Combine space c equals 4 invisible times a minus 3 space and space 3 invisible times b equals 3 invisible times a minus 6  to give an equation for c not involving a.

We are given two equations:

1. \( c = 4(a - 3) \)
2. \( 3b = 3a - 6 \)

### Step 1: Solve the second equation for \( a \)
We start with the second equation:
\[
3b = 3a - 6
\]
Divide the entire equation by 3:
\[
b = a - 2
\]
Now, solve for \( a \):
\[
a = b + 2
\]

### Step 2: Substitute \( a = b + 2 \) into the first equation
Now, substitute \( a = b + 2 \) into the first equation for \( c \):
\[
c = 4(a - 3)
\]
Substitute \( a = b + 2 \):
\[
c = 4((b + 2) - 3)
\]
Simplify the expression inside the parentheses:
\[
c = 4(b + 2 - 3)
\]
\[
c = 4(b - 1)
\]
Thus, the equation for \( c \) that does not involve \( a \) is:
\[
c = 4(b - 1)
\]

### Final Equation:
\[
c = 4b - 4
\]

Would you like any additional details or explanations?

Here are some related questions to explore:

1. What is the geometric interpretation of the equation \( c = 4(b - 1) \)?
2. Can you derive a similar equation if the constants were different in the given system?
3. How would you express \( b \) in terms of \( c \) from the final equation?
4. What happens to \( c \) as \( b \) increases linearly?
5. How can you verify the solution by plugging values into the original equations?

**Tip:** Always simplify equations as much as possible before substituting expressions!

Combine c = 4(a - 3) and 3b = 3a - 6 to give an equation for c not involving a.

Learn how to combine the equations c = 4(a - 3) and 3b = 3a - 6 to find an equation for c that doesn't involve a. This step-by-step solution explains algebraic substitution and linear equation techniques, ideal for students in Grades 8-10.

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