To find the solution to the system of equations formed by the line \(2x + 4y = 0\) and the line represented by the data in the table, we need to first rewrite the equation \(2x + 4y = 0\) in slope-intercept form (i.e., \(y = mx + b\)).
- Rearranging the equation: \[ 4y = -2x \ y = -\frac{1}{2}x \]
Now we have the line represented by the equation \(y = -\frac{1}{2}x\).
Next, we will check the points from the table to find the \(x\)-value where the two lines intersect.
Given table data:
- For \(x = -1\), \(y = 8\)
- For \(x = 3\), \(y = -4\)
- For \(x = 5\), \(y = -10\)
- For \(x = 6\), \(y = -13\)
We can substitute \(x\) values from the table into the equation \(y = -\frac{1}{2}x\) to find the corresponding \(y\)-values.
Let's check each \(x\)-value from the table:
-
For \(x = -1\): \[ y = -\frac{1}{2}(-1) = \frac{1}{2} \quad (\text{not equal to } 8) \]
-
For \(x = 3\): \[ y = -\frac{1}{2}(3) = -\frac{3}{2} \quad (\text{not equal to } -4) \]
-
For \(x = 5\): \[ y = -\frac{1}{2}(5) = -\frac{5}{2} \quad (\text{not equal to } -10) \]
-
For \(x = 6\): \[ y = -\frac{1}{2}(6) = -3 \quad (\text{not equal to } -13) \]
None of the \(x\)-values correspond to a point that lies on the line \(y = -\frac{1}{2}x\). Let's check if there are any intersections:
To see which of the table points might be on the line \(2x + 4y = 0\) (or \(y = -\frac{1}{2}x\)), calculate \(y\) using the table points:
For this, we check \(y = -\frac{1}{2} \cdot 6 = -3\):
- This has \(x = 6\) but \(y = -13 \neq -3\).
Thus, none of the points in the table satisfy the line equation \(2x + 4y = 0\).
Since there are no points of intersection from the given table data, there is no solution to this system of equations.
Hence, the answer to the question "What is the \(x\)-value of the solution to the system?" is that the system has no solution. Therefore, no \(x\)-value exists for the solution.