Additional Mathematics 0606 / Simultaneous equations and quadratics / 3.23.2: Exponential simultaneous equations and reciprocal exponential0606/11/M/J/24 — Question 9 · 9 marksMark as done · Save for later · Show all solutions(a)Solve the following simultaneous equations.ex+y×e3x−2y=1e^{x + y} \times e^{3x - 2y} = 1ex+y×e3x−2y=1x2y=256x^{2}y = 256x2y=256[5]▸ Answer(b)Solve the equation 10e2x−1−11=6e1−2x10\mathrm{e}^{2x - 1} - 11 = 6\mathrm{e}^{1 - 2x}10e2x−1−11=6e1−2x, giving your answer in exact form.[4]▸ Answer▸ Official mark scheme← 3.1: Sine equation with phase shift in radians3.13: Completing the square for a maximum value →
