Non linear equation using Matrix

[markraz]

Hi I have a couple equations like this:


I want to solve this system using linear algebra matrix. Can this be done?
do I just put this in a matrix (see below) and solve Ax=b ??
or do the exponents have to be setup some special way?
thanks in advance

 

[Harry_the_cat]

I would set it up the LHS as


and solve for \(\displaystyle x^2\) and \(\displaystyle y^2\), then for x and y.

 

[markraz]

I would set it up the LHS as

View attachment 21986
and solve for \(\displaystyle x^2\) and \(\displaystyle y^2\), then for x and y.

Thanks appreciate it. This makes perfect sense. Just curious, is this the only way this can be done?? Thanks

 

[Harry_the_cat]

If you use the process of elimination and subtract the second equation from the first, you are left with y^2 = 4, ie, y=2 or -2. Sub back in and you can work out x. Much easier than using matrices in this case,

 

[markraz]

If you use the process of elimination and subtract the second equation from the first, you are left with y^2 = 4, ie, y=2 or -2. Sub back in and you can work out x. Much easier than using matrices in this case,

thanks so I would set it up like this:
3x^2 + 3y^2 = 27
3x^2 + 2y^2 = 23

3x^2 + 3y^2 - 27 = 3x^2 + 2y^2 - 23

move around and combine terms:

3x^2 + 3y^2 = 3x^2 + 2y^2 - 23 + 27
3x^2 - 3x^2+ 3y^2 - 2y^2= - 23 + 27

y^2 = 4

is this how to do it??
thanks appreciate it

 

[Harry_the_cat]

thanks so I would set it up like this:
3x^2 + 3y^2 = 27
3x^2 + 2y^2 = 23

3x^2 + 3y^2 - 27 = 3x^2 + 2y^2 - 23

move around and combine terms:

3x^2 + 3y^2 = 3x^2 + 2y^2 - 23 + 27
3x^2 - 3x^2+ 3y^2 - 2y^2= - 23 + 27

y^2 = 4

is this how to do it??
thanks appreciate it

Yeah that's one way. Don't forget that means that y=2 or y= -2.

 

[markraz]

Yeah that's one way. Don't forget that means that y=2 or y= -2.

Yeah that's one way. Don't forget that means that y=2 or y= -2.

thanks appreciate it
so when I try to use Gauss I get [5,4] ?? is that right??
thanks

 

[Steven G]

thanks so I would set it up like this:
3x^2 + 3y^2 = 27
3x^2 + 2y^2 = 23

3x^2 + 3y^2 - 27 = 3x^2 + 2y^2 - 23

move around and combine terms:

3x^2 + 3y^2 = 3x^2 + 2y^2 - 23 + 27
3x^2 - 3x^2+ 3y^2 - 2y^2= - 23 + 27

y^2 = 4

is this how to do it??
thanks appreciate it

Why not simply subtract?
3x^2 + 3y^2 = 27
-
3x^2 + 2y^2 = 23
aaaaaaay^2 = 4

 

[Steven G]

thanks appreciate it
so when I try to use Gauss I get [5,4] ?? is that right??
thanks

View attachment 22017

You solved for x^2 and y^2 correctly. Now solve for x and y!

 

[Harry_the_cat]

thanks appreciate it
so when I try to use Gauss I get [5,4] ?? is that right??
thanks

View attachment 22017

Yes x^2=5 and y^2=4, so x=? And y=?

 

[markraz]

Yes x^2=5 and y^2=4, so x=? And y=?

thanks Harry_the_cat, do I take the square root of 5 and 4??
thanks

 

[Steven G]

Not exactly. Which number, numbers, if any when you square them will give you 5? 4?

 

[Harry_the_cat]

If y^2=4, what is y? (Note: There are 2 solutions)

If x^2=5, what is x? (2 solutions again)

 

[markraz]

Not exactly. Which number, numbers, if any when you square them will give you 5? 4?

Thanks if I square 2 or -2 that gives me 4 if I square 2.237 or -2.237 that gives me 5 ??
Thanks

 

[Steven G]

Sorry but if you square 2.2360679774997896964091736687313 you do not get 5. Sqrt(5) is an irrational number and can not be expressed as an ending decimal number.

Harry_the_cat clearly stated that that x^2 = 4 has two solutions yet you stated only one solution. Can you think of the other solution? There are two numbers that when you square them give you 4. Yes, 2 is one of them. What is the other?

There are two numbers that what you square them give you 5. One of them is -sqrt(5). What is the other number?

 

[markraz]

Sorry but if you square 2.2360679774997896964091736687313 you do not get 5. Sqrt(5) is an irrational number and can not be expressed as an ending decimal number.

Harry_the_cat clearly stated that that x^2 = 4 has two solutions yet you stated only one solution. Can you think of the other solution? There are two numbers that when you square them give you 4. Yes, 2 is one of them. What is the other?

There are two numbers that what you square them give you 5. One of them is -sqrt(5). What is the other number?

thanks sqrt(5) and sqrt(-5) ??

thanks

 

[Harry_the_cat]

Think about it! Is the second answer sqrt(-5) as you have said or is it -sqrt(5) ? They are very different.

 

[markraz]

Think about it! Is the second answer sqrt(-5) as you have said or is it -sqrt(5) ? They are very different.

oops would it bb
sqrt(5) and -sqrt(5) ??

thanks

 

[markraz]

Sorry but if you square 2.2360679774997896964091736687313 you do not get 5. Sqrt(5) is an irrational number and can not be expressed as an ending decimal number.

Harry_the_cat clearly stated that that x^2 = 4 has two solutions yet you stated only one solution. Can you think of the other solution? There are two numbers that when you square them give you 4. Yes, 2 is one of them. What is the other?

There are two numbers that what you square them give you 5. One of them is -sqrt(5). What is the other number?

thanks appreciate it,

So

if x^2 = 5 then x=sqrt{5} and x=-sqrt{5}
ify^2= 4 then y=2 y=-2

is that correct??
thanks

 

[Steven G]

So

if x^2 = 5 then x=sqrt{5} and x=-sqrt{5}
ify^2= 4 then y=2 y=-2

is that correct??
thanks

Yes, that is correct. Now what are the solutions to your system of equations?

Please do not write that x=sqrt{5} and x=-sqrt{5} as x can NOT be both those numbers at the same time. What should you write instead?