We are given the following system of equations:

1. $16b = 6f$
2. $c = b + 4a$
3. $d = b + 2a$
4. $e = 4b + 3a$
5. $a = f$ (but $f \neq a$, so this equation is only helpful when considering the value of $f$)
6. $c = 130$
7. $d = 30$
8. $e = 134$

Step 1: Use the known values for $c$, $d$, and $e$

From equations (2), (3), and (4), we can substitute the known values.

• $c = b + 4a = 130$ → Equation (2) gives: $b + 4a = 130 \quad \text{(Equation 2)}$

• $d = b + 2a = 30$ → Equation (3) gives: $b + 2a = 30 \quad \text{(Equation 3)}$

• $e = 4b + 3a = 134$ → Equation (4) gives: $4b + 3a = 134 \quad \text{(Equation 4)}$

Step 2: Solve the system of equations

We have the system:

$b + 4a = 130 \quad \text{(Equation 2)}$ $b + 2a = 30 \quad \text{(Equation 3)}$ $4b + 3a = 134 \quad \text{(Equation 4)}$

First, subtract Equation (3) from Equation (2):

$(b + 4a) - (b + 2a) = 130 - 30$ $2a = 100$ $a = 50$

Step 3: Substitute $a = 50$ into the equations

Substitute $a = 50$ into Equation (3) to find $b$:

$b + 2(50) = 30$ $b + 100 = 30$ $b = 30 - 100 = -70$

Step 4: Find $f$

Now substitute $b = -70$ and $a = 50$ into Equation (1) to solve for $f$:

$16b = 6f$ $16(-70) = 6f$ $-1120 = 6f$ $f = \frac{-1120}{6} = -\frac{560}{3} \approx -186.67$

Final Answer:

• $a = 50$
• $b = -70$
• $f = -\frac{560}{3} \approx -186.67$

Let me know if you need any more clarification or details!

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Related Questions:

1. How would the solution change if $c = 120$ instead of 130?
2. What happens if $e = 140$ instead of 134?
3. How can you verify that the value of $f$ is correct?
4. What is the relationship between $f$ and $a$ given the condition $f \neq a$?
5. How do you handle systems of equations when one equation involves a non-integer result?

Tip: To efficiently solve systems of equations, use substitution or elimination methods depending on the structure of the equations.