How do you solve #-3y-2 = abs(6y + 25)#?

Question

1529 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-16T23:32:29

#color(red)(y = -3)# and # color(red)(y = -23/3)#.

We need to write two different equations without the absolute value symbols and solve for #y#.

These equations would be

(1) #−3y−2 = (6y+25)#
(2) #−3y−2= -(6y+25)#

Solve Equation 1:

#−3y−2 = 6y+25#

Add #3y# to each side.

#-2 = 9y + 25#

Subtract #25# from each side.

#-27 = 9y#

Divide each side by #3#.

#y = -3#

Solve Equation 2.

#−3y−2= -(6y+25)#

Remove parentheses.

#−3y−2= -6y-25#

Add #6y# to each side.

#3y-2 = -25#

Add 2 to each side.

#3y = -23#

Divide each side by #3#.

#y = -23/3#

The solutions are #y = -3# and #y = -23/3#.

Check:

If #y = -3#,

#−3y−2=|6y+25|#
#-3(-3) -2 = |6(-3) + 25|#
#9-2 = |-18+25|#
#7 =|7|#
#7=7#

If #y= -23/3#,

#−3y−2=|6y+25|#
#-3×(-23/3) -2 = |6×(-23/3) + 25|#
#23-2 = |-46+25|#
#21 =|-46 +25|#
#21= |-21|#
#21 = 21#