(15–7xy^2)–(y^3–y^2x–15)
2 answers:
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Given
We have to set the restraint
because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:
The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.
Similarly, we have
So, we have to impose
Squaring both sides, we have
The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.