Vectors: system of equations

Plane 1: 3(x-1) + 2y + (z+1) = 0
Plane 2: (x-1) + 4y - (z+1) = 0

The book asks to find the line that is the intersections of the two planes.

I saw online how they did it by using the cross product of the normals, but I want to know how (if it's possible) to solve it just using the system of equations.

I imagine it should be possible.

Thanks!

I would guess that if you just add the equations, the z+1 will go away, and then you will have an equation with x-s and y's that should define a line..

3(x-1) + 2y + (z+1)+ (x-1) + 4y - (z+1) = 0

y=(2-2x)/3

So y = -2/3x+2/3 should be the equation for your intersect line.

I found my second reference where someone claims that you can find the line of intersection of two planes algebraically, and he gave an example. Try my second reference:

Does that equation match what they got?

(Luis: We do not want to solve the system to a point, we want to solve it to a line, eliminating only one variable. I believe that two equations is enough to do that. What I am not sure of is how to convert the vector form of the equation that you get from the cross product of the normals, (which is, I believe, {6,-4,-10} from cross({1,4,-1}, {3,2,1})) to a Cartesian equation.

Err.. I meant:
Using the second reference I got:
x= -3/5*t + 2/5
y= 2/5*t + 2/5
z= t

Sorry.

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3(x-1)+2y+(z+1)=0_(x-1)+4y-(z+1)=0

This system of equations cannot be solved because there are more unique variables (3) than there are equations (2).
The system cannot be solved.