Question

What are the x-coordinates of the solutions to this system of equations?

x2 + y2 = 36
y = x - 6

Answer 1

the second equation, pick a number for x and the results will be for why, and from their. Keep doing that until u have like 3 ordered pairs and draw a line for it .
y = x - 6

Answer 2

for the first one, change the equation and solve for y.
x2 + y2 = 36 change it to y^2 =36 + x^2

Answer 3

how do i do it with the exponent

Answer 4

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

Answer 5

x + y = 6
y = x - 6

x+ (x-6) = 6
2x - 6 = 6
2x = 12
x = 6
substitute that into y = x- 6
to get the value for y :)

Answer 6

i think i did something wrong...
Does the lines intersect,so the y-value will be the same for each?

Answer 7

thank you!

Answer 8

zale ,
x^2 + y^2 = 36
is the same as
x + y = 6

Answer 9

and you did
y^2 = 36 + x^2
it would be
y^2 = 36 - x^2

Answer 10

oh, thx for the correction

Answer 11

Shamil98...INCORRECT! You said that x^2 + y^2 = 36 is the same as x + y = 6. That is FALSE!!

Answer 12

-_-
if you square it then yes ..

Answer 13

x + y = 6 is simplified form

Answer 14

Arishagee......watch

Answer 15

Shamil...NO!

Answer 16

x = 6 , y = 0

Answer 17

Shamil..last time..you are giving incorrect statements!

Answer 18

Arishagee...you there?

Answer 19

@Easyaspi314 explain how's he wrong?

Answer 20

omg, will you leave it?
the second pair of solutions is
(0,6)
if thats what ur lookin for

Answer 21

To solve that original syatem of equations:

First note that we take y = x - 6 and substiutute that into y in the first equation,

So the FIRST equation becomes x^2 + (x-6)^2 = 36
Simplify the above, you get x^2 + x^2 - 12x + 36 = 36
2x^2 - 12 x = 0
2x(x - 6) = 0
x = 0 or x = 6
If x = 0, y = -6; if x = 6, y = 0

Final solutions: x = 0 and y = -6 AND x = 6 and y= 0

Answer 22

-6* yes i forgot the negative in the secoond pair

Answer 23

what did i do wrong?.

Answer 24

Shai...what you did was wrong..you CANNOT make a statement that x^2 + y^2 = 36 is the same as x + y = 6!!

Answer 25

the question is closed, leave it . I made a false statement, okay? My bad? Jeez

Answer 26

Thats nonsense. The x^2 + y^2 = 36 is a circle whose radius is 6 and center is at the origin, while x+ y = 6 is a straight line. Two different animals!

Answer 27

Shamil..I must and needed to correct you, as others may walk away assuming that if they see x^2 + y^2 = 36 it would be the same as x+ y = 6, and it is NOT the same. Besides, I have two pairs of solutions that work correctly. I did it mathematically correct.

Answer 28

The algebraic way:

Use substitution: \[\begin{split}
x^2+y^2&=36\\
x^2+(x-6)^2&=36&\quad &y=x-6\\
\end{split}
\]Now simplify you have a quadratic function in the ends.

Answer 29

wio..thats what I just showed them

Answer 30

that was the easiest method ever!

Answer 31

the second equation was y=x-6, i should've plugged it to y of the first equation -.-

Answer 32

Anyway:\[
(x+y)^2=6^2\\
x^2+2xy+y^2=36
\]Using elimination you get:\[
2xy=0
\]Meaning there are solutions when \(x=0\) and \(y=0\).

Answer 33

Zale...Good Morning...thanks for smelling the coffee.

Answer 34

I showed a simple method, I made an incorrect statement, leave it at that.

Answer 35

keep convos in pm >.>

Answer 36

Why you get 2 medals?

Answer 37

Zale and the asker gave me one each lol

Answer 38

shamil, ur answer was explained perfectly, that's why i gave u that medal xD

Answer 39

I figured out a way to do it with elimination and substitution. That is pretty boss.

Answer 40

I did it by making it simple :D

Answer 41

You didn't do it right though. You were wrong.

Answer 42

The solutions were right ._.

Answer 43

-.-

Answer 44

i'm keep getting notifications :\

Answer 45

i*

Answer 46

yep >.>

Answer 47

Close this question @arishagee, it's solved and done

Answer 48

it is closed o.o