10. \( \begin{array}{l}\frac{y}{2}-\frac{x+3}{4}=\frac{3}{2} \\ \frac{y+3}{4}-\frac{x}{2}=\frac{3}{2}\end{array} \)
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To solve the system of equations: 1. \(\frac{y}{2} - \frac{x + 3}{4} = \frac{3}{2}\) 2. \(\frac{y + 3}{4} - \frac{x}{2} = \frac{3}{2}\) We'll first eliminate the fractions by multiplying each equation by the least common multiple of the denominators. For the first equation, the least common multiple of 2 and 4 is 4: \[ 4 \left(\frac{y}{2}\right) - 4\left(\frac{x + 3}{4}\right) = 4\left(\frac{3}{2}\right) \] This simplifies to: \[ 2y - (x + 3) = 6 \implies 2y - x - 3 = 6 \implies 2y - x = 9 \tag{1} \] For the second equation, the least common multiple of 4 and 2 is 4: \[ 4\left(\frac{y + 3}{4}\right) - 4\left(\frac{x}{2}\right) = 4\left(\frac{3}{2}\right) \] This simplifies to: \[ y + 3 - 2x = 6 \implies y - 2x + 3 = 6 \implies y - 2x = 3 \tag{2} \] Now we have a new system of equations: 1. \(2y - x = 9\) 2. \(y - 2x = 3\) Now we can solve for \(y\) in terms of \(x\) using equation (2): \[ y = 2x + 3 \] Substituting \(y\) back into equation (1): \[ 2(2x + 3) - x = 9 \] Expanding it, we get: \[ 4x + 6 - x = 9 \implies 3x + 6 = 9 \implies 3x = 3 \implies x = 1 \] Now substitute \(x = 1\) back into the equation for \(y\): \[ y = 2(1) + 3 = 2 + 3 = 5 \] Thus, the solution to the system of equations is: \[ \boxed{(1, 5)} \]