How do you solve #5y+3=3(y-3)-2(y+2)#? Algebra Linear Equations Equations with Variables on Both Sides 1 Answer t0hierry · EZ as pi Jan 31, 2017 #y = -4# Explanation: #5y + 3 = 3y - 9 - 2y - 4 = y - 13# thus #4y = -16# #y = -4# check #5y + 3 = -20 + 3 = -17# #3y -9 -2y - 4 = -12 - 9 + 8- 4 = -17# 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 1330 views around the world You can reuse this answer Creative Commons License