Given:
The inequality is:
[tex]6x+\dfrac{1}{4}(4x+3)>12[/tex]
To find:
The values for x.
Solution:
We have,
[tex]6x+\dfrac{1}{4}(4x+3)>12[/tex]
Using distributive property, we get
[tex]6x+\dfrac{1}{4}(4x)+\dfrac{1}{4}(3)>12[/tex]
[tex]6x+x+\dfrac{3}{4}>12[/tex]
[tex]7x>12-\dfrac{3}{4}[/tex]
[tex]7x>\dfrac{48-3}{4}[/tex]
On further simplification, we get
[tex]7x>\dfrac{45}{4}[/tex]
Divide both sides by 7.
[tex]x>\dfrac{45}{7\times 4}[/tex]
[tex]x>\dfrac{45}{28}[/tex]
Therefore, the required solution is [tex]x>\dfrac{45}{28}[/tex].
Note: All options are incorrect.
