solve the system of linear equations separate the x- and y- values with a comma. -13x = -54 - 20y and -10x= 60 + 20y

[tex]\bf \begin{cases} -13x=-54-20y\\ -10x=60+20y \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{-13x=-54-20y}\implies -13x+20y=-54\implies \boxed{20y}=13x-54 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting on the 2nd equation}}{-10x=60+\left( \boxed{13x-54} \right)}\implies -10x=6+13x\implies -10x-6=13x[/tex][tex]\bf -6=23x\implies \blacktriangleright -\cfrac{6}{23}=x \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting on the 1st equation}}{-13\left( -\cfrac{6}{23} \right)=-54-20y}\implies \cfrac{78}{23}=-54-20y[/tex][tex]\bf \stackrel{\textit{multipying both sides by }\stackrel{LCD}{23}}{23\left( \cfrac{78}{23} \right)=23(-54-20y)}\implies 78=-1242-460y\implies 1320=-460y \\\\\\ \cfrac{1320}{-460}=y\implies \blacktriangleright -\cfrac{66}{23}=y \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left( -\frac{6}{23}~,~-\frac{66}{23} \right)~\hfill[/tex]