\( \left. \begin{array} { c } { 15 , \frac { 3 } { 2 } x + 2 y = 12 } \\ { x - 2 y = 6 } \end{array} \right. \)
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Beyond the Answer
To solve the system of equations given, we have: 1. \( \frac{3}{2}x + 2y = 12 \) 2. \( x - 2y = 6 \) Let’s start with the second equation, which is straightforward to work with. We can express \( x \) in terms of \( y \): \[ x = 2y + 6 \] Now, we can substitute this expression for \( x \) into the first equation: \[ \frac{3}{2}(2y + 6) + 2y = 12 \] Expanding this gives us: \[ 3y + 9 + 2y = 12 \] Combining like terms results in: \[ 5y + 9 = 12 \] Now, solving for \( y \): \[ 5y = 12 - 9 \] \[ 5y = 3 \] \[ y = \frac{3}{5} \] Next, we substitute \( y \) back into the expression for \( x \): \[ x = 2\left(\frac{3}{5}\right) + 6 \] Calculating this yields: \[ x = \frac{6}{5} + 6 = \frac{6}{5} + \frac{30}{5} = \frac{36}{5} \] So, the solution to the system of equations is: \( x = \frac{36}{5} \) and \( y = \frac{3}{5} \). Therefore, the final answer is: \( \left( x, y \right) = \left( \frac{36}{5}, \frac{3}{5} \right). \)