Transition Matrix from B to C 
If $B=\{ 2+x,1+2x\}$ and $C=\{ 1+x, 1-x\}$ are 2 basis for $P_1$, and $v=-3x+4$ find $[v]_B$, $_BP_C$ and $[v]_C$. 

my attempt:
Since $B$ is a basis for $V$, then any $v\in B$ can be written "uniquely" as a linear combination of the vectors of $B$.
then $v=-3x+4$ can be written uniquely  in the form:
$-3x+4=c_1(2+x)+c_2(1+2x)$ where $c_1$ and $c_2$ are scalars.
$-3x+4=c_1(2+x)+c_2(1+2x)$
$=2c_1+c_2+(c_1+2c_2)x$
Then
$2c_1+c_2=4$
$c_1+2c_2=-3$
and by solving this system, we have
$c_1=\frac{11}{3}$ and $c_2=\frac{-10}{3}$
so, $[v]_B=$ $\left[ {\begin{array}{c}
\frac{11}{3} \\
\frac{-10}{3}
\end{array} } \right]
$
The transition matrix from $B$ to $C$ is 
$_BP_C=$ $\left[ {\begin{array}{cc}
[2+x]_C & [1+2x]_C
\end{array} } \right]
$
we can find $[2+x]_c$ as follows
$2+x=\alpha_1(1+x)+\alpha_2(1-x)$
$=(\alpha_1+\alpha_2)+(\alpha_1-\alpha_2)x$
and hence
$\alpha_1+\alpha_2=2$
$\alpha_1-\alpha_2=1$
and by solving this system we get that
$\alpha_1=\frac{3}{2}$ and $\alpha_1=\frac{1}{2}$
Similarly, 
$1+2x=\alpha_1(1+x)+\alpha_2(1-x)$
$=(\alpha_1+\alpha_2)+(\alpha_1-\alpha_2)x$
and hence
$\alpha_1+\alpha_2=1$
$\alpha_1-\alpha_2=2$
and by solving this system we get that
$\alpha_1=\frac{3}{2}$ and $\alpha_1=\frac{-1}{2}$
$_CP_B=\left[ {\begin{array}{cc}
\frac{3}{2} & \frac{3}{2} \\
\frac{-1}{2} & \frac{1}{2}
\end{array} } \right]
$
or,
$_CP_B=\left[ {\begin{array}{cc}
\frac{3}{2} & \frac{3}{2} \\
\frac{1}{2} & \frac{-1}{2}
\end{array} } \right]
$
what is the true?
and we know that
$[v]_C=_BP_C[v]_B$
$[v]_C=\left[ {\begin{array}{cc}
\frac{3}{2} & \frac{3}{2} \\
\frac{1}{2} & \frac{-1}{2}
\end{array} } \right]$
$\left[ {\begin{array}{c}
\frac{11}{3} \\
\frac{-10}{3}
\end{array} } \right]
$
$= \left[ {\begin{array}{c}
\frac{1}{2} \\
\frac{7}{2}
\end{array} } \right]
$
and if I use $_CP_B=\left[ {\begin{array}{cc}
\frac{3}{2} & \frac{3}{2} \\
\frac{-1}{2} & \frac{1}{2}
\end{array} } \right]
$
I got that
$= \left[ {\begin{array}{c}
\frac{1}{2} \\
\frac{7}{2}
\end{array} } \right]
$
what is the true please? Thanks.

$b_1 = 2+x = a_{11} c_1 + a_{21} c_2\\
2+x = a_{11}(1+x) + a_{21}(1-x)\\
a_{11}+a_{21} = 2\\
a_{11}-a_{21} = 1\\
a_{11} = \frac 32, a_{21} = \frac 12$
and 
$b_2 = 1+2x = a_{12} c_1 + a_{22} c_2\\
a_{12} + a_{22} = 1\\
a_{12} - a_{22} = 2\\
a_{12} = \frac 32, a_{22} = - \frac 12$
$_CP_B = \begin{bmatrix} \frac 32 &\frac 32\\\frac 12&-\frac 12\end {bmatrix}$
Here is a different way to think about it.
$B = \begin {bmatrix} 2&1\\1&2 \end{bmatrix}$
Will transform vectors represented in the basis $B$ into the standard basis.
$C = \begin {bmatrix} 1&1\\1&-1 \end{bmatrix}$
Will transform vectors represented in the basis $C$ into the standard basis.
$C^{-1} = \begin {bmatrix} \frac {1}{2}&\frac {1}{2}\\\frac {1}{2}& -\frac {1}{2} \end{bmatrix}$
Will reverse that and take vectors in the standard basis to the basis $C.$
$C^{-1}B$ will take a vector in $B$, translated to the standard basis and then transform that into the basis $C.$
$\begin {bmatrix} \frac {1}{2}&\frac {1}{2}\\\frac {1}{2}& -\frac {1}{2} \end{bmatrix}\begin {bmatrix} 2&1\\1&2 \end{bmatrix}= \begin {bmatrix} \frac 32 & \frac 32 \\ \frac 12 & -\frac 12 \end{bmatrix}$
