Question

Giải hệ phương trình\(\left\{{}\begin{matrix}\text{3(3y+2)+4(5x-4)=0}\\\dfrac{5x-4}{15x-2}=\dfrac{3y+2}{9y+4}\end{matrix}\right.\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\dfrac{2}{15}\\y\ne-\dfrac{4}{9}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}9y+6+20x-16=0\\\left(5x-4\right)\left(9y+4\right)=\left(3y+2\right)\left(15x-2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}20x+9y=10\\5x+15y=-6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{5}\\y=-\dfrac{2}{3}\end{matrix}\right.\)