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explain how the top equation transforms into the bottom, solving for A and C 

The image you uploaded contains two algebraic equations:

1. \( 1 = -4A + (2c + 1) \)
2. \( 2A + c = 2 \)

We'll solve this system of equations to find the values of \( A \) and \( c \).

First, let's rearrange the second equation to solve for \( c \) in terms of \( A \):

\[ c = 2 - 2A \]

Next, substitute this expression for \( c \) into the first equation:

\[ 1 = -4A + (2(2 - 2A) + 1) \]

Simplifying the equation:

\[ 1 = -4A + (4 - 4A + 1) \]
\[ 1 = -4A + 4 - 4A + 1 \]
\[ 1 = -8A + 5 \]

Now, solve for \( A \):

\[ -8A = 1 - 5 \]
\[ -8A = -4 \]
\[ A = \frac{-4}{-8} \]
\[ A = 0.5 \]

Now that we have \( A \), substitute it back into the equation for \( c \):

\[ c = 2 - 2 \times 0.5 \]
\[ c = 2 - 1 \]
\[ c = 1 \]

So, the solution to the system is \( A = 0.5 \) and \( c = 1 \).

Do you need further explanation on any step, or have other questions?

Here are some follow-up questions related to this topic:

1. How does the solution change if the coefficients in the equations are altered?
2. What happens to the solution if a third variable, such as \( B \), is introduced into the system?
3. Could you solve similar equations using matrix methods?
4. How does a change in the constant term (like the '1' becoming '2' in the first equation) affect the solution?
5. What are some applications of solving systems of linear equations in real-life scenarios?

Tip: In solving systems of linear equations, always consider checking your solution by substituting back into the original equations to confirm accuracy.

Explain how the top equation transforms into the bottom, solving for A and C

This solution explains how to solve a system of linear equations using the substitution method. It involves transforming two equations to find the values of A and C. Suitable for high school students in Grades 9-12.

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