What is the solution to the following system?
mc014-1.jpg
x = 10, y = 2, z = 5
x = –4, y = 2, z = 5
x = 0, y = 2, z = 5
x = 4, y = 6, z = 5
Answers
Answered by
11
Answer:
The answer is x=0, y=2, and z=5
Step-by-step explanation:
mc014-1.jpg is as follows
x+2y+z=9 (1)
x-y+3z=13 (2)
2z=10 (3)
Equation 2 can be divided by 2 to get z = 5
plug in z=5 in to all equations to get rid of the z variable temporarily
x+2y+5=9
x-y+15=13
if we input any other number for x or y we don't get 9 or 13, therefore by proccess of elimination they have to be 0, and 2
Answered by
0
Answer:
ddddddddddddd
Step-by-step explanation:
ddddddddddddd
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