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!!
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
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 \(\Rightarrow\) \( x = 9\)
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
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!!.
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