Question

if x+y is 13 and x*y is 24 then what is the distance from the point to the origin

Answer 1

So we have
x+y=13
xy=24

We have to solve for x and y, can you solve that?

Answer 2

Or, I suggest graphing it

Answer 3

hm how would one graph this

Answer 4

Yea my fault, youc an't xd

Answer 5

:( then what do you reccomened doing? should u try to solve for x and y?

Answer 6

Use substitution

Answer 7

ok so x=13-y and then y(13-y)=13y-y^2=24

Answer 8

So that would make
y^2 -13y +24

Answer 9

should i do the same for x

Answer 10

Yea

Answer 11

so 13-x=y x(13-x)=13x-x^2 so x is x^2-13x+24

Answer 12

then since they equal the same thing i compare the to sides to get x and plug in one to y then i find the distance

Answer 13

\(\color{#0cbb34}{\text{Originally Posted by}}\) @crispyrat
so 13-x=y x(13-x)=13x-x^2 so x is x^2-13x+24
\(\color{#0cbb34}{\text{End of Quote}}\)
wait im a bit confused cus if they were the same then x and y would be the same xp

Answer 14

I think i'm more confused than you-

Answer 15

so i tried doing thesame thing with x and got x+y=13 and so 13-x=y x(13-x)=13x-x^2 so x is x^2-13x+24

Answer 16

yes yes, ik, that's why i'm confused at that too-

Answer 17

@darkknight

Answer 18

@Angle

Answer 19

\(\color{#0cbb34}{\text{Originally Posted by}}\) @crispyrat
so i tried doing thesame thing with x and got x+y=13 and so 13-x=y x(13-x)=13x-x^2 so x is x^2-13x+24
\(\color{#0cbb34}{\text{End of Quote}}\)

From this step, you can use the quadratic formula to solve for the x value.
Then use the x value to find the y value.
\(x = \huge \frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Answer 20

Here are some other resources you can take a look at:
https://www.wolframalpha.com/input/?i=x%2By%3D13+%3B+xy%3D24

And Desmos to figure out where the two equations intersect: