solve the system: 1.) y= 2/3x-3 , y=-5/6x+7

First, I am not sure if your equations are supposed to be
y =(2/3)x -3 or y = 2/(3x) -3.  I am going to solve the equations as though they are the
first y = (2/3)x - 3.

Tosolve a system of equations you can
use either the substition or elimination method.  Many students dislike solving
equations that involve fractions. If you are one of these, then the first thing you want
to do is eliminate the fraction by multiplying both by the number in the
denominator.

Equation 1: y = (2/3)x
-3 

3y = 2x - 9                    1. multiply every term
in the equation by 3

Equation 2: y = (-5/6)x +
7

6y = -5x + 42               2. multiply every term in the
equation by 6

The next step for this set of equations is to
use the elimination method.  To do this you need to multiply one, or both equations, by
a number that will result in a common coeffiecent for one of the variables.  In this
case it will be easiest to multiply equation 1 by
2. 

Equation 1:   3y = 2x -9         multiplied: 6y = 4x
-18

Then you will subtract equation 2 from equation
1:

 Equation 1:       6y = 4x -
18

- Equation 2:    -(6y =-5x +
42)

answer                 0 = 9x -
60

solve for x            -9x = - 60             x = 6
2/3

you would then substitute 6 for x into one of the
equations

y = 2/3 (6 2/3) -3  = 40/9 - 3 = 4 4/9 -3 = 1
4/9

To check your answer substitute the 6 2/3  for x and
the 1 4/9 for y in the other equation.

y =-5/6 x +
7

1 4/9 = -5/6 (6 2/3) + 7

1
4/9 = -100/18 + 7

1 4/9 = -5 5/9 +
7

1 4/9 = 1 4/9  check