SAT Mathematics Practice Test 6

The intersection points of the graphs of $=3{\mathit{x}}^2-14\mathit{x}$ and $\mathit{y}=\mathit{x}$ can be found by solving the system consisting of these two equations. To solve the system, substitute $\mathit{x}$ for $\mathit{y}$ in the first equation. This gives $\mathit{x}=3{\mathit{x}}^2-14\mathit{x}$. Subtracting x from both sides of the equation gives $0=3{\mathit{x}}^2-15\mathit{x}$. Factoring 3x out of each term on the left-hand side of the equation gives 0 =3x(x − 5). Therefore, the possible values for $\mathit{x}$ are $0$ and $5$. Since $\mathit{y}=\mathit{x}$, the two intersection points are $\left(0,0\right)$ and $\left(5,5\right)$. Therefore, $\mathit{a}=5$.