Find Jue truts set of the following \( \operatorname{Simultonears~equation:~} \) \( \frac{5}{6} x-\frac{3}{4} y=2 \) and \( \frac{1}{2} x+\frac{2}{3} y=\frac{5}{2} \)

To find the solution set for the simultaneous equations 1) \( \frac{5}{6} x - \frac{3}{4} y = 2 \) 2) \( \frac{1}{2} x + \frac{2}{3} y = \frac{5}{2} \), we can solve these equations together using substitution or elimination. First, let's eliminate the fractions by multiplying the entire first equation by 12 (the least common multiple of 6 and 4): \[ 12 \left(\frac{5}{6} x - \frac{3}{4} y\right) = 12 \cdot 2 \] This simplifies to: \[ 10x - 9y = 24 \quad (1) \] Now, for the second equation, multiply by 6 (the least common multiple of 2 and 3): \[ 6 \left(\frac{1}{2} x + \frac{2}{3} y\right) = 6 \cdot \frac{5}{2} \] This simplifies to: \[ 3x + 4y = 15 \quad (2) \] Now we have the equations: 1) \( 10x - 9y = 24 \) 2) \( 3x + 4y = 15 \) Next, we can use the elimination method. Let's multiply the second equation by 3 to align the coefficients of \(x\): \[ 3(3x + 4y) = 3(15) \] Which gives us: \[ 9x + 12y = 45 \quad (3) \] Now we can multiply the first equation by 1 to keep it the same: - From (1): \( 10x - 9y = 24 \) Now we have: 1) \( 10x - 9y = 24 \) 3) \( 9x + 12y = 45 \) Now let's isolate \(x\) from the first equation: \[ 10x = 24 + 9y \] \[ x = \frac{24 + 9y}{10} \] Now we can substitute \(x\) in the third equation (4). Substituting \(x\): \[ 9\left(\frac{24 + 9y}{10}\right) + 12y = 45 \] Multiply through by \(10\) to eliminate the fraction: \[ 9(24 + 9y) + 120y = 450 \] Expanding: \[ 216 + 81y + 120y = 450 \] Combining like terms: \[ 216 + 201y = 450 \] Subtract 216: \[ 201y = 234 \] Divide by 201: \[ y = \frac{234}{201} = \frac{78}{67} \approx 1.16 \] Now substitute \(y\) back to find \(x\): Using \(x = \frac{24 + 9y}{10}\): Substituting \(y\): \[ x = \frac{24 + 9\left(\frac{234}{201}\right)}{10} \] Calculating this gives us \(x \approx 2.59\). Thus, the solution set (truth set) of the simultaneous equations is \((x, y) \approx (2.59, 1.16)\).