Question

tìm x, y

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)

Answer 1

Ta có : \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)

Nên : \(\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)

<=> 6x = 12

=> x = 2 . 

Answer 2

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}=\frac{\left(2x+1\right)+\left(3y-2\right)}{5+7}=\frac{2x+3y-1}{12}\)

\(\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)