Q:

How many solutions does the equation have, x1 + x2 + x3 = 10 , where x1 , x2, and x3 are non-negative integers?

Accepted Solution

A:
Non-negative integers are positive integers or zero. 1. When [tex]x_1=0,[/tex] then there are such possible cases for [tex]x_2[/tex] and [tex]x_3:[/tex][tex]x_2=0,\ x_3=10;[/tex][tex]x_2=1,\ x_3=9;[/tex][tex]x_2=2,\ x_3=8;[/tex][tex]x_2=3,\ x_3=7;[/tex][tex]x_2=4,\ x_3=6;[/tex][tex]x_2=5,\ x_3=5;[/tex][tex]x_2=6,\ x_3=4;[/tex][tex]x_2=7,\ x_3=3;[/tex][tex]x_2=8,\ x_3=2;[/tex][tex]x_2=9,\ x_3=1;[/tex][tex]x_2=10,\ x_3=0.[/tex]In total 11 solutions for [tex]x_1=0.[/tex]2. For [tex]x_1=1,[/tex] there are such possible cases for [tex]x_2[/tex] and [tex]x_3:[/tex][tex]x_2=0,\ x_3=9;[/tex][tex]x_2=1,\ x_3=8;[/tex][tex]x_2=2,\ x_3=7;[/tex][tex]x_2=3,\ x_3=6;[/tex][tex]x_2=4,\ x_3=5;[/tex][tex]x_2=5,\ x_3=4;[/tex][tex]x_2=6,\ x_3=3;[/tex][tex]x_2=7,\ x_3=2;[/tex][tex]x_2=8,\ x_3=1;[/tex][tex]x_2=9,\ x_3=0.[/tex]In total 10 solutions for [tex]x_1=1.[/tex]3. This process gives youfor [tex]x_1=2[/tex] - 9 solutions;for [tex]x_1=3[/tex] - 8 solutions;for [tex]x_1=4[/tex] - 7 solutions;for [tex]x_1=5[/tex] - 6 solutions;for [tex]x_1=6[/tex] - 5 solutions;for [tex]x_1=7[/tex] - 4 solutions;for [tex]x_1=8[/tex] - 3 solutions;for [tex]x_1=9[/tex] - 2 solutions;for [tex]x_1=10[/tex] - 1 solution.4. Add all numbers of solutions:[tex]11+10+9+8+7+6+5+4+3+2+1=66.[/tex]Answer: there are 66 possible solutions (with non-negative integer variables)