Answer:
[tex]\dfrac{7}{9}[/tex]
Step-by-step explanation:
[tex]\dfrac{x+1}{y+1}=\dfrac{4}{5}\\\Rightarrow 5x-4y=-1[/tex]
[tex]\dfrac{x-5}{y-5}=\dfrac{1}{2}\\\Rightarrow 2x-y=5[/tex]
Putting it in matrix form
[tex]\begin{bmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{bmatrix}{\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}{c_{1}}\\{c_{2}}\end{bmatrix}\\\Rightarrow\begin{bmatrix}5 & -4\\2 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-1\\ 5\end{bmatrix}[/tex]
From Cramer's rule we have
[tex]x=\dfrac{\begin{vmatrix}c_1 &b_1 \\ c_2 & b_2\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow x=\dfrac{\begin{vmatrix}-1 &-4 \\ 5 & -1\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1\end{vmatrix}}\\\Rightarrow x=\dfrac{1+20}{-5+8}\\\Rightarrow x=7[/tex]
[tex]y=\dfrac{\begin{vmatrix}a_1 &c_1 \\ a_2 & c_1\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow y=\dfrac{\begin{vmatrix}5 &-1 \\ 2 & 5\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1 \end{vmatrix}}\\\Rightarrow y=\dfrac{25+2}{-5+8}\\\Rightarrow y=9[/tex]
Verifying the results
[tex]\dfrac{7+1}{9+1}=\dfrac{8}{10}=\dfrac{4}{5}[/tex]
[tex]\dfrac{7-5}{9-5}=\dfrac{2}{4}=\dfrac{1}{2}[/tex]
Hence, the fraction is [tex]\dfrac{7}{9}[/tex].
