SAT Mathematics Practice Test 4

Since $\mathit{y}=\left(2\mathit{x}-3\right)\left(\mathit{x}+9\right)$ and $\mathit{x}=2\mathit{y}+5$, it follows that $\mathit{x}=2\left(\left(2\mathit{x}-3\right)\left(\mathit{x}+9\right)\right)+5=4{\mathit{x}}^2+30\mathit{x}-54.$ This can be rewritten as $4{\mathit{x}}^2+29\mathit{x}-54=0$ Because the discriminant of this quadratic equation,$292-\left(4\right)\left(-54\right)=292+4\left(54\right)$, is positive, this equation has $2$ distinct roots. Using each of the roots as the value of $\mathit{x}$ and finding $\mathit{y}$ from the equation $\mathit{x}=2\mathit{y}+5$ gives $2$ ordered pairs $\left(\mathit{x},\mathit{y}\right)$ that satisfy the given system of
equations. Since no other value of x satisfies $4{\mathit{x}}^2+29\mathit{x}-54=0,$ there are no other ordered pairs that satisfy the given system. Therefore, there are $2$ ordered pairs (x,y) that satisfy the given system of equations.
Choices A and B are incorrect and may be the result of either a miscalculation or a conceptual error. Choice D is incorrect because a system of one quadratic equation and one linear equation cannot have infinitely many solutions.