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

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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

I got c because when you solve theres no solution

This what I came up with

[tex]Solve the inequality. Graph the solution set. 26 + 6b ≥ 2(3b + 4)[/tex]

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

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.

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.

Distribute on the right

26+6b≥6b+8

Then subtract 6b from both sides

26≥8

True

Final All real numbers, A

B no solutions is incorrect.

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.

[tex]Solve the inequality. graph the solution set. 26 +6b> 2(3b+4)[/tex]

we have

[tex]26 + 6b\geq2(3b + 4)[/tex]

Applying the distributive property on the right side

[tex]26 + 6b\geq6b+8[/tex]

subtract [tex]6b[/tex] from both sides

[tex]26\geq 8[/tex] -------> 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