Find the values of x, y and lambda that satisfy the system of equations. Such systems arise in certain problems of calculus, and lambda is called the Lagrange multiplier. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.) 3x + lambda = 0 3y + lambda = 0 x + y - 6 = 0 (z, y, lambda) -

λ =--9

x=3

y=3

Step-by-step explanation:

3x+λ=0

3y+λ=0

x+y-6=0

Taking sum of first two equations;

3x+3y+2λ=0

3(x+y)+2λ=0 ---(a)

multiply third equation by 3

3(x+y)-18=0  (b)

Subtracting eq a from eq b

==> 2λ+18=0

==> λ = -9

Putt this value of λ in 3x+λ=0

==> 3x-9=0

==> x=3

Again put value of λ in 3y+λ=0

==> 3y -9=0

==> y=3

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square