Question

I solved everything else I just need to solve #6-10.

it says to solve it algebraically
&
to use a graph to demonstrate that my answers are correct.

I believe I already have a graph I just need help solving them algebraically.

Answer 1

I would redraw the answer to #2 to show the circles touching only at the 1 point.

Answer 2

Done. Do you think you can help me with #6-10?

Answer 3

I think i have #6 completed.

Answer 4

I think i have # 7 also. So just numbers 8-10.

Answer 5

For #6, reorder the 2nd equation to make the x term first
\[ 32x^2 +9y^2 =324\\ -x^2 +3y^2 = 3\]
I see if the bottom y^2 term had a -9 in front of it (instead of 3) we could add the two equations and eliminate y^2. So I would multiply the bottom equation by -3
all terms , both sides:
\[ 32x^2 +9y^2 =324\\ 3x^2 -9y^2 = -9\]

Answer 6

Okay yay! I have #6 correct

Answer 7

if we add the two equations we get
\[ 35x^2 +0 = 315 \\ 35x^2= 315\]
divide both sides by 35
\[ x^2= \frac{315}{35}= 9 \]
take the square root of both sides
\[ x = \pm \sqrt{9}= \pm 3 \]
so x=-3 or x=3
we now need to find the y values

Answer 8

can you find the y values using algebra?
you would pick either equation (I would choose the 2nd one because it has smaller numbers)
for x=-3 you will get two y values
and ditto for x=3

Answer 9

would it be 2?

Answer 10

y=-2 and y=2

Answer 11

using the 2nd equation, written this way
\[ 3y^2 = 3+x^2\]
if x is 3 we get
\[ 3y^2= 3+9=12\\y^2=4\\y= \pm2\]

so (-3,2) and (-3,-2) are solutions
for x=+3 we will get the same two y values, so
(3,2) and (3,-2) are the other two solutions

Answer 12

Thank you!!

Answer 13

Im not sure if i have the work right for #7

Answer 14

for #7, do you have any thoughts ?
I would multiply the first equation by -3
what do you get ?

Answer 15

I converted the 2nd equation to \[21x^2 =15y^2-60y+9\]

Answer 16

how did you get -60y ?

Answer 17

Well I aslo converted the first one \[7x^2 = 5y^2-20y+3\]

Answer 18

I see the first equation is
\[ 7y^2 -5x^2+20x=3\]
(no y term).

multiply that equation by -3

Answer 19

you should get
\[ -21y^2 +15x^2 -60x = -9\]

Answer 20

Yes

Answer 21

can you add the two equations?

Answer 22

im working on it on a separate piece of paper

Answer 23

what do you get ?

Answer 24

sorry. I'm really stuck on this one

Answer 25

to add two equations you write down the left side and then a plus sign and then the left side of the other equation
adding
\[ -21y^2 +15x^2 -60x = -9 \\ 21y^2+5x^2=209\]

you could write
\[ -21y^2 +15x^2 -60x + 21y^2+5x^2 =-9+209\]

Answer 26

you can simplify the right side, right ?

Answer 27

on the left side, combine "like terms"
notice you have \(-21y^2 +21y^2\)
21 y^2 take away 21 y^2 leaves no y^2 (they cancel out)
you also have 15 x^2 plus 5 x^2
that can be combined to get 20x^2

Answer 28

so it's y2 +3y2 so far

Answer 29

Im really bad at this :/

Answer 30

here is what we have when we add the two equations
\[ -21y^2 +15x^2 -60x + 21y^2+5x^2 =-9+209 \]
on the right side, what is -9+209 ?

Answer 31

the sum?

Answer 32

yes, what is -9+209 ?

Answer 33

uh

Answer 34

Oh my god

Answer 35

im so stupid

Answer 36

200

Answer 37

yes, 200. so we have, so far,
\[ -21y^2 +15x^2 -60x + 21y^2+5x^2 = 200\]

Answer 38

next, notice you have \(-21y^2 +21y^2 \)
do you know what that adds up to ?

Answer 39

you have 21 y squared and you take away 21 y squared.
how many y squared are left ?

Answer 40

0 or 1

Answer 41

would it just by y

Answer 42

0
21y^2 - 21 y^2 is 0
so we can ignore those two terms, and we have
\[ 15x^2 -60x +5x^2 = 200\]

Answer 43

next, notice we have two terms with an x^2
we have 15 x squared plus another 5 x squared
how many x squared do we have ?

Answer 44

20x4 or 20x2

Answer 45

20x^2

(you don't change the x^2 to x^4)

you now have
\[ 20x^2-60x= 200\]

Answer 46

you can add -200 to both sides
\[ 20x^2-60x-200 = 200-200\]
or
\[ 20x^2 -60x-200=0\]

luckily we can simplify that because we can divide each term by 20
(or if you know how to factor, we can factor out 20, like this:
\[ 20(x^2-3x-10)=0\]

Answer 47

now divide both sides by 20
\[ \frac{20}{20} (x^2-3x-10)= \frac{0}{20} \\
x^2 -3x-10= 0\]

Answer 48

so "solve for x" you should factor that equation
do you know how ?

Answer 49

no ;/

Answer 50

would it be positive and negative 2

Answer 51

it's a bit complicated to explain how to factor
you would first look at the last number 10, and write all pairs that multiply to get 10. they are:
1,10
2,5
next, we see the sign on the 10 is -10 (minus). that means the factors are different signs
the sign on the middle term -3x is -. that means the biggest factor is minus
put in the signs and add
+1-10= -9
+2-5= -3

if any add up to the middle number -3, those are the factors.
in this case +2, and -5

the factors are
(x+2)(x-5)=0

Answer 52

we now use the idea that either
x+2=0 or x-5 =0
and that means
x=-2 or x= 5

Answer 53

so would this be the answer?

(-2,3)
(-2,-3)
(5,2)
(5,-2)

Answer 54

yes, all your answers are correct. But you have to show your algebra

Answer 55

Thank you! Sorry i took so long..