Final answer:
To solve the given system of equations using substitution, solve the first equation for y, then substitute this value of y into the second equation. Simplify the resulting equation and solve for x. Finally, substitute the value of x back into the first equation to solve for y. However, in this case, the given system of equations does not have exactly one solution.
Step-by-step explanation:
To solve the given system of equations using substitution, we'll start by solving the first equation for y:
1 + y^2 - y - 3x + \frac{1}{y} = 0
y^2 - y + \frac{1}{y} = 3x - 1
Next, we'll substitute this value of y into the second equation:
2y + 8 = 12x + 10 + 4xy + 7 - 3
2y = 12x + 14 + 4xy
Substituting the expression for y from the first equation:
2\left(3x - 1 + \frac{1}{3x - 1}\right) = 12x + 14 + 4x\left(3x - 1 + \frac{1}{3x - 1}\right)
Expanding and simplifying:
6x - 2 + \frac{2}{3x - 1} = 12x + 14 + 12x - 4 + \frac{4}{3x - 1}
8 = 18x + 12x + 12x - 6 - 4
8 = 42x - 10
42x = 18
x = \frac{3}{7}
Substituting this value of x back into the first equation to solve for y:
1 + y^2 - y - 3\left(\frac{3}{7}\right) + \frac{1}{y} = 0
Simplifying:
y^2 - y + \frac{1}{y} = \frac{9}{7} - 1
y^2 - y + \frac{1}{y} = \frac{2}{7}
Multiplying by y to clear the fraction:
y^3 - y^2 + 1 = \frac{2y}{7}
y^3 - y^2 - \frac{2y}{7} + 1 = 0
Unfortunately, we are unable to find an exact solution for y in this equation. Therefore, the system of equations does not have exactly one solution.