Question

Alguém me ajuda em sistemas lineares? Me de substituição.
2x/3 - y/4 - 1/9 = 0
x - y = 1

Answer 1

$\begin{cases} \dfrac{2x}{3} - \dfrac{y}{4} - \dfrac{1}{9} =0\\ x-y=1\end{cases}\\\\\\ Em~~exemplos~~como~~esse,~~para~~resolver~~voc\^e~~pode~~simplificar\\ a~~ equa\c c\~ao~~com~~fra\c c\~oes~~usando~~mmc, ~~afim~~de~~tornar~~mais\\simples.~~Vamos~~simplificar~~a~~1\° equa\c c\~ao:\\\\ \dfrac{2x}{3} - \dfrac{y}{4} - \dfrac{1}{9} =0\to~~~~mmc~(3,4,9)=36\\\\ Quando~~ temos~~ igualdade~~desprezamos~~os~~denominadores;\\\\ 24x-9y-4=0\to~~24x-9y=4 \\\\Ent\~ao~~teremos:$



$\begin{cases} 24x-9y=4 \\ x-y=1\end{cases}\\\\\\ Por~~substitui\c c\~ao:\\\\\\x-y=1\to~~x=y+1\\\\\\ 24x-9y=4\to~~ 24(y+1)-9y=4\to~~ 24y+24-9y=4\to\\\\ 24y-9y=4-24\to~~15y=-20\to~~y=- \dfrac{20}{15} \to~~ \boxed{y=- \dfrac{4}{3}}\\\\\\ x=y+1\to~~x=- \dfrac{4}{3}+1\to~~x= \dfrac{-4+3}{3}\to~~ \boxed{x=- \dfrac{1}{3} }\\\\\\\\\\ \large \boxed{ \boxed{S=\left\{- \dfrac{1}{3}~~;~~- \dfrac{4}{3} \right\}}}$