Since you have a system of 4 equations in 2 variables, I am going to find the solution for a system of 2 of them. I will then check the solution in the other 2 equations. If the solution does not work for all 4 equations, then there is no solution.
3x = y - 3 Equation 1
2y = 6 - 3x Equation 2
2y = 6 - (y - 3) Substitute the value of 3x in equation 1 for the 3x in equation 2
2y = 6 - y + 3 Distributive property (distribute the negative)
2y = 9 - y Simplify
2y + y = 9 - y + y Add y to each side
3y = 9 Simplify
3y/3 = 9/3 Divide each side by 3
y = 3 Simplify
3x= y - 3 Equation 2
3x = 3 - 3 Substitute the value of y
3x = 0 Simplify
3x/3 = 0/3 Divide each side by 3
x = 0 Simplify
Now that we have the solution to the first two equations (0,3), we need to check the solution in the third and fourth equations.
5x = 14 - y Equation 3
5(0) = 14 - 3 Substitution
0 = 11 Simplify
The solution to the first two equations does not work in the third, so there is no solution for the system of 3 equations. If the first three equations do not have a solution, then there will not be a solution for all 4.
There is no solution to this system of 4 equations.