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Surds in simultaneous equations?

TheTechN1304

I'm really stuck on how to do this...Could someone please give me an idea of how to work out this question? :confused:

5x - 3y = 41

(7 root 2)x + (4 root 2)y = 82

(I've attached a link to the question) http://imgur.com/l5GdiVP

Thankks!

Reply 1

ChemEng

Original post by Dominicoben

I'm really stuck on how to do this...Could someone please give me an idea of how to work out this question? :confused:

5x - 3y = 41

(7 root 2)x + (4 root 2)y = 82

(I've attached a link to the question) http://imgur.com/l5GdiVP

Thankks!

I gave it a go. There's probably a faster way of doing it.

First i multiplied everything is the second equation by r2/2 (r=root) to get 7x + 4y = 41r2
then i multiplied first equation by 4 and this equation by 3 so i could cancel the ys

Spoiler

Reply 2

davros

Original post by Dominicoben

I'm really stuck on how to do this...Could someone please give me an idea of how to work out this question? :confused:

5x - 3y = 41

(7 root 2)x + (4 root 2)y = 82

(I've attached a link to the question) http://imgur.com/l5GdiVP

Thankks!

To be honest, you just treat it like any other pair of simultaneous equations with "nice" coefficients - you can either multiply both equations by new numbers to get the coefficients of x or y the same and then take one away from the other; OR rearrange one equation to give y in terms of x (or vice versa) and then substitute in the other equation.

It's just a nasty algebra slog I'm afraid!

Reply 3

BlueSam3

An alternative solution:

Spoiler

Reply 4

samir12

You could multiply the first equation by root(2), and then multiply both equations accordingly to get the same coefficients for either x or y and then proceed to add/subtract the equations to eliminate y or x.

Or multiply the second equation by root(2) to get rid of the surds on the x and y coefficients and proceed as normal.