4(3y^2 - 7y + 4) = 1

Holt McDougal Larson Algebra 1 Common Core Edition page 539 number 30

Asked by Dee Dee on 25 Mar 10:39

Last updated by anonymous on 27 Mar 14:37

Answered by CherieS on 27 Mar 14:37

12y2-28y+16=1

12y2-28y+16-1=1-1

12y2-28y+15=0

Factor:

12y2 -28y +15=0

(2y )(6y ) ( -3)( -5)

(2y-3)(6y-5)=0

Check:

(2y-3)(6y-5)=0

(2y)(6y) (2y)(-5) (-3)(6y) (-3)(-5)=0

12y2 -10y -18y +15

12y2-28y+15=0

Simplify:

(2y-3)=0

Solve:

(2y-3)=0

2y-3+3=0+3

2y=3

2y/2=3/2

y=3/2

Simplify:

(6y-5)=0

Solve:

(6y-5)=0

6y-5+5=0+5

6y=6

6y/6=5/6

y=5/6