\( 2 x^{2}-3 x y=-4 \) and \( 4^{x+y}=2^{y+4} \)

We are given the system of equations:   (1) 2x² – 3xy = –4   (2) 4^(x+y) = 2^(y+4) Step 1. Solve Equation (2): Write 4^(x+y) in terms of base 2. Since 4 = 2², we have   4^(x+y) = (2²)^(x+y) = 2^(2(x+y)). Thus, Equation (2) becomes   2^(2(x+y)) = 2^(y+4). Since the bases are the same and the exponential function is one-to-one, we can equate the exponents:   2(x+y) = y + 4. Now solve for y:   2x + 2y = y + 4   2x + 2y – y = 4   2x + y = 4   y = 4 – 2x. Step 2. Substitute y into Equation (1): Substitute y = 4 – 2x into 2x² – 3xy = –4:   2x² – 3x(4 – 2x) = –4. Expand and simplify:   2x² – 12x + 6x² = –4   (2x² + 6x²) – 12x = –4   8x² – 12x = –4. Bring –4 to the left side:   8x² – 12x + 4 = 0. Divide the entire equation by 2 to simplify:   4x² – 6x + 2 = 0. Step 3. Solve the Quadratic Equation: Use the quadratic formula for ax² + bx + c = 0. Here, a = 4, b = –6, and c = 2.   x = [–b ± √(b² – 4ac)] / (2a)   x = [6 ± √((–6)² – 4·4·2)] / (2·4)   x = [6 ± √(36 – 32)] / 8   x = [6 ± √4] / 8   x = [6 ± 2] / 8. Thus, the two solutions for x are:   x = (6 + 2)/8 = 8/8 = 1   x = (6 – 2)/8 = 4/8 = 1/2. Step 4. Find the Corresponding y Values: Recall y = 4 – 2x. For x = 1:   y = 4 – 2(1) = 4 – 2 = 2. For x = 1/2:   y = 4 – 2(1/2) = 4 – 1 = 3. Step 5. Verify the Solutions: For (x, y) = (1, 2):   Equation (1): 2(1)² – 3(1)(2) = 2 – 6 = –4 ✓   Equation (2): 4^(1+2) = 4³ = 64 and 2^(2+4) = 2⁶ = 64 ✓ For (x, y) = (1/2, 3):   Equation (1): 2(0.5)² – 3(0.5)(3) = 2(0.25) – 4.5 = 0.5 – 4.5 = –4 ✓   Equation (2): 4^(0.5+3) = 4^(3.5) = (2²)^(3.5) = 2^(7) = 128 and 2^(3+4) = 2^(7) = 128 ✓ The solutions satisfy both equations. Final Answer:   x = 1, y = 2  or  x = 1/2, y = 3.