System of equations solving for a b c d

Question

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?

Answer 1

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".

Answer 2

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.