Non linear simultaneous equation

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Deano7

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Hi, trying to brush up on this math to help my eldest. But I keep end up not being able to factor. Any help please?

X squared + y squared = 36
X=2y + 6

I substituted the linear into the nonlinear. Rearranged to equal 0
Then factorise.

But struggling. Any help please
 
Why not demonstrate?

(2y+6)^2 + y^2 = 36

(4y^2 + 24y + 36) + y^2 = 36

Is that where you landed?

x^2 + y^2 = 6^2 -- Do you recognize this as a circle?
 
Last edited by a moderator:
Please show what you got after substituting, so we can see what you are trying to factor. (What I get is easy to factor.)

Also, note that you can write x2 + y2 = 36 using the superscript button, or just write x^2 + y^2 = 36.
 
Deano7 said:
Hi, trying to brush up on this math to help my eldest. But I keep end up not being able to factor. Any help please?

X squared + y squared = 36
X=2y + 6

I substituted the linear into the nonlinear. Rearranged to equal 0
Then factorise.

But struggling. Any help please
Click to expand...
Instead of trying to factorise you could always use the quadratic formla. Google it if you can't remember it, although your child should know it.
 
Deano7 said:
Hi, trying to brush up on this math to help my eldest. But I keep end up not being able to factor. Any help please?
X squared + y squared = 36
X=2y + 6
Click to expand...
\(\displaystyle x-2y=6\) is the equation of a line with \(\displaystyle y\text{-intercept: }(0,-3)~\&~\text{ slope } m=\frac{1}{2}\)
\(\displaystyle x^2+y^2=(6)^2\) is a circle centred at \(\displaystyle (0,0)\) with radius \(\displaystyle r=6\).
Is there a graphical solution?
Here
 
x^2 + y^2 = (x-2y)^2
x^2 + y^2 =x^2 -4xy + 4y^2
3y^2-4xy=0
y(3y-4x)=0
y=0 so x=6
OR
3y-4x=0 so y = 4x/3
x-2y=6
x-8x/3 =6
-5x/3 =6
x=-18/5, y= -24/5
 
tkhunny said:
Why not demonstrate?

(2y-6)^2 + y^2 = 36

(4y^2 - 24y + 36) + y^2 = 36

Is that where you landed?

x^2 + y^2 = 6^2 -- Do you recognize this as a circle?
Click to expand...
Hi and thanks for all the replies

The question states substitute the linear equation into the non linear.

so I have (2y+6)^2 +y^2=36

then it says rearrange so it equals zero
So I have (2y+6)^2 + y^2 - 36 = 0

Then it says factorise and solve for the first variable and this where I fall over. What's the next step please or have I made a mistake above
 
Deano7 said:
Hi and thanks for all the replies

The question states substitute the linear equation into the non linear.

so I have (2y+6)^2 +y^2=36

then it says rearrange so it equals zero
So I have (2y+6)^2 + y^2 - 36 = 0

Then it says factorise and solve for the first variable and this where I fall over. What's the next step please or have I made a mistake above
Click to expand...
You are repeating your original post!

Did you even read the 6 responses after your original post? Are you working with pencil and paper or just staring at the screen?
 
Subhotosh Khan said:
You are repeating your original post!

Did you even read the 6 responses after your original post? Are you working with pencil and paper or just staring at the screen?
Click to expand...
The 6 replies are not clear to me. So I thought I would work through the question step by step so I understand it.
 
HallsofIvy said:
The next step is to actually do the square! What is \(\displaystyle (2y+ 6)^2\)?
Click to expand...
Thanks so (2y+6)^2 solved is 4y^2+24y+36

So I landed at 4y^2+24y+36+y^2-36=0

is this correct so far? I can see I could simplify this but before I do the question specifically asked to factorise and solve the first variable. What's the next step please.
 
Deano7 said:
The 6 replies are not clear to me. So I thought I would work through the question step by step so I understand it.
Click to expand...
Let's pick response #2:
Why not demonstrate?

(2y+6)^2 + y^2 = 36

(4y^2 + 24y + 36) + y^2 = 36

Is that where you landed?

x^2 + y^2 = 6^2 -- Do you recognize this as a circle?
Click to expand...
Please tell us exactly where you are getting lost?
 
Deano7 said:
Thanks so (2y+6)^2 solved is 4y^2+24y+36

So I landed at 4y^2+24y+36+y^2-36=0

is this correct so far? I can see I could simplify this but before I do the question specifically asked to factorise and solve the first variable. What's the next step please.
Click to expand...
First simplify - then factorize.
 
Deano7 said:
Thanks so (2y+6)^2 solved is 4y^2+24y+36

So I landed at 4y^2+24y+36+y^2-36=0

is this correct so far? I can see I could simplify this but before I do the question specifically asked to factorise and solve the first variable. What's the next step please.
Click to expand...
Yes, that is correct. The "next step" is to simplify as you say! Once you do that, you should find it very easy to "factorize".
 
Thanks, it's this step I fall on ?.
so if I simplify by removing the 36s so 4y^2+24y+y^2=0. Was this the correct step?
Now I'm stuck. Don't know how to factorise from here? Appreciate any help
 
Deano7 said:
Thanks so (2y+6)^2 solved is 4y^2+24y+36

So I landed at 4y^2+24y+36+y^2-36=0

is this correct so far? I can see I could simplify this but before I do the question specifically asked to factorise and solve the first variable. What's the next step please.
Click to expand...
I know you are a long way from studying algebra, but one of the first recommendations is to gather like terms.

[MATH]4y^2 + 24y + 36 + y^2 - 36 = 0 \implies [/MATH]
[MATH](4y^2 + y^2) + 24y + (36 - 36) = 0 \implies[/MATH]
[MATH]5y^2 + 24y = 0 \implies[/MATH]
[MATH]y(5y + 24) = 0.[/MATH]
You have factored. Now what?
 
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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.
 
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?
Click to expand...
The whole point of factoring is to consider _all_ factors!
When is ab=0? When a is 0 OR b is 0.
y(5y+24)=0
One factor is y. So the product is 0 if y=0.
What's the other factor?
 
lev888 said:
The whole point of factoring is to consider _all_ factors!
When is ab=0? When a is 0 OR b is 0.
y(5y+24)=0
One factor is y. So the product is 0 if y=0.
What's the other factor?
Click to expand...
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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