tex]
This is equivalent to writing:
[tex]$$
r \leq \frac{-9}{20}.
$$[/tex]

Thus, the solution to the inequality is:
[tex]$$
\boxed{r \leq -\frac{9}{20}}.
$$[/tex]

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