From the first equation x=3y-4 and we can substitute this into the second equation to give y^3-9y^2+24y-16=0. You might just be lucky enough to spot that y=1 is a solution by substituting this value into this equation. That means (y-1) is a factor and we can divide (y-1) into the cubic equation using algebraic long division. y goes into y^3 y^2 times so multiply (y-1) by y^2 and we get y^3-y^2. Subtract this frm the cubic and we get -8y^2+24y-16. This time we multiply by -8y to get -8y^2+8y. Subtract this and we get 16y-16, so the last part of the quotient is 16. So (y-1) divided into the cubic is y^2-8y+16. Does this factorise? Yes! It's (y-4)^2, so the cubic factorised is (y-1)(y-4)^2=0. For the values of y we get the corresponding values of x by putting them into the equation x=3y-4 or y^3-x^2=0. So the solution is given by (x,y) and is (-1,1) and (8,4). Note that x=-1 satisfies the equation y^3-x^2=0 when y=1, but not the other equation.