Solve $\displaystyle 3/4x - 5 = 4$

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.

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- Dec 27th 2007, 10:07 AMelliotfslMultiplicative Inverse?
Solve $\displaystyle 3/4x - 5 = 4$

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. - Dec 27th 2007, 10:17 AMPlato
$\displaystyle \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}$ - Dec 27th 2007, 10:17 AMDINOCALC09
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 - Dec 27th 2007, 10:43 AMelliotfsl