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.
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