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stiv31 [10]
3 years ago
9

Solve the compound inequality 6b < 36 or 2b + 12 > 6.

Mathematics
2 answers:
ss7ja [257]3 years ago
5 0

Answer:

all values of b

Step-by-step explanation:

6b < 36 or 2b + 12 > 6.

First solve the one on the left

6b < 36

Divide by 6

6b/6 < 36/6

b  <6

Then solve the one on the right

2b + 12 > 6

Subtract 12 from each side

2b+12-12 >6-12

2b >-6

Divide by 2

2b/2 >-6/2

b >-3

b<6 or b >-3

Rewriting

b>-3  or b<6

b > -3 is an open circle at -3 with a line going to the right

b < 6 is an open circle at 6 with a line going to the left

The or means we add the lines together

We have a line going from negative infinity to infinity

all values of b

Murrr4er [49]3 years ago
4 0

Answer:

All real numbers  b ∈ (-\infty, \infty)

Step-by-step explanation:

First we solve the following inequality

6b < 36

Divide by 6 both sides of the inequality

b

The set of solutions is:

(-\infty, 6)

Now we solve the following inequality

2b + 12 > 6

Subtract 12 on both sides of the inequality

2b + 12-12 > 6-12

2b> -6

Divide by 2 on both sides of the inequality

\frac{2}{2}b> -\frac{6}{2}

b> -3

The set of solutions is:

(-3, \infty)

<em>Finally, the set of solutions for composite inequality is:</em>

(-\infty, 6)  ∪ (-3, \infty)

This is: All real numbers (-\infty, \infty)

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