For this case we have the following inequality:
[tex]6- \frac {2} {3} x <x-9[/tex]
We add[tex]\frac {2} {3} x[/tex] to both sides of the inequality:
[tex]6 <\frac {2} {3} x + x-9\\6 <\frac {2 * 1 + 3 * 1} {3} x-9\\6 <\frac {2 + 3} {3} x-9\\6 <\frac {5} {3} x-9\\[/tex]
We add 9 to both sides of the inequality:
[tex]6 + 9 <\frac {5} {3} x\\15 <\frac {5} {3} x[/tex]
We multiply by 3 on both sides of the inequality:
[tex]15 * 3 <5x\\45 <5x[/tex]
We divide between 5 on both sides of the inequality:
[tex]\frac {45} {5} <x\\9 <x[/tex]
Thus, the solution is given by all values of x greater than 9.
Answer:
[tex]x> 9[/tex] is equivalent to the given inequality.
