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Need help with this problem. Solve ∛(3x+8) + 6 = 1
Need help with this problem. Divide √(15xy^3)/√3x
1. ∛(3x+8) + 6 = 1, subtract 6 from both sides: ∛(3x+8) = -5
. now cube both sides: (∛(3x+8))^3 = (-5)^3
. which gives: 3x+8 = -125, now solve for x
. subtract 8 from both sides: 3x = -133
. divide by 3 to get: x= -133/3
2. √(15xy^3)/√3x, I'll assume the "x" in the denominator is also
. in the square root...yes? Thus, we get
. √[(15xy^3)/(3x)] = √(5y^3) = √(5y^2*y)
. = y√(5y) if y is geater than or equal to zero.
Abe
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| Rating(1-10) | Knowledgeability = 10 | Clarity of Response = 10 | Politeness = 10 |
| Comment | Abe, Thank you very much for the positive feedback. I did not know that you would reply so quickly. I did solve the first problem however, on the second problem I did not put the y on the outside I only had 5y. thanks again I look forward to your assistance in the future. Now that I see that you reply so quickly. | ||
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Answers by Expert:
Abe Mantell
Expertise
Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!
Experience
Over 15 years teaching at the college level.
Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.
Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook
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