Question

Help please (will medal) THANK YOU!

Answer 1

\[\left| x^2-4y^2+x+3y-1 \right|+(2x-y-1)^2=0\]

Answer 2

here is a website i think this is what your looking for but i dont know this stuff ......

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffsoldirectory/

Answer 3

Do you need help solving or what?

Answer 4

who are you talking to? btw i dont wanna say it in a mean way

Answer 5

your fine! @wcrmelissa2001 what you need help with?

Answer 6

oh ok well ima go..

Answer 7

@NicoNeedsAlotOfHelp thanks anways!
@Qwertty123 I'm not sure how to solve this question (and eyyy ur back!)

Answer 8

Everytime I expand it I end up with lot of terms and no idea how to simplify it:
y^2, y, x, x^2, xy

Answer 9

I am back! Lets take a look here! :D

Answer 10

@phi

Answer 11

what techniques do you know to tackle this type of problem ?

Answer 12

Off the top of my head it would be graphs (not sure how that can be used), expansion??

Answer 13

both terms are either 0 or positive (assuming real numbers)
the only way to get the sum of two positive numbers to be 0 is to require both to be zero.

Answer 14

yup

Answer 15

but once you remove the modulus sign, the first term can be negative too am i right?

Answer 16

yes. I'll have to think about this one.

Answer 17

thanks :D

Answer 18

graphing both terms (but ignoring the absolute value), we have this

Answer 19

to solve this analytically, we start by requiring
2x-y-1= 0
from which we get y= 2x -1

using that in the second term, we have
\[ x^2 -4(2x-1)^2 + x +3(2x-1) -1 = 0\]
if we expand that out we get
\[ -15x^2 +23x -8=0 \\ 15x^2 -23x +8 =0\]
the zeros occur at
\[ x= \frac{23 \pm 7}{30} \]

Answer 20

we get x= 1 (and y = 2*1 -1 = 1)
and x= 16/30 = 8/15
and the corresponding y= 1/15

Answer 21

wait sorry how would u graph it out? like what are the 2 equations

Answer 22

we require 2x-y-1=0 --> y= 2x-1
so the solution has to lie on that line

Answer 23

we also require x^2-4y^2 +x+3y-1 =0
that is a hyperbola, and the solution has to lie on it.

so the solution has to lie on the intersection of the two curves.

Answer 24

because we insist on the term being zero, we can ignore the absolute value operator because | 0 | = 0

Answer 25

I get it :D Except for how to get the graph from x^2-4y^2 +x+3y-1 =0. You can't exactly key that in to a graphic calculator. And there is y^2 and y terms which is hard to make into equation of y

Answer 26

are you sure? It's a hyperbola. If your calculator does circles e..g. x^2+y^2=1
then it should do hyperbolas. Or use geogebra (free download) or wolfram

Answer 27

I'm not sure because I thnk my calculator can only do y= blah blah functions but oh well. Thank you anywawy! :D

Answer 28

you can "complete the square" on both the x and y terms and put the equation into a more standard form
http://www.purplemath.com/modules/hyperbola.htm

Answer 29

ahhhh thanks :D

Answer 30

@wcrmelissa2001

Answer 31

thank you

Answer 32

i mean your welcome by the "thanks anyways"