Question 728564
>>...A machine caps a 1000 bottles in 10 minutes,...<< 
<pre>
So the slower machine's bottle capping rate is 1000 bottles 
per 10 minutes or

{{{1000_bottles/10_min}}} = {{{1000/10}}}{{{bottles/min}}} = 100{{{bottles/min}}}
</pre>
another machine caps 1000 bottles in 8 minutes.
<pre>
So the faster machine's bottle capping rate is 1000 bottles 
per 8 minutes or

{{{1000_bottles/8_min}}} = {{{1000/8}}}{{{bottles/min}}} = 125{{{bottles/min}}}
</pre>
>>...If these machines were together how much time will it take to cap 1000 bottles?...<<
<pre>
Let that time be x minutes

So their combined bottle capping rate is 1000 bottles 
per x minutes or

{{{1000_bottles/x_min}}} = {{{1000/x}}}{{{bottles/min}}}

The equation comes from:

            {{{(matrix(4,1,

SLOWER,"MACHINE'S",BOTTLING,RATE))}}}{{{""+""}}}{{{(matrix(4,1,

FASTER,"MACHINE'S",BOTTLING,RATE))}}}{{{""=""}}}{{{(matrix(4,1,

THEIR,COMBINED,BOTTLING,RATE))}}}

                100{{{bottles/min}}} + 125{{{bottles/min}}}{{{""=""}}}{{{1000/x}}}

                           100 + 125 = {{{1000/x}}}
                                 225 = {{{1000/x}}}
                                225x = 1000
                                   x = {{{1000/225}}}
                                   x = {{{40/9}}{}
                                   x = {{{4&4/9}}} minutes

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</pre>
b) if t1 and t2 are the time in the above problem. Determine the time taken by both machines working together?   
<pre>

>>...A machine caps a 1000 bottles in t1 minutes,...<< 
<pre>
So the slower machine's bottle capping rate is 1000 bottles 
per t<sub>1</sub> minutes or

{{{1000_bottles/t[1]min}}} = {{{1000/t[1]}}}{{{bottles/min}}}
</pre>
>>...another machine caps 1000 bottles in t2 minutes...<<
<pre>
So the faster machine's bottle capping rate is 1000 bottles 
per t<sub>2</sub> minutes or

{{{1000_bottles/t[2]min}}} = {{{1000/t[2]}}}{{{bottles/min}}}
</pre>
>>...If these machines were together how much time will it take to cap 1000 bottles?...<<
<pre>
Let that time be x minutes

So their combined bottle capping rate is 1000 bottles 
per x minutes or

{{{1000_bottles/x_min}}} = {{{1000/x}}}{{{bottles/min}}}

The equation comes from:

            {{{(matrix(4,1,

SLOWER,"MACHINE'S",BOTTLING,RATE))}}}{{{""+""}}}{{{(matrix(4,1,

FASTER,"MACHINE'S",BOTTLING,RATE))}}}{{{""=""}}}{{{(matrix(4,1,

THEIR,COMBINED,BOTTLING,RATE))}}}

                {{{1000/t[1]}}}{{{bottles/min}}} + {{{1000/t[2]}}}{{{bottles/min}}} = {{{1000/x}}}{{{bottles/min}}}

                           {{{1000/t[1]}}} + {{{1000/t[2]}}} = {{{1000/x}}}

The LCD is t<sub>1</sub>t<sub>2</sub>x

                           1000t<sub>2</sub>x + 1000t<sub>1</sub>x = 1000t<sub>1</sub>t<sub>2</sub>
                 
Divide through by 1000

                                   t<sub>2</sub>x + t<sub>1</sub>x = t<sub>1</sub>t<sub>2</sub>

Factor out x on left
              
                                  x(t<sub>2</sub> + t<sub>1</sub>) = t<sub>1</sub>t<sub>2</sub>

Divide both sides by (t<sub>2</sub> + t<sub>1</sub>)

                                    {{{x(t[2]+t[1])/((t[2]+t[1]))}}} = {{{(t[1]t[2])/((t[2]+t[1]))}}}

                                    {{{x(cross(t[2]+t[1]))/((cross(t[2]+t[1])))}}} = {{{(t[1]t[2])/((t[2]+t[1]))}}}

                                     x = {{{(t[1]t[2])/(t[2]+t[1])}}} minutes 
 
That's the answer.
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So the formula is "the product of the times over the sum of the times":

We could have used that formula to solve (a) if we had had it, with

t<sub>1</sub> = 10 and t<sub>2</sub> = 8

                                     x = {{{10*8/(10+8)}}}
                                     x = {{{80/18}}}
                                     x = {{{40/9}}}
                                     x = {{{4&4/9}}}minutes  

Edwin</pre>