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# Hello, this is juriemagic,

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Hello, juriemagic here. Normally I do not have problems with solving simultaneous equations, but this one has got me a bit. I did work it out, but still feels un-easy about the answers. Would someone please help me with this one?:

Equation 1: x^2-y^2=0

Equation 2: x^2-4x-9=4y

All and any help will be greatly appreciated!!

Guest Nov 9, 2015

#2
+91039
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Hi Juriemagic,

It is good to see you again

Equation 1: $$x^2-y^2=0$$

Equation 2: $$x^2-4x-9=4y$$

Mmm  This is an unusual one.

It always helps if you can picture what the graphs look like

$$x^2-y^2=0\\ x^2=y^2\\ y=\pm x\\ \mbox{This is a big X on the number plane centred on (0,0)}\\ \mbox{I can also see that the other one is a concave up parabola.}\\ \mbox{anyway that means}\\ x^2-4x-9=4x\qquad or \quad x^2-4x-9=-4x\\ x^2-8x-9=0 \qquad or \quad x^2-9=0\\ (x-9)(x+1)=0 \qquad or \quad (x-3)(x+3)=0\\ x=9\;\;\;or\;\;\;x=-1 \qquad or\qquad x=3\;\;\;or\;\;\;x=-3\\ \mbox{Now I am going to check if y should be pos or neg by subbing into equ2}\\ x=9 \qquad 81-36-9>0\;\;\;so\;\;\;y>0\;\;\;y=9\;\;\;(9,9)\\ x=-1 \qquad 1+4-9<0\;\;\;so\;\;\;y<0\;\;\;y=-1\;\;\;(-1,-1)\\ x=3 \qquad 9-12-9<0\;\;\;so\;\;\;y<0\;\;\;y=-3\;\;\;(3,-3)\\ x=-3 \qquad 9+12-9>0\;\;\;so\;\;\;y>0\;\;\;y=3\;\;\;(-3,3)\\$$

So the 4 points of intersection (simuiltaneous solutions)  are  (9,9), (-1,-1),  (3,-3),  and   (-3,3)

Here is the graph to show you what is happening

https://www.desmos.com/calculator/lvjgdscib6

Melody  Nov 9, 2015
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#1
+18715
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Equation 1: x^2-y^2=0

Equation 2: x^2-4x-9=4y

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

(x-y) = 0  $$\Rightarrow$$  x = y

(x+y) = 0  $$\Rightarrow$$ x = -y

$$y = \pm x$$

(2) x^2 -4x-9 = 4(-x)

x^2  -9 = 0

x^2 = 9

$$x = \pm 3$$

x^2 - 4x -9 = 4(x)

x^2 - 8x - 9 = 0    factor

(x+1)(x-9)= 0

x+1 = 0   $$\Rightarrow$$ $$x = -1$$

x-9 = 0  $$\Rightarrow$$  $$x = 9$$

heureka  Nov 9, 2015
#2
+91039
+5

Hi Juriemagic,

It is good to see you again

Equation 1: $$x^2-y^2=0$$

Equation 2: $$x^2-4x-9=4y$$

Mmm  This is an unusual one.

It always helps if you can picture what the graphs look like

$$x^2-y^2=0\\ x^2=y^2\\ y=\pm x\\ \mbox{This is a big X on the number plane centred on (0,0)}\\ \mbox{I can also see that the other one is a concave up parabola.}\\ \mbox{anyway that means}\\ x^2-4x-9=4x\qquad or \quad x^2-4x-9=-4x\\ x^2-8x-9=0 \qquad or \quad x^2-9=0\\ (x-9)(x+1)=0 \qquad or \quad (x-3)(x+3)=0\\ x=9\;\;\;or\;\;\;x=-1 \qquad or\qquad x=3\;\;\;or\;\;\;x=-3\\ \mbox{Now I am going to check if y should be pos or neg by subbing into equ2}\\ x=9 \qquad 81-36-9>0\;\;\;so\;\;\;y>0\;\;\;y=9\;\;\;(9,9)\\ x=-1 \qquad 1+4-9<0\;\;\;so\;\;\;y<0\;\;\;y=-1\;\;\;(-1,-1)\\ x=3 \qquad 9-12-9<0\;\;\;so\;\;\;y<0\;\;\;y=-3\;\;\;(3,-3)\\ x=-3 \qquad 9+12-9>0\;\;\;so\;\;\;y>0\;\;\;y=3\;\;\;(-3,3)\\$$

So the 4 points of intersection (simuiltaneous solutions)  are  (9,9), (-1,-1),  (3,-3),  and   (-3,3)

Here is the graph to show you what is happening

https://www.desmos.com/calculator/lvjgdscib6

Melody  Nov 9, 2015
#3
+5

My goodness!!!,

Melody and Heureka!!, you did it again!.. . It's soooo different from how I did it!. The X^2=y^2 is really what had me. Awww!, and it's really so simple!. Thank you to both of you!, you are really superb!!.

Guest Nov 9, 2015
#4
+91039
0

Thank you for all the flattery Juriemagic :))

You need to get used to using Desmos Graphing calculator (or any graphing calc) to help you understand what is happening with these problems.

Learning to anticipate what a wide variety of graphs will look like will help you with many problems.

I see that you are logged on at present.  I hoope this means that you log on problems are over

Melody  Nov 9, 2015
#5
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hi Melody, you mean you can see I'm logged on?...strange, because the reply still gives me as "guest", and I also still get the "cookie" error. ??

Guest Nov 9, 2015
#6
+91039
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Hi Juriemagic,

Yes earlier your name was displayed as being logged on.   ://    That is an enigma.

I hope this problems is fixed soon. ://

Melody  Nov 9, 2015

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