Thread: i need help with two questions <3

please help me solve these! i have an exam tomorrow! please explain the solutions ok ^_^

F = (9/5)C +32

The idea is to isolate C alone by itself, woth no attachments, on one side of the equation.
32 is with the (9/5)C in the righthand side (RHS), so remove the 32 from there. Transpose the 32 to the lefthand side (LHS), or subtract 32 from both sides,
F -32 = (9/5)C
The C is attached to the fraction 9/5. Remove that 9/5 by dividing both sides by 9/5,
(F -32) / (9/5) = C
(5/9)(F -32) = C
C = (5/9)(F -32) ------------answer.

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6 -(n +(1/9)x) = 7n

Again, try anything to isolate n on one side of the equation.

6 -n -x/9 = 7n
6 -x/9 = 7n +n
6 -x/9 = 8n
Divide both sides by 8,
(6 -x/9) / 8 = n
((6*9 -x) /9) / 8 = n
((54 -x) / 9) *1/8 = n
(54 -x)*1 / (9*8) = n
n = (54 -x) / 72 -------------answer.