Math Problem Solver

True, $\left( 1, \ 10\right)$ is a solution of $y > 6 x + 1 \text{ and } y \leq x + 9$.

$$\begin{gathered}y > 6 x + 1\\10 \stackrel{?}{\gt} 6 \left(1\right) + 1\\10 \gt 7\end{gathered}$$$$\begin{gathered}y \leq x + 9\\10 \stackrel{?}{\le} 1 + 9\\10 \le 10\end{gathered}$$