Solve the following system.y=(-)x2 + 2x - 1 and 3x - y=1The solutions are _and_

Since the coefficient of the quadratic variable is missing, I obtained it from a similar question.The system is:y = (1/2)x² + 2²x - 1 and 3x - y = 1Answer:The two solutions are (0, -1) and (2, 5)Explanation:1. Write the system[tex]y=(1/2)x^2+2x-1\\ \\ 3x-y=1[/tex]2. Clear y from the second equation to solve by substitution[tex]y= 3x-1[/tex]3. Substitute in the first equation and solve for x[tex]3x-1=(1/2)x^2+2x-1\\ \\ (1/2)x^2+2x-3x=0\\ \\ (1/2)x^2-x=0\\ \\ x(x/2-1)=0[/tex][tex]x=0[/tex][tex]x/2-1=0\\ \\ x/2=1\\ \\ x=2[/tex]4. Subsitute both values into the equation y= 3x - 1y = 3(0) - 1 = - 1 ⇒ solution (0, -1)y = 3(2) - 1 = 5 ⇒ solution (2, 5)