Ta có:
\(\text{x + y = 10 }\)(1)
\(\frac{x-3}{y+7}=\frac{3}{4}\)(2)
Từ (2) suy ra: \(4\left(x-3\right)=3\left(y+7\right)\)
=> \(4x-12=3y+21\)
=> \(4x=3y+21+12\)
=> \(4x=3y+33\)
=> \(4x-3y=33\)(3)
Lấy (3) - 4.(1), vế theo vế, ta có:
\(4x-3y-4\left(x+y\right)=33-4.10\)
=> \(4x-3y-4x-4y=33-40\)
=> \(\left(4x-4x\right)+\left(-3y-4y\right)=-7\)
=> \(-7y=-7\)
=> \(y=1\)
Thế y = 1 vào (1), ta có:
\(x+1=10\)
=> \(x=9\)