Answer:
B. Substitute for x in the first equation
Step-by-step explanation:
The solution to a system of equations in which one is quadratic and the other is linear will necessarily involve solving a quadratic equation. Generally, we want to arrive at that quadratic as early as possible in the solution process.
Here, we have a linear expression for x, so it is pretty simple to use that to substitute for x in the first equation:
(2y -1)² -(2y -1) +3y = 5 . . . . substitute for x
4y² -4y +1 -2y +1 +3y = 5 . . . . eliminate parentheses
4y² -3y -3 = 0 . . . . quadratic in y in standard form
This does not have integer solutions, so it is perhaps best solved using the quadratic formula.
[tex]y=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(4)(-3)}}{2(4)}=\dfrac{3\pm\sqrt{57}}{8}[/tex]
__
Additional comment
A best first step might be to input both equations to a graphing calculator and let it show you the solutions.
