(ii) 2x + y + 4z = 6

(iii) x – 4y + 5z = 9

i’ll try to cancel from (ii) & (iii) first, because the values are numbers =)

(ii) 2x + y + 4z = 6

(iii) x – 4y + 5z = 9

__multiple (iii) by 2, to cancel ‘x’

(ii) 2x + y + 4z = 6

(iii) 2x – 8y + 10z = 18

__________________ –

(a) 9y – 6z = -12

–> (a) 3y – 2z = -4

then cancel x from (i) & (ii)

(i) x + 2y + z = k

(ii) 2x + y + 4z = 6

__multiple (i) by 2, to cancel ‘x’

(i) 2x + 4y +2z = 2k

(ii) 2x + y + 4z = 6

________________ –

(b) 3y – 2z = 2k -6

then look at (a) and (b)

(a) 3y – 2z = -4

(b) 3y – 2z = 2k -6

see that the left side of both of them are the same?

so will the right side! =D

-4 = 2k – 6

2k = -4 + 6

2k = 2

k =1

hope this help! =)