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3 years ago

Solve: 48y^5=27y^3 please solve:))))))))

Mathematics

2 answers:

miss Akunina [59]3 years ago

4 0

Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

y=0,−3/4,3/4

N76 [4]3 years ago

3 0

48y^5=27y^3

subtract 27 y^3 from each side

48y^5-27y^3=0

factor out 3y^3

3y^3(16y^2 -9)=0

factor

3y^3 (4y-3) (4y+3) =0

using the zero product property

3y^3 =0 4y-3 =0 4y+3 =0

y^3=0 4y=3 4y = -3

y=0 y = 3/4 y = -3/4

Answer: y = -3/4, 0, 3/4

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