Đáp án: $ (x,y)\in\{(1, -\dfrac29), (-\dfrac6{17}, -\dfrac{14}{17})\}$
Giải thích các bước giải:
Ta có $4x-9y=6$
$\to 9y=4x-6$
$\to 3y=\dfrac{4x-6}{3}$
Lại có:
$3x^2+6xy-x+3y=0$
$\to 3x^2+2x\cdot 3y-x+3y=0$
$\to 3x^2+2x\cdot \dfrac{4x-6}{3}-x+\dfrac{4x-6}{3}=0$
$\to 3x^2\cdot \:3+2x\dfrac{4x-6}{3}\cdot \:3-x\cdot \:3+\dfrac{4x-6}{3}\cdot \:3=0$
$\to 9x^2+2x\left(4x-6\right)-3x+4x-6=0$
$\to 17x^2-11x-6=0$
$\to x\in\{1, -\dfrac6{17}\}$
$\to y\in\{-\dfrac29, -\dfrac{14}{17}\}$
$\to (x,y)\in\{(1, -\dfrac29), (-\dfrac6{17}, -\dfrac{14}{17})\}$
