How do you evaluate #1/2m - 2/3n# when #m=3# and #n=8#?
1 Answer
Jun 13, 2017
Explanation:
#"substituting in the given values for m and n"#
#rArr1/2xxcolor(red)(3/1)-2/3xxcolor(magenta)(8/1)#
#"multiply values on numerators/denominators of"#
#"both fractions"#
#=(1xxcolor(red)(3))/(2xxcolor(red)(1))-(2xxcolor(magenta)(8))/(3xxcolor(magenta)(1))#
#=3/2-16/3#
#"we require the fractions to have a "color(blue)" common denominator"#
#=3/2xx3/3-16/3xx2/2#
#=9/6-32/6#
#=-23/6#
