Find the solution to the system by the addition (elimination) method. Check your answers. \[ \begin{aligned} 15 x+7 y=14 & \text { (1) } \\ 5 x+4 y=8 & \text { (2) }\end{aligned} \] What is the solution to the system? \( \square \) (Type an ordered pair.)

To solve the system of equations using the elimination method, we can first manipulate the two equations to eliminate one of the variables. Starting with the equations: 1. \( 15x + 7y = 14 \) (1) 2. \( 5x + 4y = 8 \) (2) We can multiply the second equation by 3 to align the coefficients of \(x\): \[ 3(5x + 4y) = 3(8) \] This gives us: \[ 15x + 12y = 24 \quad (3) \] Now, we have: 1. \( 15x + 7y = 14 \) (1) 2. \( 15x + 12y = 24 \) (3) Next, we can subtract equation (1) from equation (3): \[ (15x + 12y) - (15x + 7y) = 24 - 14 \] This simplifies to: \[ 5y = 10 \] Dividing both sides by 5 gives us: \[ y = 2 \] Now, we can substitute \(y = 2\) back into one of the original equations to find \(x\). We can use equation (2): \[ 5x + 4(2) = 8 \] This simplifies to: \[ 5x + 8 = 8 \] Subtracting 8 from both sides: \[ 5x = 0 \] Dividing by 5 gives: \[ x = 0 \] Thus, the solution to the system is: \[ \boxed{(0, 2)} \] To check our answers, we can substitute \(x = 0\) and \(y = 2\) back into both original equations: For equation (1): \[ 15(0) + 7(2) = 0 + 14 = 14 \quad \text{(True)} \] For equation (2): \[ 5(0) + 4(2) = 0 + 8 = 8 \quad \text{(True)} \] Both equations are satisfied, confirming our solution is correct.