-300.21x + 45y + 25z = 0
45x - 40.21y - 65z = 0
25x - 65y - 114.21z = 0

x,y, and z must also satisfy this as well...
x^2 + y^2 + z^2 = 1

how do u solve this? U cant use A(inverse)*b = x because the answer column is 0.

I did it by solving for z in terms of x and y, and substituting that into the first two equations.

z = (1 - x^2 - y^2)

Solve for x and y, got x = -.09123 and y = -.87391

Sub those into the equation for z and got z = .47745

u mean z = (1 - x^2 - y^2)^(1/2)?

nice, these are the numbers my prof posted on the solutions.

so by subsituting z = (1 - x^2 - y^2)^(1/2) into the 1st two equations, ull have two eq and two unknown right?

This post was edited by FamilyGuyViewer on Sep 19 2015 04:58pm

Yes. At that point I didn't solve by hand but just plugged the equations into Mathcad.

but if this was on the exam where I dont have Mathcad or any other computer programs, how would i solve this with my calculator...

You could probably brute force it?

Substitute that value of z into all three equations.

Multiply the first equation by (65 / 25), so that the first and second equations have a -65z term and a positive 65z term.

Add the first two equations together, causing the "z's" to cancel out, and you will have an equation with some amount of X + some amount of Y = 0

Solve this new equation for y and you'll get Y = b * x

Substitute y into the 3rd equation.

Your third equation should now just be a function of x and looking something like

25x - (65 * b * x) - 114.21 * ( 1 - x^2 - (b*x)^2 ) ^.5 = 0

Solve this equation for x and then everything else will fall in place.


There is probably an easier way than this though..

This post was edited by RzChaos on Sep 19 2015 05:28pm

I have this so far.. how can i graph this on my Ti-83 plus?

I don't know how the graphing functions on the 83 work.

Multiple the 2nd equation by ( 65 / 25) and add those two equations together. All of the ^2 and ^.5 will cancel out and you'll have one equation...

a*X + b*Y = 0

Y = ( - a / b ) * X

Plug this into your 3rd equation and you'll have an equation with only 1 variable and it should be easy to solve.

yea i got it thanks,

dam this is very tedious.