Question 155544
<font size = 7 color="red"><b>Edwin's solution:</b></pre></font>
On the worksheet it shows the equation: 
{{{1/z-1/2z-1/5z=10/(z+1)}}} The first thing i tried to do was get rid of the fraction by mutiplying by the LCD, which i figured would be (z)(2z)(5z)(z+1). however it ended up being a jumbled mess and left me very confused and now I have no idea what to do.
<pre><font size = 4 color = "indigo"><b>
You are a little confused on how to find the LCD.
You don't need to repeat the factor z in the LCD. 
You only need it ONCE in the LCD, not three times!!!

Let's put parentheses around every term:

{{{(1/z)-(1/(2z))-(1/(5z))=(10/(z+1))}}}

The factors of the denominators are z, 2, 5, and z+1

The factor z appears 1 time in the first fraction, 1 time in the second
fraction, 1 time in the third fraction and 0 times in the fourth fraction.

The most number of times that z appears in any ONE denominator is
1 time.  Therefore z needs to appear only 1 time in the LCD.

---

The factor 2 appears 0 times in the first fraction, 1 time in the second
fraction, 0 times in the third fraction and 0 times in the fourth fraction.

The most number of times that 2 appears in any ONE denominator is
1 time.  Therefore 2 needs to appear only 1 time in the LCD.

---

The factor 5 appears 0 times in the first fraction, 0 times in the second
fraction, 1 time in the third fraction and 0 times in the fourth fraction.

The most number of times that 5 appears in any ONE denominator is
1 time.  Therefore 5 needs to appear only 1 time in the LCD.

---

The factor z+1 appears 0 times in the first fraction, 0 times in the second
fraction, 0 times in the third fraction and 1 time in the fourth fraction.

The most number of times that z+1 appears in any ONE denominator is
1 time.  Therefore z+1 needs to appear only 1 time in the LCD.

---

Therefore, the LCD is (z)(2)(5)(z+1) or 10z(z+1).  We put the LCD
over 1 like this {{{(10z(z+1))/1}}} and multiply it by every term:
  
{{{(1/z)-(1/(2z))-(1/(5z))=(10/(z+1))}}}

{{{(10z(z+1)/1)(1/z)-(10z(z+1)/1)(1/(2z))-(10z(z+1)/1)(1/(5z))=(10z(z+1)/1)(10/(z+1))}}}

Now we cancel away the denominators:

              {{{5}}}               {{{2}}}
{{{(10cross(z)(z+1)/1)(1/cross(z))-(cross(10)cross(z)(z+1)/1)(1/(cross(2)cross(z)))-(cross(10)cross(z)(z+1)/1)(1/(cross(5)cross(z)))=(10z*cross((z+1))/1)(10/cross((z+1)))}}}

{{{10(z+1)-5(z+1)-2(z+1)=10z*10}}}

{{{10z+10-5z-5-2z-2=100z}}}

{{{3z+3=100z}}}

{{{3z-100z=-3}}}

{{{-97z=-3}}}

{{{z=(-3)/(-97)}}}

{{{z=3/97}}}

That's an ugly solution, but it's correct!

Edwin</pre>