Question 919243
The idea is to get each term to have the same denominator. Then we combine the fractions. The LCD is {{{x(x+5)}}}


{{{4 + 2/x - (3x)/(x+5)}}}


{{{4/1 + 2/x - (3x)/(x+5)}}}


{{{(4x(x+5))/(1*x(x+5)) + 2/x - (3x)/(x+5)}}}


{{{(4x(x+5))/(x(x+5)) + 2/x - (3x)/(x+5)}}}


{{{(4x(x+5))/(x(x+5)) + (2(x+5))/(x(x+5)) - (3x)/(x+5)}}}


{{{(4x(x+5))/(x(x+5)) + (2(x+5))/(x(x+5)) - (3x*x)/((x+5)*x)}}}


{{{(4x(x+5))/(x(x+5)) + (2(x+5))/(x(x+5)) - (3x^2)/(x(x+5))}}}


{{{(4x(x+5)+2(x+5)-3x^2)/(x(x+5))}}}


{{{(4x^2+20x+2x+10-3x^2)/(x(x+5))}}}


{{{(x^2+22x+10)/(x(x+5))}}}


=======================================================


{{{4 + 2/x - (3x)/(x+5)}}} simplifies to {{{(x^2+22x+10)/(x(x+5))}}}



Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>
or you can visit my website here: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a>


Thanks,


Jim