Question: For each ordered pair, determine whether it is a solution to the system of equations. \[…

For each ordered pair, determine whether it is a solution to the system of equations.

\[ \begin{cases} y = -2x - 5 \\ 6x + 3y = -15 \end{cases} \]

\[ \begin{array}{|c|c|c|} \hline (x, y) & \text{Is it a solution?} & \\ \hline (4, -13) & \text{Yes} & \text{No} \\ \hline (-3, 1) & \text{Yes} & \text{No} \\ \hline (2, 7) & \text{Yes} & \text{No} \\ \hline (5, 0) & \text{Yes} & \text{No} \\ \hline \end{array} \]

Solution

To determine whether each ordered pair is a solution to the system of equations, we need to check if they satisfy both equations: \[ \begin{cases} y = -2x - 5 & \\ 6x + 3y = -15 & \end{cases} \] Let’s evaluate each pair: 1. For \((4, -13)\): Substitute \(x = 4\) and \(y = -13\) into the first equation: \[ -13 = -2(4) - 5 \] \[ -13 = -8 - 5 \] \[ -13 = -13 \quad \text{(True)} \] Now, substitute into the second equation: \[ 6(4) + 3(-13) = -15 \] \[ 24 - 39 = -15 \] \[ -15 = -15 \quad \text{(True)} \] Thus, \((4, -13)\) is a solution. 2. For \((-3, 1)\): Substitute \(x = -3\) and \(y = 1\) into the first equation: \[ 1 = -2(-3) - 5 \] \[ 1 = 6 - 5 \] \[ 1 = 1 \quad \text{(True)} \] Now, substitute into the second equation: \[ 6(-3) + 3(1) = -15 \] \[ -18 + 3 = -15 \] \[ -15 = -15 \quad \text{(True)} \] Thus, \((-3, 1)\) is a solution. 3. For \((2, 7)\): Substitute \(x = 2\) and \(y = 7\) into the first equation: \[ 7 = -2(2) - 5 \] \[ 7 = -4 - 5 \] \[ 7 = -9 \quad \text{(False)} \] Thus, \((2, 7)\) is not a solution. 4. For \((5, 0)\): Substitute \(x = 5\) and \(y = 0\) into the first equation: \[ 0 = -2(5) - 5 \] \[ 0 = -10 - 5 \] \[ 0 = -15 \quad \text{(False)} \] Thus, \((5, 0)\) is not a solution. Summary: - \((4, -13)\) is a solution. - \((-3, 1)\) is a solution. - \((2, 7)\) is not a solution. - \((5, 0)\) is not a solution.