How do you solve #-2(x + 3) = -2(x + 1) - 4#? Algebra Linear Equations Equations with Variables on Both Sides 1 Answer Mia Sep 7, 2017 #x# can be any real number #" "# #x in RR# Explanation: #-2 (x+3) = -2 (x+1) -4# #" "# #rArr-2×x-2×3 = -2×x-2×1-4# #" "# #rArr-2x-6 = -2x-2-4# #" "# #rArr-2x-6 = -2x-6# #" "# #rArr-2x+2x = -6+6# #" "# #rArr0.x = 0# #" "# #x# can be any real number #" "# #therefore x in RR# Answer link Related questions How do you check solutions to equations with variables on both sides? How do you solve #125+20w-20w=43+37w-20w#? How do you solve for x in #3(x-1) = 2 (x+3)#? Is there a way to solve for x without using distribution in #4(x-1) = 2 (x+3)#? How do you solve for t in #2/7(t+2/3)=1/5(t-2/3)#? How do you solve #5n + 34 = −2(1 − 7n)#? How do you simplify first and then solve #−(1 + 7x) − 6(−7 − x) = 36#? Why is the solution to this equation #-15y + 7y + 1 = 3 - 8y#, "no solution"? How do you solve for variable w in the equation #v=lwh#? How do you solve #y-y_1=m(x-x_1)# for m? See all questions in Equations with Variables on Both Sides Impact of this question 1579 views around the world You can reuse this answer Creative Commons License