Question

x-2/5=3-y/7;x+y=3

Mọi ngừ zúp mik vs ạ

Answer 1

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

Nên : \(\frac{x-2}{5}=1\Rightarrow x-2=5\Rightarrow x=7\)

           \(\frac{3-y}{7}=1\Rightarrow3-y=7\Rightarrow y=-4\)

Vậy x = 7 ; y = -4 

Answer 2

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

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

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

\(\Rightarrow\hept{\begin{cases}x-2=5\\3-y=7\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\y=-4\end{cases}}\)