wow i feel so stupid that never occurred to me.
and as for the first part, i was trying to get the x out of the equation completely by solving for x and plugging it in so that i would get z in terms of just y. which makes less sense now that i think about it. (also i didnt meant to put the "=0"...
well first i tried to solve for x and substitute, as stated in the original post, which gave me 3y^2+2yz-5z-2((3y2+2yz-5z-2x=0)/2)=0 which ends up being 3y^2+2yz-5z-(3y^2+2yz-5z)=0 which is 0=0. and after that i started over and tried setting it up as 3y^2-2x=5z-2yz and dividing both sides by yz...
solve the following equation for z in terms of x and y
3y^{2}+2yz-5z-2x=0
i've spent a lot of time on this question, but i keep getting something along the lines of 0=0
i tried solving for x in temrs of y and z to replace the x but i get 0=0
Homework Statement
Homework Equations
The Attempt at...
Homework Statement
An electron is accelerated from rest through a potential difference of 26000 V, which exists between plates P1 and P2, shown below. The electron then passes through a small opening into a magnetic field of uniform field strength, B. As indicated, the magnetic field is...