Math, asked by shddhjhdbfhdvshbfjkj, 6 months ago

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 ZombifiedZombie
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 jrreed1
0

Answer:

ddddddddddddd

Step-by-step explanation:

ddddddddddddd

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