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

Answer 1

Eli Assouline

#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#

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Answer 2

Julia Adams

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#

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Answer 3

Savannah Anderson

To solve ( (2y + 6)(y - 1) = 0 ), set each factor equal to zero:

( 2y + 6 = 0 )
( y - 1 = 0 )

Solve each equation for ( y ):

( 2y = -6 ) → ( y = -3 )
( y = 1 )

So, the solutions are ( y = -3 ) and ( y = 1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.