sequences

Recall that a $\text{Fibonacci sequence}$ is one in which all entries
from the third onward are the sum of the two entries before it.
If the $5th$ and $8th$ entries of a Fibonacci Sequence are $-19$ and $79$, respectively,
then what is the first entry of this sequence?

$\begin{array}{|rcrcrl|} \hline a_1 &=& &a_1& \\ a_2 &=& && &a_2\\ a_3 &=& &a_1& + &a_2 \\ a_4 &=& &a_1& + &2a_2 \\ a_5 &=& &2a_1& + &3a_2 & \qquad\mathbf{ a_5 = 2a_1+3a_2 = -19} \\ a_6 &=& &3a_1& + &5a_2 \\ a_7 &=& &5a_1& + &8a_2 \\ a_8 &=& &8a_1& + &13a_2 & \qquad\mathbf{ a_8 = 8a_1+13a_2 = 79} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 2a_1+3a_2 &=& -19 \\ 3a_2 &=& -19 -2a_1 \\ \mathbf{ a_2 } &=& \mathbf{ \dfrac{-19-2a_1}{3} } \\ \hline 8a_1+13a_2 &=& 79 \\ 8a_1 + 13 \left( \dfrac{-19-2a_1}{3} \right) &=& 79 \quad | \quad * 3 \\\\ 24a_1 +13(-19-2a_1) &=& 79*3 \\ 24a_1 -13*19-26a_1 &=& 237 \\ -2a_1 -247 &=& 237 \\ -2a_1 &=& 484 \quad | \quad :(-2) \\ \mathbf{a_1} &=& \mathbf{-242} \\ \hline a_2 &=& \dfrac{-19-2a_1}{3} \\\\ a_2 &=& \dfrac{-19-2(-242)}{3} \\\\ a_2 &=& \dfrac{-19+484}{3} \\\\ a_2 &=& \dfrac{465}{3} \\\\ \mathbf{a_2} &=& \mathbf{155} \\ \hline \end{array}$

$\begin{array}{|rcr|} \hline a_1 &=& -242 \\ a_2 &=& 155 \\ a_3 &=& -87 \\ a_4 &=& 68 \\ a_5 &=& -19 \\ a_6 &=& 49 \\ a_7 &=& 30 \\ a_8 &=& 79 \\ \hline \end{array}$