I am considering taking an existing laptop with Windows 7 on it and setting up a triple boot system: Windows 7 + MS-DOS + Windows XP. My question is aimed at the MS-DOS part of the process I found these instructions on the web: THPC: Triple-Boot (install) Windows XP and MS-DOS 7.10 and Windows 7 on your Windows 7 computer 'Install Triple-Boot of Windows 7 + XP + MS-DOS 7.10 on Windows 7 computer (Win7 installed first)' which states that, to install MS-DOS, you need to shrink the initial Win 7 partition to make space for a MS-DOS partition after the Win 7 partition. Here (DOS - EasyBCD - NeoSmart Technologies Wiki) : 'Booting into DOS with EasyBCD' it states "The DOS partition must start within the first 2GB and 1024 cylinders of the hard drive. The safest, most-assured way of doing this is to place the DOS partition before any other partitions on the hard drive, the Windows Vista partition included." which would seem to conflict with the advice above. However the 'Triple BOOT' article goes on to say that, after shrinking the Win 7 partition and creating a FAT partition for DOS, "Bootup from a Win98SE boot floppy or 98SE MS-DOS boot CD/Flash drive. Type A:\SYS C: at the Prompt and press Enter - this creates a DOS boot sector on the FAT32 partition and installs a very basic MS-DOS 7.10". It goes on to say that, using EasyBCD, a DOS menu item can be added and the rest of the MS-DOS files installed in the FAT partition. Questions: (1) is this a safe way to install MS-DOS on a pre-existing Win 7 laptop? (2) is the SYS C: process non-destructive of the pre-existing Win 7 boot sector? (3) Does the SYS C action satisfy the requirement to have the DOS partition within the first 2GB of the hard drive? TIA