Find two numbers such that their sum equalls 21, and the greater number is 3 more than twice the smaller.

x + y = 21.

x = 2y + 3

Now we solve using these two equations.

All we have to do is substitute the value for x that is giving in the second equation into the first equation.  Now we have

2y + 3 + y = 21

3y + 3 = 21

3y = 18

y = 6

If y = 6, then x must = 15.

We can check this by seeing if 15 is 3 more than twice 6.

(6*2) + 3 = 15

12 + 3 = 15

This is true so our values are correct.