Đặt \(\dfrac{x+1}{2}=\dfrac{y+3}{4}=\dfrac{z+5}{6}=t\)
=> \(\left[{}\begin{matrix}x=2k-1\\y=4k-3\\z=6k-5\end{matrix}\right.\)
Ta có: 2x + 3y + 4z = 9
=> 2(2k - 1) + 3(4k - 3) + 4(6k - 5) = 9
=> 4k - 2 + 12k - 9 + 24k - 20 = 9
=> (4k + 12k + 24k) - 31 = 9
=> 40k - 31 = 9
=> 40k = 9 + 31
=> 40k = 40
=> k = 1
*Với k = 1 ta có:
x = 2.1 - 1 = 1
y = 4.1 - 3 = 1
z = 6.1 - 5 = 1