System of equations solving for a b c d 
Im doing a problem that requires partial fractions, Im having trouble figuring out how to solve for a b c and d.
How do i solve for each letter? Can i just take 2 at a time and try to solve for a letter?
 A: You can use row-reduce an augmented matrix to RREF form and get values for $A,B,C,D$.
Assuming that your equations are correct,
$$
  \left[\begin{array}{rrrr|r}
    1 & 0 & 1 & 1 & 0 \\
    0 & 1 & 1 & -4 &8 \\
    -7& 1 &-5 & 5 & -15\\
    6 &-6 & 3 & -2 & 8\\
  \end{array}\right]
$$
$$\rightarrow   \left[\begin{array}{rrrr|r}
    1 & 0 & 0 & 0 & \frac{-7}{16} \\
    0 & 1 & 0 & 0 & \frac{-1}{4} \\
    0 & 0 & 1 & 0 & 2\\
    0 & 0 & 0 & 1 & \frac{-25}{16}\\
  \end{array}\right]$$
i.e. $A=\frac{-7}{16},B=\frac{-1}{4},C=2,D=\frac{-25}{16}$ (assuming I didn't make a mistake).
Note: Considering it involves so many factors, you might consider using the "cover-up rule".
A: If you don't know about matrices and row-reduction, you can solve the first equation for $A$ in terms of $C$ and $D$; solve the second equation for $B$ in terms of $C$ and $D$; then subtitute in for $A$ and $B$ in the third and fourth equations. That will give you two equations in the two unknowns $C$ and $D$ --- can you handle that?
EDIT: In further detail, first equation becomes $$A=-C-D;$$ second equation becomes $$B=8-C+4D;$$ third equation becomes $$-15=-7(-C-D)+(8-C+4D)-5C+5D;\tag1$$ fourth equation becomes $$8=6(-C-D)-6(8-C+4D)+3C-2D.\tag2$$ A bit of algebra now gets (1) and (2) to look like two linear equations in two unknowns, and you're on your way. 
