Question

[cocaidaubuoi]

2x-3/3 =3y-2/5=2x-3y-1/2x  . tìm x,y

Answer 1

[Nguyễn Thị Thương Hoài]

      \(\dfrac{2x-3}{3}\) = \(\dfrac{3y-2}{5}\) = \(\dfrac{2x-3y-1}{2x}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{2x-3}{3}\) = \(\dfrac{3y-2}{5}\) = \(\dfrac{2x-3y-1}{2x}\) = \(\dfrac{2x-3-3y+2-2x+3y+1}{3-5-2x}\) 

       \(\dfrac{2x-3}{3}\) = \(\dfrac{\left(2x-2x\right)+\left(-3y+3y\right)+\left(-3+2+1\right)}{\left(3-5\right)+2x}\)

       \(\dfrac{2x-3}{3}\) = \(\dfrac{3y-2}{5}\) = \(\dfrac{0+0+0}{-2-2x}\) = 0

         2\(x-3=0\) ⇒ \(x=\dfrac{3}{2}\); 3y - 2 = 0 ⇒ 3y = 2 ⇒ y = \(\dfrac{2}{3}\)

Vậy \(x=\dfrac{3}{2}\); y = \(\dfrac{2}{3}\)