Determination of Ea question: Edexcel unit 4

Hello, another maths question from me and I'm assuming a pretty dim one again!

I have been asked to work out Ea from the data in the table:
T: 293; 1/time: 0.0250; 1/T: 3.41 x 10-3; ln(1/time): -3.69
T: 303; 1/time: 0.500; 1/T: 3.30 x 10-3; ln(1/time): -3.00

You're told the initial volumes and concentrations are constant, the only difference is the change in temperature.

Also you're not allowed to draw a graph and you're given: ln rate = -Ea/R x 1/T + a constant [R]

I know that -Ea/R is equal to the gradient of the graph you aren't allowed to draw and that 1/T and ln(1/time) are the X and Y variables, but how do you know which way around to put them into the change in Y/change in X equation?

You have to use simultaneous equations as your only two unknowns are Ea and A

k = Ae^-Ea/RT

hence

lnk = lnA - Ea/RT

lnk - lnA = -Ea/RT

1/t is proportional to rate

hence

lnA - ln(1/t) = Ea/RT

substitute values from each experiment and solve for Ea

lnA - ln(1/t1) = Ea/RT1
lnA - ln(1/t2) = Ea/RT2
--------------------------------------------- subtract
[ln(1/t2) + ln(1/t1)] = Ea/RT1 - Ea/RT2

solve for Ea

Thank you for your reply. You don't have to do it like that though (and not being great at maths, I wouldn't do it like that). I was going down the route I explained and then didn't know which order to put the values into the gradient calculation.