Solve for both systems: y+7z=22 and 8y + 7z = 127

Manipulate the first equation to

y = 22-7z

Then substitute in the second to find z

8 (22-7z) + 7z = 127

176-56z+7z = 127

-49z = - 49

z = 1

So y = 22-7 (1) = 22-7 = 15

y = 15; z = 1

This is easy, one way is to subsitute

y+7z=22 and

8y+7z=127

y+7z=22

subtract y

7z=22-y

8y+7z=127

subtract 8y

7z=127-8y

subsitute

22-y=7z=127-8y

22-y=127-8y

add y to both sides

22=127-7y

subtract 127

-105=-7y

divide by 7

15=y

subsitute

15+7z=22

subtract 15

7z=7

divide 7

z=1

y=15

z=1