[tex]tx+(2t+1)y=t+1\\
(t-1)x+(2t)y=t(t+1)[/tex]
So our coefficient matrix is given by:
[tex] \left[\begin{array}{cc}t&(2t+1)\\ (t-1)& 2t\end{array}\right][/tex]
Taking the determinant of our coefficient matrix:
[tex]D=\left|\begin{array}{cc}t&(2t+1)\\ (t-1)& 2t\end{array}\right|=t(2t)-(2t+1)(t-1)=t+1[/tex]
If we take the determinant of our coefficient matrix with x-column replaced by the answer column values:
[tex]D_x= \left|\begin{array}{cc}(t+1)&(2t+1)\\ t(t+1)& 2t\end{array}\right|=(t+1)(2t)-(2t+1)t(t+1)\\
\\
D_x=-2t^3-t^2+t[/tex]
Cramer's Rule tells us:
[tex]x=\dfrac{D_x}{D}=\dfrac{-2t^3-t^2+t}{t+1}[/tex]
Applying Polynomial Long Division or Synthetic Division (with -1) should give you the result -2t^2+t if I did my calculations correctly.
This can be factored, giving us a nicer looking answer of [tex]x=t(1-2t)[/tex]
This is a fairly long problem, easy to get confused along the way. Lemme know if anything didn't make sense.
