Solve the system using the elimination method.

2x + 2y + 5z =−1

2x − y + z =2

2x + 4y − 3z =14

what x, y, and z?

Multiplying the first equation by -1 and adding it to the third, we get 2y-8z=15 and by multiplying the first equation by -1 and adding it to the second equation, we get -3y-4z=3

Using the two equations, we have 
2y-8z=15
-3y-4z=3

Multiplying the second equation by -2 and adding it to the third, we have 8y=9 and y = 9/8 by dividing both sides by 8. Plugging it back into
-3y-4z=3, we have -27/8-4z=3 and by adding 27/8 to both sides, we have -4z=51/8 and by dividing both sides by -4 we get z=-51/32.

Plugging it into 2x − y + z =2, we get 2x-9/8+5(-51/32)=2, we get x=151/64 by adding -(5(-51/32)) to both sides as well as adding 9/8 to both.

The mode of the numbers is 29.