How do you solve #|- 3- 3y | = | 2y + 1|#?

2 Answers
Apr 6, 2017

Solution : #y = -2 or y = -4/5 #

Explanation:

#| -3-3y| = |2y +1| # squaring both sides we get

#(-3-3y)^2 = (2y +1)^2 or 9(1+y)^2 = 4y^2 +4y +1 # or

#9(y^2+2y+1) = 4y^2 +4y +1 or 5y^2 +14y +8 =0 # or

#5y^2 +10y + 4y +8 =0 or 5y(y+2) +4(y+2) =0 # or

#(y+2)(5y+4) =0 :. #Either #y+2=0 :. y=-2 or 5y+4=0 :. y = -4/5#

Solution : #y = -2 or y = -4/5 # [Ans]

Apr 6, 2017

#y=-2" or " y=-4/5#

Explanation:

There are 2 possible solutions.

#-3-3y=color(red)(+-)(2y+1)#

#color(blue)"First solution"#

#-3-3y=2y+1#

#rArr-5y=4rArry=-4/5#

#color(blue)"Second solution"#

#-3-3y=-2y-1#

#rArr-y=2rArry=-2#

#color(blue)"As a check"#

Substitute these values into the equation and if left side is equal to right side then they are the solutions.

#"left side "=|-3+12/5|=|-3/5|=3/5#

#"right side " =|-8/5+1|=|-3/5|=3/5#

#"left side " =|-3+6|==|3|=3#

#"right side " =|-4+1|=|-3|=3#