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

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  1. 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. 

  2. 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.

  3. 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]

  4. 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

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