Problem to solve
Solve \[-4x+y-5z=2\], \[4x+2y+z=1\], \[2x-y+3z=4\]
Solution
STEP 1
Solve the following system:
{-4 x + y - 5 z = 2 | (equation 1)
4 x + 2 y + z = 1 | (equation 2)
2 x - y + 3 z = 4 | (equation 3)
STEP 2
Add equation 1 to equation 2:
{-(4 x) + y - 5 z = 2 | (equation 1)
0 x+3 y - 4 z = 3 | (equation 2)
2 x - y + 3 z = 4 | (equation 3)
STEP 3
Add 1/2 × (equation 1) to equation 3:
{-(4 x) + y - 5 z = 2 | (equation 1)
0 x+3 y - 4 z = 3 | (equation 2)
0 x - y/2 + z/2 = 5 | (equation 3)
STEP 4
Multiply equation 3 by 2:
{-(4 x) + y - 5 z = 2 | (equation 1)
0 x+3 y - 4 z = 3 | (equation 2)
0 x - y + z = 10 | (equation 3)
STEP 5
Add 1/3 × (equation 2) to equation 3:
{-(4 x) + y - 5 z = 2 | (equation 1)
0 x+3 y - 4 z = 3 | (equation 2)
0 x+0 y - z/3 = 11 | (equation 3)
STEP 6
Multiply equation 3 by 3:
{-(4 x) + y - 5 z = 2 | (equation 1)
0 x+3 y - 4 z = 3 | (equation 2)
0 x+0 y - z = 33 | (equation 3)
STEP 7
Multiply equation 3 by -1:
{-(4 x) + y - 5 z = 2 | (equation 1)
0 x+3 y - 4 z = 3 | (equation 2)
0 x+0 y+z = -33 | (equation 3)
STEP 8
Add 4 × (equation 3) to equation 2:
{-(4 x) + y - 5 z = 2 | (equation 1)
0 x+3 y+0 z = -129 | (equation 2)
0 x+0 y+z = -33 | (equation 3)
STEP 9
Divide equation 2 by 3:
{-(4 x) + y - 5 z = 2 | (equation 1)
0 x+y+0 z = -43 | (equation 2)
0 x+0 y+z = -33 | (equation 3)
STEP 10
Subtract equation 2 from equation 1:
{-(4 x) + 0 y - 5 z = 45 | (equation 1)
0 x+y+0 z = -43 | (equation 2)
0 x+0 y+z = -33 | (equation 3)
STEP 11
Add 5 × (equation 3) to equation 1:
{-(4 x)+0 y+0 z = -120 | (equation 1)
0 x+y+0 z = -43 | (equation 2)
0 x+0 y+z = -33 | (equation 3)
STEP 12
Divide equation 1 by -4:
{x+0 y+0 z = 30 | (equation 1)
0 x+y+0 z = -43 | (equation 2)
0 x+0 y+z = -33 | (equation 3)
STEP 13
Collect results:
Answer: | 
 | {x = 30
y = -43
z = -33