https://file.aiquickdraw.com/mathbot/file/17301275033942119wcpa.jpg

The system of equations given in the image is:

\[
\begin{cases}
2a + b + 3c = 5 \\
5a + 2b + 8c = 8 \\
a + b + c = 8
\end{cases}
\]

To solve for \( a \), \( b \), and \( c \), we can use the elimination or substitution method. Here’s the step-by-step solution:

### Step 1: Express one of the variables in terms of the others
From the third equation:
\[
a + b + c = 8 \implies a = 8 - b - c
\]

### Step 2: Substitute \( a = 8 - b - c \) into the other equations
Using this substitution in the first equation:
\[
2(8 - b - c) + b + 3c = 5
\]
Expanding and simplifying:
\[
16 - 2b - 2c + b + 3c = 5
\]
\[
16 - b + c = 5
\]
\[
-b + c = -11 \implies b - c = 11 \tag{Equation 4}
\]

Now, substitute \( a = 8 - b - c \) into the second equation:
\[
5(8 - b - c) + 2b + 8c = 8
\]
Expanding and simplifying:
\[
40 - 5b - 5c + 2b + 8c = 8
\]
\[
40 - 3b + 3c = 8
\]
\[
-3b + 3c = -32 \implies -b + c = -\frac{32}{3}
\]



Solve the system of equations:

2a + b + 3c = 5
5a + 2b + 8c = 8
a + b + c = 8

Learn how to solve a system of linear equations with three variables using substitution and elimination methods. This problem involves algebraic concepts and is suitable for high school students in Grades 10-12.

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