You can probably do this problem by trial and error,
but more complicated problems similar to this which have
more given restrictions cannot be.  So to learn the methods
we must do this algebraically and graphically by the use 
of linear inequalities.


Here's how to solve it algebraically:

Let x = the number of girls
Let y = the number of boys

Jenny has y brothers and x-1 sisters (Jenny is not her own sister,
so we must eliminate her from the girls who can be Jenny's sisters
by subtracting 1)

Jim has x sisters and y-1 brother (Jim is not his own brother, so we
must eliminate him from the boys who can be Jim's brothers, also by
subtracting 1)

Since Jenny is the 6th child, the total of the girls and boys must
be 6 or more.  Therefore, 

By Jenny's statement, 

By Jim's statement, 

So we have the system of inequalities:



We graph the boundary lines, which are the inequalities
with equality signs instead of inequality signs:





The correct solution point must be 

1.  on or above the red line,

2.  on or above the green line, and 

3.  on or below the blue line.

Therefore the solution point is either inside or on the 
little triangle bounded by the three lines.

Also, the solution point must 

4.  have whole number coordinates,
as we cannot have fractions of people!

Therefore the only point in or on the triangle
which has all four properties above is the point (4,3), 
which is marked.  Thus x=4, y=3 and there are 4 girls 
and 3 boys.

Checking: 
Jenny has 3 brothers and 3 sisters, so she has at least 
as many brothers as sisters, in fact, the same number.

Jim has 4 sisters and 2 brothers, so he has at least 
twice as many sisters as he has brothers, in fact, 
exactly twice as many.

Edwin