system of linear equations?

Question

solve the following system of linear equations:

x+y+3z=1
2x+2y+6z=2

Which one of the following statements best describes your solution:
A. There is no solution.
B. There is a unique solution.
C. There are 3 solutions.
D. There are infinitely many solutions with one arbitrary parameter.
E. There are infinitely many solutions with two arbitrary parameters.
F. There are infinitely many solutions with three arbitrary parameters.

ANSWER IS E

Enter your solution below. If a variable is an arbitrary parameter in your solution, then set it equal to itself, e.g., w = w.

x=
y=
z=

HOW DO U DO THIS SECOND PART AND WHAT IS X= please explain????

Answer 1

You have two arbitrary parameters, as you said. In this case, none of the variables has any special equality conditions on it, whcih means that your assignment of which two variables are the arbitrary parameters is, well, arbitrary. I'd suggest setting x = x and y = y. The result is that z will have a formulaic solution in terms of x and y. x + y + 3z = 1 ==> z = (1/3)(1 - x - y)

Answer 2

there are infinitely many solutions with 2 arbitrary parameters
as a1/a2=b1/b2=c1/c2
the lines arecoincidental

Answer 3

I don't think you can solve that. You'd need 3 equations to solve a three variable system.