
Olivia B.
asked 09/06/16Solve for y: x^4 + 1/ay = xy + a.
Solve for y:
x^4 + 1/ay = xy + a.
1) y = ax − 1/ ax^4 − a^2
2) y = ax^4 − a^2/ x − 1
3) y = ax^4 − a^2/ ax − 1
x^4 + 1/ay = xy + a.
1) y = ax − 1/ ax^4 − a^2
2) y = ax^4 − a^2/ x − 1
3) y = ax^4 − a^2/ ax − 1
4) y = x^4 − a^2/ x − 1
5) y = ax^4 − a/ ax − 1
5) y = ax^4 − a/ ax − 1
I am very confused on how to manipulate this for y because y is one both sides and I don't know what to do. Please help me! Thank you!
More
2 Answers By Expert Tutors
x4 + (1/a)y = xy + a
Multiply by a: ax4 + y = axy + a2
y - axy = a2 - ax4
y(1 - ax) = a2 - ax4
y = (a2 - ax4)/(1 - ax)
y = (ax4 - a2)/(ax - 1)
Note: The answer choices are missing IMPORTANT parentheses!!
Michael J.
You were actually able to interpret this equation without the presence of any parentheses?
Report
09/06/16
Mark M.
tutor
I tried 1/(ay) and (1/a)y. The first option led to a very messy equation where none of the answer choices looked feasible. The second option led to an equation that was easy to solve.
Mark M (Bayport, NY)
Report
09/07/16
Michael J. answered 09/06/16
Tutor
5
(5)
Applying SImple Math to Everyday Life Activities
You need to first move all the y terms to one side of the equation, and move all other terms to the other side of the equation.
Step one:
Move all the y terms to the left side. All the other terms to the right side. When moving terms to the opposite side, always negative the signs while doing so.
Note, it does not necessary have to be in this order. You could move all the y-terms to the right side of equation.
Step two:
Factor out a positive "y" using distributive property. "y" is the common factor on the side that contains the y-terms.
Step three:
Divide both sides of the equation by the coefficients of y. The coefficient is the value being multiplied by "y".
Still looking for help? Get the right answer, fast.
Ask a question for free
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
Mark M.
09/06/16