The solution to the given system of equation is (3, 3, 3)
Given the following system of equation expressed as:
2x-y+5z=16
X-6у +2z=-9
3x+4y- z=32
Multiply equation 2 by 2 to have:
2x-y+5z=16
2X-12у +4z=-18
Subtract
10y+z = 34 .........4
Similarly
3X-18у +6z=-27
3x+4y- z=32
Subtract
-22y+7z = -59 ........... 5
Equate 4 and 5
10y+z = 34 .........4 * 7
-22y+7z = -59 ........... 5 * 1
_______________________
70y+7z = 238
-22y+7z = -59
Subtract
92y = 297
y = 3
Recall that
10y+z = 34
10(3) + z = 34
z = 3
Since x - 6y +2z = -9
x-6(3)+2(3) = -9
x - 18 + 6 = -9
x = 3
Hence the solution to the given system of equation is (3, 3, 3)
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