simultaneous equation, help.

ironictoaster

I have being trying this for ages. I don't know where to begin.

Solve x, y and z

x + y = -1
2x - z = -2
y + 3z = 10

I have done questions where each equation has a x y and z variable but, this one has a variable missing in each equation.

Please help!

eden_my_ass

Just substitute a variable from one equation and use it in another...

x+y=-1
so
x=-1-y

now substitute (-1-y) for x in 2x-z = -2...then you're back to what you are used to doing, solve for y, then z...and finally x

Using the first equation, express y in terms of x. In the second equation, express z in terms of x. Then, sub both of these rearranged equations into the third equation, this will let you find x, and hence find y and z using the first two equations.

eden_my_ass

lumo22 wrote: »

Number one...you're wrong, I'll let you see where, and number two, isn't there a charter to this forum saying not to give complete answers...helps noone...especially incorrect ones

Edit...I wasn't talking to myself there or referring to JammyDodgers post, just in case theres any confusion, post deleted

ironictoaster

eden_my_ass wrote: »

Just substitute a variable from one equation and use it in another...

x+y=-1
so
x=-1-y

now substitute (-1-y) for x in 2x-z = -2...then you're back to what you are used to doing, solve for y, then z...and finally x

OK so,

So Let x= -1 - y

Subbing that into 2nd equation gives :

2 -2y - z = -2 then,

-2y - z = 0

Then using that with the 3rd equation

-2y - z = 0
y + 3z = 10

Eliminating the y's gives z = 5
Eliminating the z's gives y = 2

Sub in y = 2 into x + y = -1 therefore,

x = -3

x = -3, y = 2 and z = 5

Is this correct?

eden_my_ass

creggy wrote: »

-2y - z = 0
y + 3z = 10

Eliminating the y's gives z = 5

Almost there, you went slightly wrong here though....double the lower line to allow youself eliminate the ys, then you should get 5z=20.....carry on

ironictoaster

Oh ****e thanks!

Edit: So x = -3, y = 2 and z = 4 ?

eden_my_ass

creggy wrote: »

Oh ****e thanks!

Edit: So x = -3, y = 2 and z = 4 ?

Test them with your equations (try 2x - z = -2)...and yes thats my way of saying z is right, but x and y aren't....its the simple mistakes that always mess it up in the end but at least you have the idea right

ironictoaster

I cannot find these mistakes I do know what the answers are though after you confirming that z is 4.

EDIT: I'm an idiot, silly sign mistakes.

Thanks for the help, I should be back with more interesting questions from the world of mathematics!

ironictoaster

Ok, I have another one that giving me some problems.

2x - 3y = 1
x^2 - 2xy - 3y^2 = -3

I know questions with a quadratic you usually make x or y the subject in the linear and sub into the quadratic. However, 2x = 3y + 1 doesn't work when subbing into the quadratic!

Help!

Fremen

creggy wrote: »

However, 2x = 3y + 1 doesn't work when subbing into the quadratic!

Help!

You're not required to stick to whole numbers. If 2x = 3y + 1, then x = (3y + 1)/2.
You had the right idea, now proceed exactly as you would have otherwise.

Edit: you can visualise the equation here. The answer is also given, so you can check it.

ironictoaster

Fremen wrote: »

You're not required to stick to whole numbers. If 2x = 3y + 1, then x = (3y + 1)/2.
You had the right idea, now proceed exactly as you would have otherwise.

Edit: you can visualise the equation here. The answer is also given, so you can check it.

So sub in x = (3y + 1)/2 into the quadratic. Sorry for the silly questions