Non linear simultaneous equation

Deano7 said:
i will be honest.....i dont know :) i keep getting stuck/held up on it has to be zero. any help in showing the workings from here? to find the other factor
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What are we multiplying? y times what?
 
Deano7 said:
thanks.
my gut tells me the answer is y=0 in the above. as that equation needs to equal 0. and any value = y, other than zero would mean that wouldn't happen. is there any process (so i could demonstrate it in workings) other than leaping to that outcome?

If y=0 if i put that into the original equations in the question.

x^2 + y^2 = 36
x=2y +6

so y=0 means x=-6 or x=6

is this correct? or am i a mile off still?
to give some context this is revision for a mock exam for my eldest. its not an actual homework question so im interested in being able explain how to work it through so it can be applied to any different question in an exam.

thanks for all the guidance so far.
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In my previous post, we got to

[MATH]y(5y + 24) = 0.[/MATH]
Were you ok to there? That was the factoring.

The factor method of finding roots depends on this fact:

[MATH]ab = 0 \implies a = 0 \text { or } b = 0 \text { or } a = 0 = b.[/MATH]
To put it in English, the product of two non-zero numbers is also a non-zero number. So if the product of two numbers is zero, then at least one of them is zero.

[MATH]y(5y + 24)= 0[/MATH]
means that y may equal zero and therefore x may equal 6. (6, 0) is one possible answer.

But if y is not zero, then, by process of elimination, 5y + 24 = 0. So now calculate x and y if that is so. It is very frequent that non-linear systems have multiple solutions.

So one possibility is that y = 0. You saw that. But if y is not zero then 5y - 24 = 0. And that means what
 
JeffM said:
In my previous post, we got to

[MATH]y(5y + 24) = 0.[/MATH]
Were you ok to there? That was the factoring.

The factor method of finding roots depends on this fact:

[MATH]ab = 0 \implies a = 0 \text { or } b = 0 \text { or } a = 0 = b.[/MATH]
To put it in English, the product of two non-zero numbers is also a non-zero number. So if the product of two numbers is zero, then at least one of them is zero.

[MATH]y(5y + 24)= 0[/MATH]
means that y may equal zero and therefore x may equal 6. (6, 0) is one possible answer.

But if y is not zero, then, by process of elimination, 5y + 24 = 0. So now calculate x and y if that is so. It is very frequent that non-linear systems have multiple solutions.

So one possibility is that y = 0. You saw that. But if y is not zero then 5y - 24 = 0. And that means what
Click to expand...
hi jeffm, thanks
might have had our latest posts cross at the same time.
is the other y= 4.8
based on 24/5 = 4.8
 
Factors are the parts being multiplied. You have y(5y-24)=0. The factors are y and (5y-24). If (5y-24) = 0 then what would y(5y-24) equal, that is what will y*0 equal? y*0 = 0. So now figure out when 5y-24=0. Since 24 -24=0 it must be that 5y=24. Hence y=24/5.
 
Deano7 said:
hi jeffm, thanks
might have had our latest posts cross at the same time.
is the other y= 4.8
based on 24/5 = 4.8
Click to expand...
Almost.

[MATH]y(5y + 24) = 0 \implies y = 0 \text { or } (5y + 24) = 0 \implies y = 0 \text { or } 5y = -\ 24 \implies[/MATH]
[MATH]y = 0 \text { and } x = 6 \text { or } y = -\ \dfrac{24}{5} = -\ 4.8 \text { and } x = 2 * \left (-\ \dfrac{24}{5} \right ) + 6 = \dfrac{-\ 48 + 30}{5} = -\ 3.6 [/MATH]
Let's check.

[MATH]6^2 + 0^2 = 36 \text { and } 2 * 0 + 6 = 6.[/MATH]
[MATH](-\ 3.6)^2 + (-\ 4,8)^2 = 12.96 + 23.04 = 36 \text { and } 2 * (-\ 4.8)+ 6 = -\ 9.6 + 6 = -\ 3.6.[/MATH]
 
Last edited:
A*B*C*D*E = 0

A = 0
OR
B = 0
OR
C = 0
OR
D = 0
OR
E = 0
 
Do you know what "factors" are? You are given that y(5y+ 24)= 0 and say that "one factor is y". Now, what is the other factor?

The point is that "if ab= 0 then either a= 0 or b= 0 or both". Here your "ab" is y(5y+24). "a" is y. What is "b"?
 
Jomo said:
That's what they all say after I point out their mistake. For the record this is all in fun!
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For the additional record, I actually considered putting "Inclusive OR" on each one. I decided against it, hoping the reader would assume it. I was prepared for the question. :)
 
tkhunny said:
For the additional record, I actually considered putting "Inclusive OR" on each one. I decided against it, hoping the reader would assume it. I was prepared for the question. :)
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Many languages distinguish between inclusive and exclusive "or." For example, Latin uses "... vel ..." for inclusive "or" and "aut ... aut ..." for exclusive "or." English is not so clear-cut, but I view "A or B" as being best interpreted as inclusive, and "either A or else B" as being exclusive.
 
tkhunny said:
For the additional record, I actually considered putting "Inclusive OR" on each one. I decided against it, hoping the reader would assume it. I was prepared for the question. :)
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Like I said, I was just joking and being a pain. This is a math site and when someone here uses OR, I do interrupt it as what I call the mathematical OR. If it is a student and I am not sure if they know this special OR I will point it out to them.
I hope that everyone is having fun on the forum tonight!
 
Jomo said:
Like I said, I was just joking and being a pain. This is a math site and when someone here uses OR, I do interrupt it as what I call the mathematical OR. If it is a student and I am not sure if they know this special OR I will point it out to them.
I hope that everyone is having fun on the forum tonight!
Click to expand...
...and I never mind the conversation. Students may learn to care about specifics.
 
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