find all solutions of the system of equations algebraically. write your solutions as coordinate…

Explanation:

Step1: Substitute (y) from the first equation into the second equation

Substitute (y = 2x^{2}+3x - 94) into (3x + y=14). We get (3x+(2x^{2}+3x - 94)=14). Simplify the left - hand side: (2x^{2}+3x+3x - 94 = 14), which is (2x^{2}+6x-94 - 14=0). So, (2x^{2}+6x - 108 = 0). Divide the entire equation by (2): (x^{2}+3x - 54=0).

Step2: Solve the quadratic equation (x^{2}+3x - 54 = 0)

Use the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for the quadratic equation (ax^{2}+bx + c = 0). Here (a = 1), (b = 3), (c=-54). First, calculate the discriminant (\Delta=b^{2}-4ac=(3)^{2}-4\times1\times(-54)=9 + 216=225). Then (x=\frac{-3\pm\sqrt{225}}{2}=\frac{-3\pm15}{2}). Case 1: (x=\frac{-3 + 15}{2}=\frac{12}{2}=6). Case 2: (x=\frac{-3-15}{2}=\frac{-18}{2}=-9).

Step3: Find the corresponding (y) values

When (x = 6), substitute (x = 6) into (3x + y=14). So (3\times6+y=14), (18 + y=14), (y=14 - 18=-4). When (x=-9), substitute (x=-9) into (3x + y=14). So (3\times(-9)+y=14), (-27 + y=14), (y=14 + 27=41).

Answer:

((6,-4)) and ((-9,41))