Solve the system of equations.
2x + 2y + 5z = 7
6x + 8y + 5z = 9
2x + 3y + 5z = 6
a. (x = 2, y = -1,2 - 1)
b. (x = 1, y = 0, z = 2)
c. (x = 3, y = -2, z = 0)
d. (x = 4, y = -3, z = 3)
select the best answer from the choices provided

using the first two equations cancelling the z

-2x-2y-5z=-7

6x+8y+5z=9

4x +6y= 2

using the second and the third equations cancelling the z

6x+8y+5z= 9

-2x-3y-5z= -6

4x +5y = 3

4x + 6y = 2

-4x -5y = -3

y= -1

2x + 2(-1) + 5z = 7

2x - 2 + 5z = 7

2x + 5z = 9

6x + 8(-1) + 5z = 9

6x - 8 + 5z = 9

6x + 5z = 17

-2x - 5z= -9

4x = 8

x = 2

2(2) + 2(-1) + 5z= 7

4 - 2 + 5z = 7

2 + 5z = 7

5z = 5

z = 1

(2, -1, 1)

the answer is a.

Step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

what is the original number? you forgot to add it.