To solve the simultaneous equations:
[tex]\[
\begin{array}{c}
e + d + \frac{1}{f} = 0 \\
2e - d + \frac{3}{f} = -9 \\
3e + 2d - \frac{4}{f} = 19
\end{array}
\][/tex]
We need to find the values of [tex]\( e \)[/tex], [tex]\( d \)[/tex], and [tex]\( f \)[/tex] that satisfy all three equations.
### Step-by-Step Solution:
1. Express [tex]\( \frac{1}{f} \)[/tex] in terms of [tex]\( e \)[/tex] and [tex]\( d \)[/tex] from the first equation:
[tex]\[
e + d + \frac{1}{f} = 0 \implies \frac{1}{f} = -e - d \quad \text{(Equation 1.1)}
\][/tex]
2. Substitute [tex]\( \frac{1}{f} \)[/tex] from Equation 1.1 into the second equation:
[tex]\[
2e - d + \frac{3}{f} = -9 \\
2e - d + 3(-e - d) = -9 \\
2e - d - 3e - 3d = -9 \\
-e - 4d = -9 \quad \Rightarrow \quad e + 4d = 9 \quad \text{(Equation 2.1)}
\][/tex]
3. Substitute [tex]\( \frac{1}{f} \)[/tex] from Equation 1.1 into the third equation:
[tex]\[
3e + 2d - \frac{4}{f} = 19 \\
3e + 2d - 4(-e - d) = 19 \\
3e + 2d + 4e + 4d = 19 \\
7e + 6d = 19 \quad \text{(Equation 3.1)}
\][/tex]
4. We now have a system of two equations with two variables:
[tex]\[
\begin{cases}
e + 4d = 9 \quad \text{(Equation 2.1)} \\
7e + 6d = 19 \quad \text{(Equation 3.1)}
\end{cases}
\][/tex]
5. Solve this system of linear equations. First, we solve Equation 2.1 for [tex]\( e \)[/tex]:
[tex]\[
e = 9 - 4d \quad \text{(Equation 2.2)}
\][/tex]
6. Substitute Equation 2.2 into Equation 3.1:
[tex]\[
7(9 - 4d) + 6d = 19 \\
63 - 28d + 6d = 19 \\
63 - 22d = 19 \\
-22d = 19 - 63 \\
-22d = -44 \\
d = 2
\][/tex]
7. Substitute [tex]\( d = 2 \)[/tex] back into Equation 2.2 to find [tex]\( e \)[/tex]:
[tex]\[
e = 9 - 4(2) \\
e = 9 - 8 \\
e = 1
\][/tex]
8. Find [tex]\( f \)[/tex] using Equation 1.1:
[tex]\[
\frac{1}{f} = -e - d = -1 - 2 = -3 \\
f = -\frac{1}{3}
\][/tex]
### Final Solution:
The values that satisfy all three equations are:
[tex]\[
e = 1, \quad d = 2, \quad f = -\frac{1}{3}
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
\boxed{(1, 2, -\frac{1}{3})}
\][/tex]
