Doubt on a math problem

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

AVY2024

22 posts

AVY2024

#1 h Apr 8, 2025, 11:09 AM

Y by

Solve for x and y given that xy=923, x+y=84

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

fruitmonster97

2478 posts

fruitmonster97

#2 h Apr 8, 2025, 11:39 AM

Y by

We have $x=42-a$ and $y=42+a$ for some $a.$ We have $42^2-a^2=923,$ so $a=29.$ Thus, $(x,y)=\boxed{(13,71)}.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

sp0rtman00000

2 posts

sp0rtman00000

#3 h Apr 11, 2025, 11:04 AM

Y by

Yous must solve the following second degree equation: t^2-84t+932=0

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

CatsRule222

2 posts

CatsRule222

#4 h Apr 11, 2025, 2:36 PM

Y by

I agree with sp0rtman00000;
It will be easier to change it to a second degree polynomial and then use the quadratic formula, which is: (-b+-Sqrt(b^2-4ac))/2a

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

fruitmonster97

2478 posts

fruitmonster97

#5 h Apr 11, 2025, 2:56 PM

Y by

CatsRule222 wrote:

I agree with sp0rtman00000;
It will be easier to change it to a second degree polynomial and then use the quadratic formula, which is: (-b+-Sqrt(b^2-4ac))/2a

That's what I did, except instead of using the quadratic formula or factoring I used Po-Shen Loh's recently(ish) found method which combines vieta's and some common sense.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

jasperE3

11229 posts

jasperE3

#6 h Apr 11, 2025, 5:47 PM • 1 Y

Y by Solocraftsolo

fruitmonster97 wrote:

We have $x=42-a$ and $y=42+a$ for some $a.$ We have $42^2-a^2=923,$ so $a=29.$ Thus, $(x,y)=\boxed{(13,71)}.$

$(x,y)=(71,13)$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Soupboy0

334 posts

Soupboy0

#7 h Apr 11, 2025, 5:59 PM

Y by

$x = 1434$ by 1434 number theory lemma. Now, define a sigma number $\mathcal{J}$ from the set $\mathcal{S}$ and such that such that any number can now be expressed as ultra-complex, e.g. $a+bi+c\mathcal{J}$ . Now, using the chicken jockey steve theorem, we find there are $2$ distinct solutions. These solutions actually alter the given value of $x$ . The $2$ solutions are $4+37\sqrt{3}i+27\sqrt[3]{3} \mathcal{J}$ and $26.53+34\sqrt{2} i +14 \mathcal{J}.$ These $2$ solutions alter the quantum space limit, and change the values of $x$ to be $13$ and $71$ . Now, using the newly derived quintic formula yields $y=13, 71$ . Therefore, the $2$ solutions are $(13, 71)$ and $(71, 13)$

This post has been edited 1 time. Last edited by Soupboy0, Apr 11, 2025, 5:59 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

sadas123

1228 posts

sadas123

#8 h Apr 13, 2025, 1:15 AM

Y by

Soupboy0 wrote:

$x = 1434$ by 1434 number theory lemma. Now, define a sigma number $\mathcal{J}$ from the set $\mathcal{S}$ and such that such that any number can now be expressed as ultra-complex, e.g. $a+bi+c\mathcal{J}$ . Now, using the chicken jockey steve theorem, we find there are $2$ distinct solutions. These solutions actually alter the given value of $x$ . The $2$ solutions are $4+37\sqrt{3}i+27\sqrt[3]{3} \mathcal{J}$ and $26.53+34\sqrt{2} i +14 \mathcal{J}.$ These $2$ solutions alter the quantum space limit, and change the values of $x$ to be $13$ and $71$ . Now, using the newly derived quintic formula yields $y=13, 71$ . Therefore, the $2$ solutions are $(13, 71)$ and $(71, 13)$

I agree with you

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

maxamc

549 posts

maxamc

#9 h Apr 13, 2025, 3:39 AM

Y by

Since $3|7$ , $13|71$ (add $10$ to the left hand side and $1$ to the other, difference is $3 \cdot 3=9$ , then Euclidian Algorithm), and $71|13$ (commutativity of the divides relation). Thus our answers are $(13,71)$ and all other permutations.