a+f+e = 0
-5a+e = -7
-4f-e = 4
Find the values of a, f, e

We have the system of equations:

1. \( a + f + e = 0 \)
2. \( -5a + e = -7 \)
3. \( -4f - e = 4 \)

### Step 1: Express \( e \) in Terms of \( a \)
From Equation (2):  
\[
e = 5a - 7
\]

### Step 2: Express \( e \) in Terms of \( f \)
From Equation (3):  
\[
e = -4f - 4
\]

### Step 3: Equating the Expressions for \( e \)
Since both expressions represent \( e \), we set them equal to each other:  
\[
5a - 7 = -4f - 4
\]
Rearrange:  
\[
5a + 4f = 3
\]

### Step 4: Solve for \( a \) and \( f \)
Using Equation (1):  
\[
a + f + e = 0
\]
Substituting \( e = 5a - 7 \):  
\[
a + f + (5a - 7) = 0
\]
\[
6a + f = 7
\]

Now, we solve the system:

1. \( 6a + f = 7 \)
2. \( 5a + 4f = 3 \)

Using the first equation, solve for \( f \):  
\[
f = 7 - 6a
\]

Substituting into the second equation:  
\[
5a + 4(7 - 6a) = 3
\]
\[
5a + 28 - 24a = 3
\]
\[
-19a + 28 = 3
\]
\[
-19a = -25
\]
\[
a = \frac{25}{19}
\]

### Step 5: Find \( f \) and \( e \)
\[
f = 7 - 6\left(\frac{25}{19}\right)
\]
\[
f = 7 - \frac{150}{19}
\]
\[
f = \frac{133}{19} - \frac{150}{19}
\]
\[
f = \frac{-17}{19}
\]

\[
e = 5\left(\frac{25}{19}\right) - 7
\]
\[
e = \frac{125}{19} - \frac{133}{19}
\]
\[
e = \frac{-8}{19}
\]

### Final Answer:
\[
a = \frac{25}{19}, \quad f = \frac{-17}{19}, \quad e = \frac{-8}{19}
\]

Would you like a detailed verification or explanation?  

### Related Questions:
1. How can we verify the solution using substitution?
2. What happens if we change one of the equations slightly?
3. Can this system be solved using matrices?
4. How does elimination compare to substitution in solving this?
5. What are real-world applications of solving such systems?

**Tip:** Always double-check your solutions by substituting them back into the original equations!

Find the values of a, f, e given the system of equations: a + f + e = 0, -5a + e = -7, -4f - e = 4.

Learn how to solve the system of equations a + f + e = 0, -5a + e = -7, -4f - e = 4 using the substitution method. Step-by-step solution with detailed calculations, ideal for Grades 9-11 students.

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