Answer :
Given a function [tex]f(x)[/tex], we can solve for [tex]f^{-1}(y)[/tex] for some [tex]y[/tex] by solving for [tex]x[/tex] in the equation [tex]f(x)=y[/tex].
I'm not sure whether your function [tex]f(x)[/tex] is [tex]2x-\frac{7}{3}[/tex] or [tex]\frac{2x-7}{3}[/tex], so I'll solve it separately for each.
In the first case, we have [tex]2x-\frac{7}{3}=3[/tex], so adding [tex]\frac{7}{3}[/tex] to both sides, [tex]2x=\frac{16}{3}[/tex]. Dividing by two gives [tex]x=\frac{8}{3}[/tex].
In the second, we have [tex]\frac{2x-7}{3}=3[/tex], so multiplying by three gives [tex]2x-7=9[/tex], and adding seven gives [tex]2x=16[/tex]. Dividing by two gives [tex]x=8[/tex].
I'm not sure whether your function [tex]f(x)[/tex] is [tex]2x-\frac{7}{3}[/tex] or [tex]\frac{2x-7}{3}[/tex], so I'll solve it separately for each.
In the first case, we have [tex]2x-\frac{7}{3}=3[/tex], so adding [tex]\frac{7}{3}[/tex] to both sides, [tex]2x=\frac{16}{3}[/tex]. Dividing by two gives [tex]x=\frac{8}{3}[/tex].
In the second, we have [tex]\frac{2x-7}{3}=3[/tex], so multiplying by three gives [tex]2x-7=9[/tex], and adding seven gives [tex]2x=16[/tex]. Dividing by two gives [tex]x=8[/tex].