How do you add or subtract #(x+6)/(5x+10) - (x-2)/(4x+8)#?

1 Answer
George C.
May 25, 2015

Notice that the denominator of each of these terms is a simple multiple of #(x+2)#, so we only need to multiply by constants to make the denominators of the two terms the same:

#(x+6)/(5x+10)-(x-2)/(4x+8)#

#=(x+6)/(5(x+2))-(x-2)/(4(x+2))#

#=(4*(x+6))/(4*5(x+2))-(5*(x-2))/(5*4(x+2))#

#=(4(x+6))/(20(x+2))-(5(x-2))/(20(x+2))#

#=(4(x+6)-5(x-2))/(20(x+2))#

#=(4x+24-5x+10)/(20(x+2))#

#=(14-x)/(20(x+2))#

#=(14-x)/(20x+40)#

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