# Solve the inequality. graph the solution set. 26+6b (> =) 2(3b+4 a. all real numbers b. b> or

Solve the inequality. graph the solution set. 26+6b (> =) 2(3b+4 a. all real numbers b. b> or less than 1/2 c. no solutions

## This Post Has 9 Comments

1. maggiestevens5321 says:

I got c because when you solve theres no solution

2. queen6931 says:

This what I came up with
$Solve the inequality. Graph the solution set. 26 + 6b ≥ 2(3b + 4)$

3. yeontan says:

Ci think but i'm not entirely sure so..

4. sihamabdalla591 says:

The inequality is : 26 + 6b= 2(3b + 4)
On opening the bracket we get,
or 26 + 6b= 6b + 8
Now, bring the variable terms to right hand side,
or 26 + 6b - 6b = 8
or 26 = 8, which is not possible because 26 can never be equal to 8.
Thus, this inequality has no solutions.

5. 22chandlerlashley says:

Step-by-step explanation:

26+6b≥2(3b+4)

25+6b≥6b+8

25≥8

which is true as 25>8

so b can have any real value.

6. skateyo2553 says:

Distribute on the right
26+6b≥6b+8

Then subtract 6b from both sides
26≥8
True

Final All real numbers, A

7. dbenitezmontoya3 says:

B no solutions is incorrect.

8. rileyeddins1010 says:

True for all b

Interval notation; (-∞, ∞)

Step-by-step explanation:

We have been given the following inequality;

26+6b>2(3b+4)

The first step is to open the brackets on the right hand side;

26+6b>6b+8

26-8>6b-6b

18>0

Since 18 is actually greater than 0, the solution set to the inequality is;

True for all b.

$Solve the inequality. graph the solution set. 26 +6b> 2(3b+4)$

9. AmazingColor says:

we have

$26 + 6b\geq2(3b + 4)$

Applying the distributive property on the right side

$26 + 6b\geq6b+8$

subtract $6b$ from both sides

$26\geq 8$ -------> is true

for all real numbers the inequality is true

therefore

the graph is a shaded area everywhere.

the answer is

the solution is all real numbers