How do you solve $13\leq 1+ 2b$ ?

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Answer 1 · 2017-12-18T13:31:23

See a solution process below:

First, subtract $color(red)(1)$ from each side of the inequality to isolate the $b$ term while keeping the inequality balanced:

$13 - color(red)(1) <= 1 - color(red)(1) + 2b$

$12 <= 0 + 2b$

$12 <= 2b$

Now divide each side of the inequality by $color(red)(2)$ to solve for $b$ while keeping the inequality balanced:

$12/color(red)(2) <= (2b)/color(red)(2)$

$6 <= (color(red)(cancel(color(black)(2)))b)/cancel(color(red)(2))$

$6 <= b$

We can reverse or "flip" the entire inequality to state the solution in terms of $b$ :

$b >= 6$

How do you solve 13\leq 1+ 2b13\leq 1+ 2b?