How can I solve this linear transformation? If given $T_1(x,y) = (2x + y, 4x+ 15y)$ How would I solve $T_1(x,y) = (-4,-1)$ for x and y?. 
I tried forming a matrix and finding the RREF to get the solution, but the numbers I got did not make sense, and I am not sure how to solve it.
 A: $$2x+y = -4\\ 
4x+15y=-1$$
Solving using elimination gives $(x,y)=(-59/26,7/13)$
A: "I tried forming a matrix and finding the RREF to get the solution, but the numbers I got did not make sense, and I am not sure how to solve it."
It should have worked.
$(2,1,-4)$ and $(4,15, -1)$ should give us:
$(4,2,-8)$ and $(4, 15, -1)$ should give us:
$(0, 13, 7)$ and $(4, 15, -1)$ should give us:
$(0, 1, \frac 7{13})$ and $(\frac 4{15}, 1, -\frac 1{15})$ should give us
$(0, 1, \frac 7{13})$ and $(\frac 4{15}, 0, -\frac 1{15}-\frac 7{13}$ should give us:
$(0, 1, \frac 7{13})$ and $(1, 0 , (-\frac 1{15}-\frac 7{13})\frac {15}4)$ which means 
$y = \frac 7{13}$ and $x =  (-\frac 1{15}-\frac 7{13})\frac {15}4=-\frac 14 -\frac {7*15}{4*13}= -\frac {13+7*15}{4*13} = -\frac {105+13}{52}=\frac {118}{52}=-\frac {59}{26}$.
A: If it's a linear transformation, you would want to turn it into a matrix then interpret it from there.
Like so.
$\displaystyle \begin{pmatrix} 2 & 1\\ 4 & 15 \end{pmatrix}\begin{pmatrix}  x \\ y \end{pmatrix} =\begin{pmatrix}  -4 \\ -1 \end{pmatrix}$
Then Cramer's rule can take over from there. But you can still go easy with this one. The function is well defined anyway so:
$2x+y=-4$
$4x+15y=-1$
$4x+2y=-8$
$13y=7$
$y=\dfrac{7}{13}$
$x=\dfrac{-4-\dfrac{7}{13}}{2}$
$=\dfrac{-59}{26}$
I think that should be it. But just because you mentioned linear algebra you can check out what I was saying https://youtu.be/jBsC34PxzoM there
