How to solve #1/2 +x/3 +x^2/4# where #x=1/2#? Please show steps. Algebra Expressions, Equations, and Functions Variable Expressions 1 Answer Alan P. Sep 8, 2015 #1/2+x/3+x^2/4# with #x=1/2# #color(white)("XXXXXXXXXXXXXXXX")=35/48# Explanation: Given #1/2+x/3+x^2/4# with #x=color(red)(1/2)# #1/2+color(red)(1/2)/3+(color(red)(1/2))^2/4# #=1/2+color(blue)((1/2)/3)+color(green)((1/4)/4)# #=1/2+color(blue)(1/2*1/3)+color(green)((1/4)*1/4)# #=1/2+color(blue)(1/6) + color(green)(1/16)# least common multiple is #3xx16 = 48# #color(blue)(1/2)+color(orange)(1/6) + color(green)(1/64)# #=color(blue)(24/48)+color(orange)(8/48) + color(green)(35/48)# #=35/48# Answer link Related questions How do you write the variable expression for: a quotient of 2 and the sum of a number and 3 ? What are variables? What are variable expressions? How do you write variable expressions? How do you evaluate variable expressions? How do you simplify the expression #3x-x+4#? How do you write a quotient of a number and 6 as an expression? How do you evaluate the expression #2x+1# for #x=1#? How do you write a product of a number and 2 as an expression? How do you write 5 less than 2 times a number as a variable expression? See all questions in Variable Expressions Impact of this question 1699 views around the world You can reuse this answer Creative Commons License