How do you simplify #(6g)/(g+5)-(g-2)/(2g)#?
1 Answer
Mar 12, 2017
Explanation:
Before we can subtract the fractions we require them to have a
#color(blue)"common denominator"# This is achieved by multiplying the numerator/denominator of
#(6g)/(g+5)" by 2g and the numerator/denominator of " (g-2)/(2g)#
by (g + 5)
#rArr((6g)/(g+5)xx(2g)/(2g))-((g-2)/(2g)xx(g+5)/(g+5))#
#=(12g^2)/(2g(g+5))-((g-2)(g+5))/(2g(g+5))# Now that the fractions have a common denominator we can subtract the numerators, leaving the denominator as it is.
#=(12g^2-(g^2+3g-10))/(2g(g+5))#
#=(12g^2-g^2-3g+10)/(2g(g+5))#
#=(11g^2-3g+10)/(2g(g+5))#
