Here we are given three equations we have to solve to arrive at the values of x, y and z.

x + 2y = 10… (1)

9y – 3z = - 6… (2)

6x – 5z = 2… (3)

We can write (1) as x = 10-2y

We can also write (3) as -5z = 2 -6x

=> 5z= 6x-2

=> 5z= 6*(10-2y) -2

=>5 z = 60- 12y -2

=> 5z = 58 – 12y

=> z = (58 – 12y)/5

Now substitute these in (2)

9y – 3z = - 6

=> 9y – 3[(58 – 12y)/5] = -6

=> 45y -174+ 36y = -30

=> 81y = 144

=> y = 144/81

=> y = 16/9

Now x = 10-2y = 10 - 2*16/9= 58/9

Also z= (58 – 12y)/5= (58 – 12*16/9)/5= 22/3

**Therefore we get x= 58/9, y = 16/9 and z = 22/3**

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