Final answer:
To find the solutions to the system of equations y=x²-5x and 6y=-4x, we can solve them simultaneously by substituting the value of y from the second equation into the first equation and then factoring the resulting quadratic equation. The solutions are (0, 0) and (13/3, -16/3).
Step-by-step explanation:
To find the solutions to the system of equations y=x²-5x and 6y=-4x, we will solve them simultaneously.
1. Start by rearranging the second equation to isolate y: y = -4/6x = -2/3x.
2. Substitute this value of y into the first equation: x²-5x = -2/3x.
3. Multiply through by 3 to clear the fraction: 3x²-15x = -2x.
4. Combine like terms and set the equation equal to 0: 3x²-13x = 0.
5. Factor the equation: x(3x-13) = 0.
6. Set each factor equal to 0 and solve for x: x = 0 or x = 13/3.
7. Substitute these values of x back into either of the original equations to solve for y.
8. When x = 0, y = 0²-5(0) = 0, so one solution is (0, 0).
9. When x = 13/3, y = (13/3)²-5(13/3) = -16/3, so another solution is (13/3, -16/3).
Therefore, the solutions to the system of equations are (0, 0) and (13/3, -16/3).