Bài 4: Cho $\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}$. Tìm $x,y.$

Giải

Ta có: $\frac{1+5y}{5x}=\frac{1+7y}{4x}$$\Leftrightarrow$ $\left\{ \begin{align}& x\ne 0 \\ & 4(1+5y)=5(1+7y) \\ \end{align} \right.$$\Leftrightarrow$$\left\{ \begin{align}& x\ne 0 \\ & 15y=-1 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x\ne 0 \\ & y=-\frac{1}{15} \\ \end{align} \right.$

$\frac{1+3y}{12}=\frac{1+5y}{5x}$$\Leftrightarrow$$\left\{ \begin{align}& x\ne 0 \\ &x=\frac{12(1+5y)}{1+3y}=4.5.\frac{3+15y}{5+15y}=4.5.\frac{3-1}{5-1}=10 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x\ne 0 \\ & y=-\frac{1}{15} \\ \end{align} \right.$

Vậy $\left( x,y \right)=\left( 10,-\frac{1}{15} \right)$