Respuesta :
so first off, let's simplify both equations, starting off by multiplying both sides by the LCD of all fractions, to do away with the denominators.
[tex]\bf \cfrac{10(x-y)-4(1-x)}{3}=y\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{10(x-y)-4(1-x)=3y} \\\\\\ 10x-10y-4+4x=3y\implies \boxed{14x-13y=4} \\\\[-0.35em] ~\dotfill\\\\ 7+x-\cfrac{x-3y}{4}=2x-\cfrac{y+5}{3}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( 7+x-\cfrac{x-3y}{4} \right)=12\left( 2x-\cfrac{y+5}{3} \right)} \\\\\\ 84+12x-3(x-3y)=24x-4(y+5) \\\\\\ 84+12x-3x+9y=24x-4y-20\implies \boxed{-15x+13y=-124}[/tex]
now, let's do some elimination on those two simplified equations.
[tex]\bf \begin{array}{cllcl} 14x&-13y&=&4\\ -15x&+13y&=&-124\\\cline{1-4} -x&&=&-120 \end{array}~\hfill x=\cfrac{-120}{-1}\implies \blacktriangleright x=120 \blacktriangleleft \\\\\\ \stackrel{\textit{substituting on the 1st equation}}{14(120)-13y=4}\implies 1680-13y=4\implies 1680-4=13y \\\\\\ 1676=13y\implies \blacktriangleright \cfrac{1676}{13}=y \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left( 120~,~\frac{1676}{13} \right)~\hfill[/tex]
