Year 12s: what will be the first way you research your future options? >>

Simulation equations with a quadratic

username4426374

2x^2-y^2=31
x+3y=1

So have I rearranged the second equation correctly? I have done: 3y=1-x

Reply 1

RDKGames

Study Forum Helper

Original post by Steve445

2x^2-y^2=31
x+3y=1

Ok, so you know what the first step is?

Reply 2

RDKGames

Study Forum Helper

Original post by Steve445

So have I rearranged the second equation correctly? I have done: 3y=1-x

You're not wrong, but you're also not quite there with this.

Whenever we solve two equations simultaneously, we are effectively looking for a solution(s)

(x,y)

which satisfy both equations at once. Graphically, these are the coordinates where the two equations intersect each other. At the point of intersection, the y-coordinates are the same, and the x-coordinates are the same.

Since we know that the y-coordinates are the same, this means that

y

2x^2-y^2=31

is exactly the same as the

y

x+3y = 1

. This means if you can rearrange the second equation for

y

on its own, then you can substitute that into the first equation and obtain a relation that

x

solely must satisfy.

So far, you have found what

3y

is, but you need to know what

y

on its own is. So what's one more step??

Also, as a side note, you may be used to 'rearranging for y and subbing it into the other equation' type of approach, but I also said that the x-coordinates are the same. So there is nothing stopping you from rearranging the second equation for

x

instead and subbing that into the first equation. In fact, this is a much simpler alternative than that which you are doing because we avoid fractions this way and the algebra is less messy hence making less room for error.

Reply 3

username4426374

Original post by RDKGames

Ok, so you know what the first step is?

I know how to do the rest, but was my first bit correct

Edit: Nvm

Original post by Steve445

...

That's wrong.

You substituted in

3y = 1-x

for

y

when you should've substituted in

y = \dfrac{1-x}{3}

instead.

Reply 6

username4426374

Original post by RDKGames

That's wrong.

You substituted in

3y = 1-x

for

y

when you should've substituted in

y = \dfrac{1-x}{3}

instead.

Oh, okay... So what I did was completely wrong?

Reply 7

RDKGames

Study Forum Helper

Original post by Steve445

Oh, okay... So what I did was completely wrong?

No, you just subbed in the wrong thing. The method is fine.

Reply 8

idontkn

Why not rearrange and get x to be the subject? And then plug that in to the quadratic.

Reply 9

xAikx

Would of been easier here to make x the subject of the second equation.

So x = 1 - 3y.

You can then substitute this expression for x in your first equation and solve for your possible values for x.

From there you can find y.