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