We have #y <= x# and #x + y >= 8#. To solve the system by graphing, and then graph the solution, you'll want to make both expression look like equations of a line, except they are actually inequalities. In other words, try to transform the expressions into something that looks like:

#y# (#<=# or #<# or #># or #>=#) #mx+b#.

In our case: #y <= x# and #x + y >= 8 => y >= -x + 8#.

In case this is still hard to understand, it's best to have #y# and #x# on the other sides of the expression, with the constant terms on the #x# side. This way, it looks like an equation of a line, so you can then graph the line as if it was an equation, then find what area around that line satisfies the inequality. Let's graph the lines

#y color(red)(=) x# and #y color(red)(=) -x + 8#

#y = x# is:

graph{y=x [-10, 10, -5, 5]}

and #y = -x + 8# is:

graph{y=-x+8 [-10, 10, -5, 5]}

Now remember that we don't want the equality. This time,

#y color(blue)(<=) x# means that we need to take the area **below** the line, and the points exactly on the line as possible solutions.

#y color(blue)(>=) -x + 8# means that we need to take the area **above** the line, and the points exactly on the line as possible solutions.

In the end, we have to see which solutions are common between the two cases, so we mark for acceptable solutions:

All points that are below or on #y=x# **and** above or on #y = -x + 8# at the same time.