7- In triangle \( P Q R \), the measure of angle \( P \) is \( (3 x+5)^{\circ} \). The measure of angle \( Q \) is \( (2 x+9)^{\circ} \), and the measure of angle \( R \) is \( (4 y+5)^{\circ} \). If side \( Q R \) is extended through point \( R \) to point \( S \), and the measure of angle \( P R S \) is \( (x+y)^{\circ} \). What is the value of \( x+y \) ?

In triangle \( PQR \), the sum of the angles is given by: \[ \angle P + \angle Q + \angle R = 180^\circ \] Substituting the expressions for the angles, we have: \[ (3x + 5) + (2x + 9) + (4y + 5) = 180 \] Combining like terms: \[ 3x + 2x + 4y + 5 + 9 + 5 = 180 \] This simplifies to: \[ 5x + 4y + 19 = 180 \] Subtracting 19 from both sides gives: \[ 5x + 4y = 161 \quad \text{(1)} \] Next, we consider the angle \( PRS \). Since \( QR \) is extended to \( S \), we have: \[ \angle PRS = 180^\circ - \angle R \] Substituting the expression for \( \angle R \): \[ \angle PRS = 180^\circ - (4y + 5) = 175 - 4y \] We are given that: \[ \angle PRS = (x + y)^\circ \] Setting the two expressions for \( \angle PRS \) equal to each other: \[ 175 - 4y = x + y \] Rearranging this equation gives: \[ 175 = x + 5y \quad \text{(2)} \] Now we have a system of equations: 1. \( 5x + 4y = 161 \) 2. \( x + 5y = 175 \) We can solve this system using substitution or elimination. Let's solve equation (2) for \( x \): \[ x = 175 - 5y \] Substituting this expression for \( x \) into equation (1): \[ 5(175 - 5y) + 4y = 161 \] Expanding this gives: \[ 875 - 25y + 4y = 161 \] Combining like terms results in: \[ 875 - 21y = 161 \] Subtracting 875 from both sides: \[ -21y = 161 - 875 \] \[ -21y = -714 \] Dividing both sides by -21: \[ y = \frac{714}{21} = 34 \] Now substituting \( y = 34 \) back into the expression for \( x \): \[ x = 175 - 5(34) \] \[ x = 175 - 170 = 5 \] Finally, we find \( x + y \): \[ x + y = 5 + 34 = 39 \] Thus, the value of \( x + y \) is: \[ \boxed{39} \]