Maths Question (Simultaneous Equations), Help needed

PKdude

Hi, I've been trying to do this maths question for a while now and I've had no luck, I don't know how to do it. It's Simultaneous Equations which I'm usually good at. Here it is:

2r - s = 10
rs - s(to the power of 2) = 12

Peregrine

Substitute one into the other. More specifically, substitute the linear equation into curve/circle.

[latex]
2r = 10 + S[/latex]

[latex]r = 5 + \frac{S}{2}
[/latex]


[latex](5 + \frac{S}{2})(S) - S^2 = 12[/latex]

Solve for two values of [latex]S[/latex]
Put them back into your linear equation and get two values for [latex]r[/latex]

Mr Pseudonym

Nimrod 7 wrote: »

Substitute one into the other. More specifically, substitute the linear equation into curve/circle.

[latex]
2r = 10 + S[/latex]

[latex]r = 5 + \frac{S}{2}
[/latex]


[latex](5 + \frac{S}{2})(S) - S^2 = 12[/latex]

Solve for two values of [latex]S[/latex]
Put them back into your linear equation and get two values for [latex]r[/latex]

Very fancy - Nimrod went all Latex on your ass! Ofc, you could also have substituted [latex]{S} = 2r - 10[/latex]

You can compare your answers with this.

PKdude

I got it now. Thanks very much guys, I would have been stuck on that one for ages before giving up! Thanks again!

Peregrine

Mr Pseudonym wrote: »

Very fancy - Nimrod went all Latex on your ass! Ofc, you could also have substituted [latex]{S} = 2r - 10[/latex]

:pac:

True, it makes no odds which one you substitute.

If you substitute [latex]r[/latex], you'd have to square it. And if you substitute [latex]S[/latex], you'd have a fraction so you'd have to mutiply the whole line by 2.