This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Write the given system of equations. 2iz + (3-2i) = 1+i (1) (1-2i)z + (3+2i) = 5+6i (2) Step 2: Eliminate z. Multiply Equation (1) by (1-2i) and Equation (2) by 2i. Multiply Equation (1) by (1-2i): (1-2i)(2iz) + (1-2i)(3-2i) = (1-2i)(1+i) (2i - 4i^2)z + (3 - 2i - 6i + 4i^2) = (1 + i - 2i - 2i^2) (4+2i)z + (-1-8i) = (3-i) (3) Multiply Equation (2) by 2i: (2i)(1-2i)z + (2i)(3+2i) = (2i)(5+6i) (2i - 4i^2)z + (6i + 4i^2) = (10i + 12i^2) (4+2i)z + (-4+6i) = (-12+10i) (4) Step 3: Subtract Equation (4) from Equation (3). [(4+2i)z + (-1-8i)] - [(4+2i)z + (-4+6i)] = (3-i) - (-12+10i) (-1-8i+4-6i) = 3-i+12-10i (3-14i) = 15-11i Step 4: Solve for . = (15-11i)/(3-14i) Multiply the numerator and denominator by the conjugate of the denominator, 3+14i: = ((15-11i)(3+14i))/((3-14i)(3+14i)) = (15(3) + 15(14i) - 11i(3) - 11i(14i))/(3^2 - (14i)^2) = (45 + 210i - 33i - 154i^2)/(9 - 196i^2) = (45 + 177i + 154)/(9 + 196) = (199 + 177i)/(205) = (199)/(205) + (177)/(205)i Step 5: Substitute into Equation (1) to solve for z. 2iz + (3-2i) = 1+i 2iz = 1+i - (3-2i) 2iz = 1+i - (3-2i)((199+177i)/(205)) 2iz = (205(1+i) - (3-2i)(199+177i))/(205) First, calculate (3-2i)(199+177i): (3-2i)(199+177i) = 3(199) + 3(177i) - 2i(199) - 2i(177i) = 597 + 531i - 398i - 354i^2 = 597 + 133i + 354 = 951 + 133i Now substitute this back: 2iz = (205+205i - (951+133i))/(205) 2iz = (205+205i - 951 - 133i)/(205) 2iz = (-746 + 72i)/(205) Step 6: Solve for z. z = (-746 + 72i)/(205 × 2i) z = (-746 + 72i)/(410i) Multiply the numerator and denominator by -i: z = ((-746 + 72i)(-i))/(410i(-i)) z = (746i - 72i^2)/(-410i^2) z = (746i + 72)/(410) z = (72)/(410) + (746)/(410)i Simplify the fractions: z = (36)/(205) + (373)/(205)i z = (36)/(205) + (373)/(205)i 3 done, 2 left today. You're making progress.