Solve the following simultaneous equations. ex + ey = 5, 2ex-3ey = 8

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

1. [M1] Solves 3eˣ + 3eʸ = 15 and 2eˣ - 3eʸ = 8 oe by elimination as far as 3eˣ + 2eˣ = 23 or substitutes eʸ = 5-eˣ into 2eˣ - 3eʸ = 8 oe OR Solves 2eˣ + 2eʸ = 10 and 2eˣ - 3eʸ = 8 oe by elimination as far as 2eʸ + 3eʸ = 2 or substitutes eˣ = 5-eʸ into 2eˣ - 3eʸ = 8 oe
2. [A1] eˣ = 23/5 or eʸ = -2/5 oe
3. [A1] x = ln4.6[=1.53] oe or y = ln 0.4[=-0.916] oe If M0 scored SC1 for using their expression of the form ceˣ = d to give x = ln(d/c) provided d/c > 0
4. [M1] FT their eʸ or eˣ Finds the other value, eʸ or eˣ, by substituting their eˣ or eʸ
5. [A1] y = ln0.4[=-0.916] oe or x = ln4.6[=1.53] oe
6. [M1] e^(2t-1-(5t-3)) = 5 or e^(5t-3-(2t-1)) = 1/5 oe
7. [A1] e^(2-3t) = 5 or e^(3t-2) = 1/5
8. [M1] FT their e^(a-bt) = 5 or their e^(ct-d) = 1/5 where a, b, c and d are positive integers 2-3t = ln 5 or 3t-2=ln(1/5)
9. [A1] t = (2-ln5)/3 or t = (2+ln 0.2)/3 or 0.13[0] oe