Question $4b + 18 \le -12b - 14 \le 14 - 5b$

Studdy Solution

STEP 1

What is this asking? We need to find the values of $b$ that satisfy both inequalities at the same time! Watch out! Don't forget to flip the inequality sign when multiplying or dividing by a negative number.
Also, keep track of both inequalities as you solve!

STEP 2

1. Solve the left inequality.
2. Solve the right inequality.
3. Combine the solutions.

STEP 3

We start with $4b + 18 \le -12b - 14$ .
Let's get those $b$ terms together!

STEP 4

Adding $12b$ to both sides gives us $4b + 12b + 18 \le -12b + 12b - 14$ , which simplifies to $16b + 18 \le -14$ .
We added $12b$ to both sides to eliminate the $-12b$ term on the right.

STEP 5

Subtracting $18$ from both sides gives us $16b + 18 - 18 \le -14 - 18$ , simplifying to $16b \le -32$ .
We're subtracting $18$ to isolate the term with $b$ .

STEP 6

Dividing both sides by $16$ gives us $\frac{16b}{16} \le \frac{-32}{16}$ , which simplifies to $b \le -2$ .
Awesome! We divided by $16$ to solve for $b$ .

STEP 7

Now let's tackle $-12b - 14 \le 14 - 5b$ .
Let's wrangle those $b$ terms again!

STEP 8

Adding $12b$ to both sides gives us $-12b + 12b - 14 \le 14 - 5b + 12b$ , which simplifies to $-14 \le 14 + 7b$ .
Adding $12b$ helps us move the $b$ terms to one side.

STEP 9

Subtracting $14$ from both sides gives us $-14 - 14 \le 14 - 14 + 7b$ , simplifying to $-28 \le 7b$ .
Subtracting $14$ isolates the term with $b$ .

STEP 10

Dividing both sides by $7$ gives us $\frac{-28}{7} \le \frac{7b}{7}$ , which simplifies to $-4 \le b$ .
We divided by $7$ to finally isolate $b$ .
We can also write this as $b \ge -4$ .

STEP 11

We found $b \le -2$ and $b \ge -4$ .
Now, we need to find the values of $b$ that satisfy both of these conditions.

STEP 12

Think of these inequalities on a number line!
The first inequality, $b \le -2$ , includes all numbers to the left of and including $-2$ .
The second inequality, $b \ge -4$ , includes all numbers to the right of and including $-4$ .
The overlap is where both are true, which is between $-4$ and $-2$ , inclusive!

STEP 13

So, the solution is $-4 \le b \le -2$ .
All the numbers between $-4$ and $-2$ , including $-4$ and $-2$ , will make both of our original inequalities true!

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