SOLUTION: The given equation is either linear or equivalent to a linear equation. Solve the equation.
z/6=((3/10)z)+2
So far I have tried ...
0=((3/10)z)+2-(z/6)
0=z+2-(z/6)
-1
Algebra.Com
Question 758052: The given equation is either linear or equivalent to a linear equation. Solve the equation.
z/6=((3/10)z)+2
So far I have tried ...
0=((3/10)z)+2-(z/6)
0=z+2-(z/6)
-12=z-z
I've also tried ...
z=((3/10)z)+12
0=((3/10)z)+12-z
-12=((3/10)z)-z
-12/(-7/10)=((-7/10)z)/(-7/10)
-120/-7=z
Found 2 solutions by oscargut, stanbon:
Answer by oscargut(2103) (Show Source): You can put this solution on YOUR website!
Solution
z/6=((3/10)z)+2
Multiply by 6
z = (18/10)z+12
(18/10-1)z = -12
(8/10)z = -12
z = -12(10)/8
z = -15
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Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
z/6=((3/10)z)+2
----
Just a suggestion::
When you have fractions, get rid of them:
---
Multiply thru by 60 to get:
10z = 6*(3z) + 120
----
10z = 18z + 120
---
6z = -120
---
z = -20
====================
Cheers,
Stan H.
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