Question

Hey peeps (: Need a little help with this question for Simultaneous Equations. 3s square + 2t square = 11 and 3s + 2t = 1. Need a working too please (: Thanks!

Answer 1

3s^2+2t^2=11
3s+2t=1--->s=(1-2t)/3
replace it in the 1st eq

Answer 2

Ooh, okay (: Then t would be...? Could I have a full working too? Thanks!

Answer 3

3s square means 3s^2 or 9s^2?

Answer 4

3s^2 (:

Answer 5

What will the full working be?

Answer 6

U got the Wolfram link and a method, just follow it....

Answer 7

Doesn't show workings tho :\

Answer 8

Need some more help. All I need is just a full working right now. Thanks! (:

Answer 9

U don't need help, u just need someone to do it all for u, u mean.

Answer 10

... Erm, no, I've already done the working. Just need to compare it with some elses', that's all.. (:

Answer 11

I guess Jimmy's solving it

Answer 12

i'll start u off with the next step
3s^2+2t^2=11
s=(1-2t)/3
now plug this value of s into the first equation:
3 [(1-2t)/3]^2 + 2t^2 = 11
3 ( 1 - 4t + 4t^2) / 9 + 2t^2 - 11 = 0
(1/3)(1-4t+4t^2) + 2t^2 - 1 = 0
multiply through by 3
1 - 4t + 4t^2 + 6t^2 - 33 = 0
10t^2 - 4t - 32 = 0
5t^2 - 2t - 16 = 0
solve this equation to get t
then substitute for t in the second of your original equatios to find s.

Answer 13

So.. s = -1 and t = 2?

Answer 14

u have to solve 5t^2 - 2t - 16 = 0 this 1st and from the values u'll get replace and see who makes the eq true