The solution to the system of equations is x=2,y=3,z=1.
Let's solve the system of linear equations:
−x+4y−5z=5
y=−x+5
−3x+5z=−1
We can substitute the expression for y from equation (2) into equations (1) and (3) to eliminate y.
Substitute y=−x+5 into equation (1):
−x+4(−x+5)−5z=5
Simplify:
−x−4x+20−5z=5
Combine like terms:
−5x−5z+20=5
Subtract 20 from both sides:
−5x−5z=−15
Divide both sides by -5:
x+z=3
Now, substitute y=−x+5 into equation (3):
−3x+5z=−1
Replace y with −x+5:
−3x+5z=−1
Now we have two equations:
x+z=3
−3x+5z=−1
Let's solve this system. We can use the first equation to express x in terms of z:
From equation (1):
x=3−z
Now substitute this into equation (2):
−3(3−z)+5z=−1
Distribute the -3:
−9+3z+5z=−1
Combine like terms:
8z=8
Divide both sides by 8:
z=1
Now substitute z=1 back into
x=3−z:
x=3−1=2
Now substitute z=1 into y=−x+5:
y=−2+5=3
So, the solution to the system of equations is x=2,y=3,z=1.