Bosko C. answered • 04/16/15
Tutor
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CSUSM Teaching Associate with degrees in Mathematics and Engineering.
9y-12*3y+27=0
(32)y-12*3y+27=0
(3y)2-12*3y+27=0
Now, we apply substitution:
t=3y
So:
t2-12t+27=0
(t-3)(t-9)=0
t1=3 ⇒ 3y=3 ⇒ 3y=31 ⇒ y1=1
t2=9 ⇒ 3y=9 ⇒ 3y=32 ⇒ y2=2
