Question

Question #f2e0e

Answer 1

Photo of question is incomplete.
No solution is possible.

Answer 2

Not enough information to pinpoint the exact values of the variables.

Explanation:

First, let's simplify the second equation.
We have, `(ax)/b-(by)/a=(a^4-b^4)/(ab)`
We use the common denominator `ab` to get:
`(ax xxa)/(ab)-(b xxby)/(ab)=(a^4-b^4)/(ab)`
`(a^2x)/(ab)-(b^2y)/(ab)=(a^4-b^4)/(ab)`
`ab[(a^2x)/(ab)-(b^2y)/(ab)=(a^4-b^4)/(ab)]`
`a^2x-b^2y=a^4-b^4`
We can quite safely say that `a^2x=a^4` and `b^2y=b^4`
When we simplify this, we get: `x=a^2` and `y=b^2`

And we see that both of the equations hold true when we plug these values into them when `a!=0` and `b!=0`. As long as the values you use fit these requirements, the equations will always hold true. Without enough information, there is infinitely many solutions you could have.