d) \( 3 x+2 y=60 ; \quad 3 x-3 y=45 \)

To solve the system of equations given, we can use the substitution or elimination method. Let's eliminate \(x\) first. By multiplying the first equation by 1 and the second by 1, we can align the coefficients: 1. \( 3x + 2y = 60 \) 2. \( 3x - 3y = 45 \) Now, subtract the second equation from the first to eliminate \(x\): \[ (3x + 2y) - (3x - 3y) = 60 - 45 \] This simplifies to: \[ 5y = 15 \] Dividing both sides by 5 gives: \[ y = 3 \] Now, substitute \(y = 3\) back into one of the original equations to find \(x\). We'll use the first equation: \[ 3x + 2(3) = 60 \] \[ 3x + 6 = 60 \] Subtract 6 from both sides: \[ 3x = 54 \] Now divide by 3: \[ x = 18 \] So the solution to the system of equations is \(x = 18\) and \(y = 3\).