That's some nice handwriting :)
Eliminate [tex]k[/tex] and solve for [tex]j[/tex].
[tex]3 (6j - k) + (3k + 6j) = 3\cdot23 + 11 \\\\ \implies 18j - 3k + 3k + 6j = 69 + 11 \\\\ \implies 24j = 80 \\\\ \implies \boxed{j = \dfrac{80}{24} = \dfrac{10}3}[/tex]
Now solve for [tex]k[/tex].
[tex]6j - k = 23 \implies k = 6j - 23 \\\\ \implies k = 6\cdot\dfrac{10}3 - 23 \\\\ \implies \boxed{k = -3}[/tex]
