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Sagot :
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]
[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]
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