How do I solve for F in F = 9/5C + 32?

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Discussion Overview

The discussion revolves around solving the equation F = 9/5C + 32 for the variable F. Participants explore various algebraic manipulations and interpretations of the equation, focusing on the correct application of mathematical operations.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests multiplying both sides of the equation by 5/9 as a method to isolate F.
  • Another participant presents a series of transformations, questioning the correctness of their steps and proposing an alternative form of the equation.
  • A subsequent reply indicates that the transformation presented is incorrect and provides a corrected version of the equation.
  • Several participants reiterate the original equation F = 9/5C + 32 without further elaboration.

Areas of Agreement / Disagreement

There is no clear consensus on the correct method to isolate F, as participants present differing approaches and corrections without settling on a single solution.

Contextual Notes

Some participants' steps involve assumptions about algebraic manipulation that may not be universally accepted, and there are unresolved mathematical transformations throughout the discussion.

alg
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How do I solve for F in

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frctl said:
How do I solve for F in

Hint: Multiply both sides of the equation by 5/9.

-Dan
 
Is the following correct?

9/5C = 9/5F - 9/5(1/32)
C = F - 1/32
F = C + 1/32
 
frctl said:
Is the following correct?

9/5C = 9/5F - 9/5(1/32)
C = F - 1/32
F = C + 1/32

not quite ...

$C = \dfrac{9}{5}(F-32)$

as suggested by topsquark ...

${\color{red}\dfrac{5}{9}} \cdot C = {\color{red}\dfrac{5}{9}} \cdot \dfrac{9}{5}(F-32)$

$\dfrac{5}{9} \cdot C = \cancel{\dfrac{5}{9}} \cdot \cancel{\dfrac{9}{5}} ( F - 32)$ (note that any number times its reciprocal equals 1)

$\dfrac{5}{9} \cdot C = F - 32$

can you finish ?
 
F = 9/5C + 32

Thank you!
 
frctl said:
F = 9/5C + 32

correction ...

$F=\dfrac{5}{9} \cdot C + 32$
 

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