Bài 3:
a, \(\dfrac{x}{4}=\dfrac{y}{3};2x+3y=10\)
Ta có: \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{2x}{8}=\dfrac{3y}{9}=\dfrac{2x+3y}{8+9}=\dfrac{10}{17}\)
* \(\dfrac{x}{4}=\dfrac{10}{17}=>x=\dfrac{40}{17}\)
* \(\dfrac{y}{3}=\dfrac{10}{17}=>y=\dfrac{30}{17}\)
Vậy (x; y) ∈{ \(\dfrac{40}{17};\dfrac{30}{17}\)}
Mình ko chắc đúng hoàn toàn, bạn kiểm tra lại nhé