How do you solve #(2y+6)(y-1)=0#?

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Answer 1 · 2017-06-02T14:20:07

#y=-3# and #y=1#

Set each factor equal to #0# and solve for #y#.

1st Factor:

#2y+6=0#

#2y=-6#

#y=-6/2=-3#

2nd Factor

#y-1=0#

#y=1#

Answer 2 · 2017-06-02T14:23:05

See a solution process below:

Using the zero product principle we solve this problem by equating each of the terms in parenthesis to #0# and then solving for #y#:

Solution 1)

#2y + 6 = 0#

#2y + 6 - color(red)(6) = 0 - color(red)(6)#

#2y + 0 = -6#

#2y = -6#

#(2y)/color(red)(2) = -6/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = -3#

#y = -3#

Solution 2)

#y - 1 = 0#

#y - 1 + color(red)(1) = 0 + color(red)(1)#

#y - 0 = 1#

#y = 1#

The solution is: #y = -3# and #y = 1#