markraz
Full Member
- Joined
- Feb 19, 2014
- Messages
- 338
Thanks appreciate it. This makes perfect sense. Just curious, is this the only way this can be done?? ThanksI 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 so I would set it up like this: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,
Yeah that's one way. Don't forget that means that y=2 or y= -2.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?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
You solved for x^2 and y^2 correctly. Now solve for x and y!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=?thanks appreciate it
so when I try to use Gauss I get [5,4] ?? is that right??
thanks
View attachment 22017
thanks Harry_the_cat, do I take the square root of 5 and 4??Yes x^2=5 and y^2=4, so x=? And y=?
Thanks if I square 2 or -2 that gives me 4 if I square 2.237 or -2.237 that gives me 5 ??Not exactly. Which number, numbers, if any when you square them will give you 5? 4?
thanks sqrt(5) and sqrt(-5) ??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?
oops would it bbThink about it! Is the second answer sqrt(-5) as you have said or is it -sqrt(5) ? They are very different.
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?
Yes, that is correct. Now what are the solutions to your system of equations?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