Ta có: \(\frac{5+x}{7+y}=\frac{5}{7}\)\(\Leftrightarrow\)7(5+x) = 5(7+y)\(\Leftrightarrow\)35+7x = 35+5y\(\Leftrightarrow\)7x=5y\(\Leftrightarrow\)\(\frac{x}{7}=\frac{y}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{7}=\frac{y}{5}=\frac{x+y}{7+5}=\frac{24}{12}=2\)
\(\Rightarrow\hept{\begin{cases}x=2\cdot7=14\\y=2\cdot5=10\end{cases}}\)
b) Ta có: \(\frac{x-1}{2}=\frac{3}{y+1}\Leftrightarrow\)\(\left(x-1\right)\cdot\left(y+1\right)=6\)
=> \(\left(x-1;y+1\right)\in\left\{\left(-1;-6\right);\left(-6;-1\right);\left(1;6\right);\left(6;1\right);\left(-3;-2\right);\left(-2;-3\right);\left(3;2\right);\left(2;3\right)\right\}\)
=> \(\left(x;y\right)\in\left\{\left(0;-7\right);\left(-5;-2\right);\left(2;5\right);\left(7;0\right);\left(-2;-3\right);\left(-1;-4\right);\left(4;1\right);\left(3;2\right)\right\}\)