Find the linear-to-linear function whose graph passes through the given three points Find the linear-to-linear function whose graph passes through the points 
$(1, 1)$, $(4, 2)$ and $(30, 3)$.
So by using the 
$$f(x)=\frac{ax +b}{x+d}$$ 
I got my final answer to be
$$f(x)=\frac{\frac{75}{23}x + \frac{64}{23}}{x + \frac{12}{23}}$$
and it is wrong.
I have to do a bunch of problems like this and I can't seem to figure it out. I am off somewhere because I did
$f(4)=(4a+b)/(4+d)= 2$
$f(30)= (30a+b)/(30+d) = 3$
and $f(1)= (1a+b)/(1 +d) = 1$
then went on the cancel out $d$ and get $a$ and $b$ then find $c$. 

So we know that $f(1) = 1$, $f(4) = 2$, and  $f(30) = 3$.
Since $f(x)$ is a linear to linear function, we know that:
$f(x) = (ax + b)/(x + c)$
Substituting, we have:
$f(1) = (a + b)/(1 + c)$ = 1, or $a + b = 1 + c$
$f(4) = (4a + b)/(4 + c)$ = 2, or $4a + b = 8 +2c$
$f(30) = (30a + b)/(30 + c)$ = 3, or $30a + b = 90 +3c$
Writing these equations so the variables are on the left gives us:
$a + b - c = 1$
$4a + b - 2c =8$
$30a + b - 3c = 90$
You can solve this system however you need to, I used this site.
Solving gives us $a = 75/23$, $b = 12/23$, and $c = 64/23$.
Thus the final result is:
$f(x) = (75/23x + 12/23)/(x + 64/23)$
It could be that you accidentally switched around some of the variables while doing calculations.
