To determine the conditions under which the system of equations has more than one solution, we need to assess when the two equations are dependent (which means they represent the same line).
The two equations are:
- \( x + 2y = -a \)
- \( -2x + by = -14 \)
We can rearrange these equations into the slope-intercept form to find conditions for dependency.
From the first equation, we can express \(y\):
\[ 2y = -a - x \implies y = -\frac{1}{2}x - \frac{a}{2} \]
The slope (m) of this line is \(m_1 = -\frac{1}{2}\) and the y-intercept (b) is \(-\frac{a}{2}\).
Now, we manipulate the second equation:
\[ -2x + by = -14 \implies by = 2x - 14 \implies y = \frac{2}{b}x - \frac{14}{b} \]
The slope of this line is \(m_2 = \frac{2}{b}\) and the y-intercept is \(-\frac{14}{b}\).
For the lines to be dependent (i.e., have more than one solution), their slopes must be equal. Therefore, we set the slopes equal to each other:
\[ -\frac{1}{2} = \frac{2}{b} \]
Cross multiplying gives:
\[ -b = 4 \implies b = -4 \]
Next, since the lines are identical, the y-intercepts must be the same as well. We set the y-intercepts equal:
\[ -\frac{a}{2} = -\frac{14}{b} \]
Substituting \(b = -4\):
\[ -\frac{a}{2} = -\frac{14}{-4} \implies -\frac{a}{2} = \frac{14}{4} = \frac{7}{2} \]
Multiplying both sides by -1 gives:
\[ \frac{a}{2} = -\frac{7}{2} \]
Multiplying both sides by 2 yields:
\[ a = -7 \]
Now we have discovered that \(a = -7\) and \(b = -4\).
To find \(a - b\):
\[ a - b = -7 - (-4) = -7 + 4 = -3 \]
Thus, the value of \(a - b\) is
\[ \boxed{-3}. \]