Question

Use back-substitution to solve the system. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set z = t and solve for x and y in terms of t.) −x + y − z = 0 2y + z = 5 1 5 z = 0

Answer 1

I think 29

Explanation:

29

Answer 2

(x,y,z) = (5/2, 5/2, 0)

If z = t,

(x,y,z) = ((5-3t)/2, (5-t)/2, t)

Explanation:

-x + y - z = 0

2y + z = 5

(1/5)z = 0

From eqn 3, z = 0

2y + z = 5

Substitute for z in eqn 2

2y + 0 = 5

y = 5/2

substituting for y and z in eq 1

-x + (5/2) - 0 = 0

x = (5/2)

(x,y,z) = (5/2, 5/2, 0)

In terms of t, if z = t,

eqn 2 becomes 2y + t = 5

2y = 5 - t

y = (5 - t)/2

Eqn1 becomes

-x + (5-t)/2 - t = 0

-x + (5/2) - (t/2) - t = 0

-x + (5/2) - (3t/2) = 0

x = (5-3t)/2

(x,y,z) = ((5-3t)/2, (5-t)/2, t)