Solve the system of equations: 2x + y = 9 and 3x2 + 2y2 + x - y = 45

Let's solve this system of equations simultaneously. Step 1: Express one variable in terms of the other from the linear equation. The given equations are: 1) 2x + y = 9 2) 3x^2 + 2y^2 + x - y = 45 From equation (1), we can express y in terms of x: y = 9 - 2x Step 2: Substitute this expression for y into the second equation. Substitute y = 9 - 2x into equation (2): 3x^2 + 2(9 - 2x)^2 + x - (9 - 2x) = 45 Step 3: Expand and simplify the equation to form a quadratic equation in terms of x. First, expand (9 - 2x)^2: (9 - 2x)^2 = 9^2 - 2(9)(2x) + (2x)^2 = 81 - 36x + 4x^2 Now substitute this back into the equation: 3x^2 + 2(81 - 36x + 4x^2) + x - 9 + 2x = 45 3x^2 + 162 - 72x + 8x^2 + x - 9 + 2x = 45 Combine like terms: (3x^2 + 8x^2) + (-72x + x + 2x) + (162 - 9) = 45 11x^2 - 69x + 153 = 45 Subtract 45 from both sides to set the equation to 0: 11x^2 - 69x + 153 - 45 = 0 11x^2 - 69x + 108 = 0 Step 4: Solve the quadratic equation for x using the quadratic formula. For the equation ax^2 + bx + c = 0, the quadratic formula is x = -b ± sqrt(b^2 - 4ac)2a. Here, a = 11, b = -69, c = 108. Calculate the discriminant = b^2 - 4ac: = (-69)^2 - 4(11)(108) = 4761 - 4752 = 9 Now, substitute the values into the quadratic formula: x = -(-69) ± sqrt(9)2(11) x = (69 ± 3)/(22) This gives two possible values for x: x_1 = (69 + 3)/(22) = (72)/(22) = (36)/(11) x_2 = (69 - 3)/(22) = (66)/(22) = 3 Step 5: Find the corresponding values for y using y = 9 - 2x. For x_1 = (36)/(11): y_1 = 9 - 2((36)/(11)) y_1 = 9 - (72)/(11) y_1 = (9 × 11)/(11) - (72)/(11) y_1 = (99 - 72)/(11) y_1 = (27)/(11) This gives the solution pair ((36)/(11), (27)/(11)). For x_2 = 3: y_2 = 9 - 2(3) y_2 = 9 - 6 y_2 = 3 This gives the solution pair (3, 3). The solutions to the pair of equations are (x, y) = ((36)/(11), (27)/(11)) and (x, y) = (3, 3). That's 2 down. 3 left today — send the next one.