Answer:
x=1
y=−2
z=5
(heres how i got the answer)
Explanation:
x+2y+3z=12
x−3y+4z=27
−x+y+2z=7
Solve x+2y+3z=12 for x.
x=−2y−3z+12
Substitute −2y−3z+12 for x in the second and third equation.
−2y−3z+12−3y+4z=27
−(−2y−3z+12)+y+2z=7
Solve equations for y and z respectively.
y=−3+
5
1
z
z=
5
19
−
5
3
y
Substitute −3+
5
1
z for y in the equation z=
5
19
−
5
3
y.
z=
5
19
−
5
3
(−3+
5
1
z)
Solve z=
5
19
−
5
3
(−3+
5
1
z) for z.
z=5
Substitute 5 for z in the equation y=−3+
5
1
z.
y=−3+
5
1
×5
Calculate y from y=−3+
5
1
×5.
y=−2
Substitute −2 for y and 5 for z in the equation x=−2y−3z+12.
x=−2(−2)−3×5+12
Calculate x from x=−2(−2)−3×5+12.
x=1
The system is now solved.
x=1
y=−2
z=5