Don't skip
solve the system of equations.

                2x + 3y = 1

                y = 3x + 15

There's not much you can do with the first equation, because it has
two variables in it ... 'x'  and  'y' .  No matter how much you move them
around, you'll never be able to get either one equal to just a number. 
Is there any way you could get rid of one of the variables in the first
equation, and have just 1 letter in it to solve for ?

Absolutely !  The second equation tells you something that 'y' is equal to,
(3x + 15).  "EQUAL" is very powerful.  It means that wherever you see 'y',
you can put (3x + 15) in its place, and you won't change anything or
upset anything.  One thing you can do is take that (3x + 15) from the
2nd equation, and put it right into the first equation in place of 'y'. 
You'll see how that helps as soon as you do it.

             First equation:    2x + 3y = 1

         Substitute for 'y' :    2x + 3(3x + 15) = 1

 Remove parentheses:    2x + 3(3x) + 3(15) = 1
                                        2x +  9x    +    45  =  1

Combine the terms with 'x' in them:    11x + 45 = 1

Look what you have now !  An equation with only one variable in it !

Subtract  45  from each side:    11x = -44

Divide each side by  11 :            x = -4

You're more than halfway there.  Now you know what 'x' is,
and you can use it with either equation to find what 'y' is.

-- If you use it with the first equation:      2x + 3y = 1

       Put in the value of 'x':    2(-4) + 3y = 1

Remove the parentheses:      -8 + 3y = 1

        Add  8  to each side:              3y = 9

    Divide each side by  3 :               y = 3

-- If you use it with the 2nd equation:    y = 3x + 15
 
            Put in the value of  'x' :      y = 3(-4) + 15

       Remove the parentheses:      y = -12 + 15

Add numbers on the right side:     y = 3  (same as the other way)

So there's your solution for the system of two equations:

         x = -4
         y = 3

Hey mate where is the graph