How do you solve #5(4x-1)=2(x+3)#? Algebra Linear Equations Equations with Variables on Both Sides 1 Answer Shwetank Mauria Aug 18, 2016 #x=11/18# Explanation: #5(4x-1)=2(x+3)# #hArr5×4x-5×1=2×x+2×3# #hArr20x-5=2x+6# or #20x-2x=6+5# or #18x=11# or #x=11/18# 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 1305 views around the world You can reuse this answer Creative Commons License