Multiplicative Inverse?

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December 27th 2007, 10:07 AM
elliotfsl

Multiplicative Inverse?

Solve $3/4x - 5 = 4$

What I don't understand, is if I have to multiply each side by the reciprocal how do I do it if the reciprocal is infinite like 3/4, which ends up being 1.3333333 etc. etc.
December 27th 2007, 10:17 AM
Plato

$\begin{array}{rcl} \frac{3}{4}x - 5 & = & 4 \\ \frac{3}{4}x & = & 9 \\ \left( {\frac{4}{3}} \right)\frac{3}{4}x & = & 9\left( {\frac{4}{3}} \right) \\ x & = & 12 \\ \end{array}$
December 27th 2007, 10:17 AM
DINOCALC09

to obtain a reciprical, just switch the numerator and denominator.

the reciprical of 3/4 is 4/3

* the purpose of a reciprical is to get rid of any coefficients in front of your variable

so if you have 3/4 x -5 = 4

first you add 5 to both sides to get:

3/4 x = 9

now multiple by the reciprical of 3/4 which is 4/3

x = 12

try some practice problems...

find the reciprical of

1/12

34/2

0.5

0.1

NOTE: to find the recipricals of decimals you will first need to find its fractional equivalent.

the example you provided:

1.333333333333

its fraction is 4/3.

now swithc numerator and denominator to get 3/4
December 27th 2007, 10:43 AM
elliotfsl

Quote:

Originally Posted by Plato

$\begin{array}{rcl} \frac{3}{4}x - 5 & = & 4 \\ \frac{3}{4}x & = & 9 \\ \left( {\frac{4}{3}} \right)\frac{3}{4}x & = & 9\left( {\frac{4}{3}} \right) \\ x & = & 12 \\ \end{array}$

Thank you very much both of you for your help, it makes sense now. :)