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Nikitich [7]
3 years ago
9

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

Mathematics
2 answers:
7nadin3 [17]3 years ago
5 0
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
Zielflug [23.3K]3 years ago
5 0
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
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