How do you solve #(x+5)-2(4x-1)=0#? Algebra Linear Equations Distributive Property for Multi-Step Equations 1 Answer Don't Memorise Jun 29, 2016 #x=1# Explanation: #(x+5) - 2 ( 4x - 1) = 0# #(x+5) - 2 * ( 4x - 1) = 0# #x+5 + (-2) * (4x) + ( -2) * (- 1) = 0# #x+5 -8x +2 = 0# #x-8x +2 + 5= 0# #-7x + 7= 0# #-7x = - 7# #x = (-7)/(-7)# #x=1# Answer link Related questions How do you solve multi step equations with distributive property? Do you always have to use the distributive property or can you just divide by the number? How do you solve #3(x - 1) - 2(x + 3) = 0#? How do you solve for w in #7(w + 20) - w = 5#? How do you solve for r in #- \frac{59}{60} = \frac{1}{6} \(- \frac{4}{3} r-5 )#? How do you solve #(c+3)-2c-(1-3c)=2#? What are the two ways to solve for x and get rid of the parentheses in #2(5x+9)=78#? How do you solve #-(m+4)=-5#? How do you solve #8(1+7m)+6=14#? How do you simplify and solve the equation #5m-3[7-(1-2m)]=0#? See all questions in Distributive Property for Multi-Step Equations Impact of this question 18349 views around the world You can reuse this answer Creative Commons License