Question

Solve this system of conic section equations
7y^2+x^2=64
x+y=4

Answer 1

First solve the equation x+y = 4 for y
\[\Large x+y=4\]
\[\Large y=4-x\]

Now plug this into the other equation so you boil things down to one variable only
\[\Large 7y^2+x^2=64\]
\[\Large 7(4-x)^2+x^2=64\]
\[\Large 7(16-8x+x^2)+x^2=64\]
\[\Large 112-56x+7x^2+x^2=64\]
\[\Large 112-56x+8x^2=64\]
\[\Large 112-56x+8x^2-64=0\]
\[\Large 8x^2-56x+48=0\]
From this point, you need to solve for x. To do so, I recommend using the quadratic formula. I'll let you do this part. Tell me what x values you get as the solutions.

Answer 2

I get x=1 and x=6

Answer 3

me too

Answer 4

if x = 1, then y = ???

Answer 5

sweet and then just plug that into on of the original equations to solve for y

Answer 6

yes or plug into y = 4-x since that already has y isolated

Answer 7

so like y=4-1 and y=4-6

Answer 8

y=3 and y=-2

Answer 9

correct, so the solutions are these two ordered pairs

(1,3) and (6,-2)

and this confirms it

http://www.wolframalpha.com/input/?i=7y^2%2Bx^2%3D64%2Cx%2By%3D4

Answer 10

wanna help me with another one?

Answer 11

sure

Answer 12

what's your question?

Answer 13

x^2+y^2+2x+2y=0
x^2+y^2+4x+6y+12=0
solve this system of equations

Answer 14

ok at first this looks really complicated, but we can eliminate quite a bit here

notice how each equation has x^2+y^2 in it

so we can subtract the equations (either equation1 - equation2 or equation2-equation1) to eliminate the x^2+y^2 terms

what do you get when you subtract?

Answer 15

2x+4y+12=0

Answer 16

Now let's solve 2x+4y+12=0 for x


2x+4y+12=0

2x+4y+12-12=0-12

2x+4y = -12

2x+4y-4y = -12 - 4y

2x = -4y - 12

2x/2 = (-4y-12)/2

x = -2y - 6

Answer 17

now plug that into the other equation

Answer 18

Next, plug x = -2y - 6 into either original equation. I'll pick the first equation


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

(-2y-6)^2+y^2+2(-2y-6)+2y=0 ... replace x with -2y-6

(4y^2+24y+36)+y^2+2(-2y-6)+2y=0

4y^2+24y+36+y^2-4y+12+2y=0

5y^2+22y+48=0


solve that for y (use the quadratic formula). Tell me what you get

Answer 19

yes correct

Answer 20

hmm I messed up, let me fix

Answer 21

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

(-2y-6)^2+y^2+2(-2y-6)+2y=0 ... replace x with -2y-6

(4y^2+24y+36)+y^2+2(-2y-6)+2y=0

4y^2+24y+36+y^2-4y-12+2y=0 ... it should be -12, not +12

5y^2+22y+24=0

Answer 22

well I tried and I got the b^2-4(a)(c) = -476
you cant take a square root of a negative number

Answer 23

I fixed my mistake and got 5y^2+22y+24=0

Answer 24

oh gotcha let me try that

Answer 25

y=-2 and y=-2.4

Answer 26

-2.4 or -12/5

Answer 27

now use each y value to find the corresponding x value

Answer 28

-12/5? how did you get that?

Answer 29

-22+ or =- the sqrt of 4

Answer 30

over 10

Answer 31

I used the quadratic formula

or you can convert -2.4 to fraction form

-2.4 = -2.4*(10/10) = -24/10 = -12/5

Answer 32

makes sense

Answer 33

so now we plug in the y values to get x

Answer 34

correct

Answer 35

you can use x = -2y - 6

Answer 36

x=-10.8 and x=-10

Answer 37

incorrect on both

Answer 38

if y = -2, then


x = -2y - 6

x = -2(-2) - 6

x = 4 - 6

x = -2

Answer 39

if y = -2.4, then


x = -2y - 6

x = -2(-2.4) - 6

x = 4.8 - 6

x = -1.2

Answer 40

so the solutions are (-2,-2) and (-1.2, -12/

Answer 41

correct and the graph confirms it

Answer 42

I think you meant to say -12/5

Answer 43

if you use -12/5 then convert -1.2 to fraction form as well

-1.2 = -1.2*(10/10) = -12/10 = -6/5

Answer 44

Yeah I did mean -12/5 thanks:))

Answer 45

np