\( \left. \begin{array} { l } { 2 a = 4 = } \\ { 3 a + b : 8 } \\ { a + b + c = 4 } \end{array} \right. \)
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The Deep Dive
To solve this system of equations, we first evaluate the first equation \(2a = 4\), which leads to \(a = 2\). Next, we substitute \(a = 2\) into the second equation \(3a + b = 8\). This gives us \(3(2) + b = 8\), simplifying to \(6 + b = 8\), and therefore \(b = 2\). Now we substitute \(a = 2\) and \(b = 2\) into the third equation \(a + b + c = 4\). This leads to \(2 + 2 + c = 4\), so \(c = 0\). In conclusion, the solution to the system is: \[ a = 2, \quad b = 2, \quad c = 0. \]