<p>there are 2 equations</p>
<p>1.) (x^4) - 5(x^2)(y^2) + 4(y^4) = 0</p>
<p>2.) 2y - x = 0</p>
<p>i need to give the coordinates of a point that is on the line of the 2nd equation but not on the graph of the 1st equation.</p>
<p>how do i go about doing this? when i set the equations equal to each other i cannot solve for intersections pts because there are 2 variables (x and y)?</p>
<p>any help is appreciated on solving this problem?</p>
<p>I don't know how to do what you're asking (pt on 2nd equation but not first equation) other than perhaps guess and check, but I know how to find the intersection.</p>
<p>In equation 2, solve for x: x=2y
In equation 1, plug in 2y wherever there is an x.
So, (2y)^4-5(2y)^2<em>y^2 + 4y^4
=16y^4-20y^2</em>y^2 + 4y^4
=20y^4-20y^4=0
y=0
x=2y=2*0=0
So, P(0,0)</p>
<p>yes (0,0) is a pt of intersection but it is only one of many pts of intersections, the way this problem needs to be solved is i need to find ALL the pts of intersections(i dunno how to do this), then i know those pts of intersection are not my answer so i can use any other pt on my second equation as my answer.</p>
<p>anyone else willing to atleast try this problem?</p>
<p>1.) x^4 - 5x^2 y^2 + 4y^2 = 0
(x^2 - 4y^2)(x^2 - y^2) = 0
(x + 2y)(x - 2y)(x + y)(x - y) = 0
x = y or x = -y or x = 2y or x = -2y</p>
<p>2.) 2y - x = 0
x = 2y</p>
<p>Therefore, it is not possible to find any points that satisfy the second equation but not the first.</p>
<p>First, solve for x: </p>
<p>2y-x=0
-2y -2y
-x=-2y
x=2y <--Substitute This Equation In The First Equation</p>
<p>(2y)^4 - 5(-2y^2)(y^2)+4(y^4) = 0
2y^4 + -10y^2 (y^2) + 4(y^4) = 0
2y^4 + -10y^2 + 4y^4 = 0
6y^4 + -10y^2 = 0
1296y + 100y = 0
1396y = 0
y = 0</p>
<p>x=2y
x=2(0)
x=0</p>
<p>P = (0,0)</p>
<p>You guys are understimating the complexity of this problem
i did what chocolateluvr88 and Murrah08 did, i also got P (0,0) however there MUST be many more pts of intersection, just try plugging in 1 for x and 1/2 for y and then you got another pt of inersection (1,1/2), there must be a way to find all the pts of intersections, my teacher said this problem involves some algebraic trick (which i apparently dont know :()<br>
i hope someone can figure this challenging problem out, i would really appreciate it</p>
<p>"im_blue" your answer seems right, can anyone check this please?</p>
<p>Im_blues answer is correct</p>
<p>wait, i thought you said he was wrong
"Im_blues eye is slightly off, the step where he factored is wrong. if
you reverse the steps you will get a y^4 which doesn't exist in the
original equation." i got that message from my email</p>
<p>In the original problem it is y^4, but when he re-wrote it, he wrote it wrong so as I followed his work, not refering back to the original problem, it appeared to be initially wrong.</p>
<p>I made a typo in my answer, which is corrected below:
1.) x^4 - 5x^2 y^2 + 4y^4 = 0
(x^2 - 4y^2)(x^2 - y^2) = 0
(x + 2y)(x - 2y)(x + y)(x - y) = 0
x = y or x = -y or x = 2y or x = -2y</p>
<p>2.) 2y - x = 0
x = 2y</p>
<p>Any points which satisfy the 2nd equation also satisfy the 1st, so it is not possible to find any points that satisfy the second equation but not the first.</p>
<p>editied, im blue already solved</p>
<p>is there any way to graph equation 1, any online graphing tool to graph a multivariable function as in equation 1??</p>
<p>you can just use any TI calculator to graph the function, it isn't multi-variable</p>