Question

Solve each sysytem. In your work, identify the shape of the graph of each equation.
1.)
{7y^2+x^2=64
x+y+4

2.)
{x^2+y^2+2x+2y=0
x^2+y^2+4x+6y+12=0

Answer 1

Use this site
https://mathway.com/

Answer 2

I don't think it does this kinda question but thanks

Answer 3

it will trust me...just click the link and type in tour problem..

Answer 4

You'll need to learn the general equations for geometric shapes such as circle, parabola, ellipse, hyperbola, and so on. Look up "equations of conic sections." Try this, and then post some questions of your own, so that someone can help you get on track.

Note that x+y=4 is a linear function; its graph is a straight line.
The other functions, the ones with x^2 and y^2 in them, are curves. It's your job to determine the names and shapes of these curves.

Answer 5

Alright thanks!

Answer 6

need help

Answer 7

Yes

Answer 8

five years later but ok

Answer 9

so do u post to work it out or something

Answer 10

im not sure :)

Answer 11

Yea

Answer 12

is this like terms

Answer 13

TAG ME WHEN U GET BACK

Answer 14

BYE

Answer 15

I'm here... @alivejeremy

Answer 16

All it says is to solve the system

Answer 17

{7y^2+x^2=64x+y+4

Answer 18

it post to look like that

Answer 19

or nah

Answer 20

Like that, just 1 big bracket

Answer 21

oh kool

Answer 22

It's ok if you can't help

Answer 23

i can im working it out

Answer 24

whats the name of this math tho

Answer 25

pre-cal

Answer 26

oh kool

Answer 27

look at that

Answer 28

Ok?

Answer 29

u get it

Answer 30

thats number 1

Answer 31

number 1. 16 + y^2 - 8y

Answer 32

16 + y^2 - 8y

Answer 33

do you still need help?

Answer 34

Yes, with the second one

Answer 35

{x^2+y^2+2x+2y=0
x^2+y^2+4x+6y+12=0

Answer 36

suggestion
complete the square for each equation then set them equal

Answer 37

x^2+y^2+2x+2y=0 |completing squares

x^2+2x+ y^2+2y =0

(x+1)^2 -1 + (y+1)^2 -1 = 0

(x+1)^2 + (y+1)^2 = 2



x^2+y^2+4x+6y+12=0 |completing squares

x^2+4x + y^2+6y +12=0

(x+2)^2 -4 + (y+3)^2 -9 + 12 = 0

(x+2)^2 + (y+3)^2 -9 = 1

Algebraically:

x^2+y^2+2x+2y = x^2+y^2+4x+6y+12