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SUppose h(4z+3)=(5+2z)/(4-z)
h(z)=?
h^-1(4)=?

if u=4z+3
then z=(u-3)/4
see if you can use this to write h(4z+3)=(5+2z)/(4-z) in terms of u

im getting (-4+8u)/(28-4u)

you could distribute on top and bottom
and then combine like terms

but anyways you can find h(z) just by now replacing u with z

the top part wouldnt be 8(u-3)?

how did you get that

oh wait nevermind unless it was 2+(u-3/4), it will remain a 2 i see what i did wrong

sooo final answer=
(14+2u)/(19-u)?

that is what i have for h(u)
so this means
\[h(z)=\frac{14+2z}{19-z}\]

\[h^{-1}(4)=z \\ h(z)=4 \text{ so solve } 4=\frac{14+2z}{19-z} \text{ for } z\]

quick question why is u replaced by z?

because we wanted to find h(z)

as for the next question why is h^-1(4)=z

oh, im starting to see a substitution pattern here

im still a bit confused on where u got 2x+14/19-x, is that just the inverse?

\[\text{ remember we got } h(z)=\frac{2z+14}{19-z}\]

so we have
\[h^{-1}(4)=a \text{ which means } h(a)=4\]

what I'm saying is I don't see a way around it

but that doesn't mean there isn't a way around it

it doesn't take too long with your method, ill stick with it. anyhow, the answer is 10 (1/3)?

you know if you wanted to skip finding h(z)

i'll look into that later. thanks for your help.

np

by the way...