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$\frac{55.3}{P_1}=1.32$

I am confused as to whether I can cross multiply; thus get

$\frac{55.3}{1}=\ P_11.32$

then divide both sides by 1.32 to get 41.9

Or these steps correct. I am utilizing any illegal math?

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  • $\begingroup$ 'Cross multiplying' is just multiplying both sides by $P_{1}$, so no problem there, as long as you were allowed to divide by $P_{1}$ $\endgroup$ – krirkrirk Sep 12 '15 at 19:46
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$\require{cancel}$Recall that you're actually multiplying/dividing both sides of the initial equality by a number. $$\frac{55.3}{P_1} = 1.32 \implies \frac{55.3}{\cancel{P_1}}\cdot \cancel{P_1} = 1.32 \cdot P_1 \implies 55.3 = 1.32 \,P_1.$$Now: $$55.3 = 1.32 \,P_1 \implies 55.3 \cdot \frac{1}{1.32} = \cancel{1.32}\,P_1 \cdot\frac{1}{\cancel{1.32}} \implies P_1 = \frac{55.3}{1.32} = 41.9.$$

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  • $\begingroup$ So no cross multiplying? $\endgroup$ – Sunny Sep 12 '15 at 19:48
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    $\begingroup$ I am explaining what is behind "cross-multiplying". You can do this, yes. $\endgroup$ – Ivo Terek Sep 12 '15 at 19:49
  • $\begingroup$ When I cross multiply which side do the products appear for the factors? $\endgroup$ – Sunny Sep 12 '15 at 19:51
  • $\begingroup$ I'm not sure I get it, but: $$\frac{A}{B} = \frac{C}{D} \implies AD = BC.$$I think you should stick with the procedure I did above until you convince yourself that it is the same thing, though. At least you'll be able to understand each step. $\endgroup$ – Ivo Terek Sep 12 '15 at 19:54
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    $\begingroup$ If you're starting an algebra class, recommendation is to let go of "cross-multiplying" because it's a fairly limited concept, and make sure that you understand how inverse operations work to transform expressions and equations. $\endgroup$ – Daniel R. Collins Sep 13 '15 at 6:23

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