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

The system of equations y= 2/3x-3 and y=-5/6x+7 has to be solved for x and y.

From y= 2/3x-3, substitute for y in y=-5/6x+7

2/3x - 3 = -5/6x + 7

(2/3 + 5/6)*x = 7 + 3

(3/2)*x = 10

x = 20/3

y = 2/3x - 3 = (2/3)*(20/3) - 3 = 40/9 - 3 = 13/9

The solution of the given set of equations is x = 20/3 and y = 13/9

Given:

y = 2/3x - 3

And:

y = -5/6x + 7

Since left hand side of both equation is same, the right hands side of these equation will be equal. Thus:

2/3x - 3 = -5/6x + 7

==> 2/3x + 5/6x = 7 + 3

==> (4 + 5)/6x = 10

==> 9/6x = 10

==> 6x = 9/10

==> x = 9/(10*6) = 9/60 = 3/20

Answer:

x = 3/20