Sure! Let's solve the equation step by step:
Given:
[tex]\[ 25x^2y^3 = 25xy \][/tex]
1. Divide both sides by 25:
[tex]\[ x^2y^3 = xy \][/tex]
2. Isolate the variable term on one side:
[tex]\[ x^2y^3 - xy = 0 \][/tex]
3. Factor out the common term [tex]\((xy)\)[/tex]:
[tex]\[ xy(xy^2 - 1) = 0 \][/tex]
4. Set each factor equal to zero:
- For the first factor:
[tex]\[ xy = 0 \][/tex]
- This gives us two possibilities:
- [tex]\( x = 0 \)[/tex]
- [tex]\( y = 0 \)[/tex]
- For the second factor:
[tex]\[ xy^2 - 1 = 0 \][/tex]
[tex]\[ xy^2 = 1 \][/tex]
[tex]\[ x = \frac{1}{y^2} \][/tex]
So, the solution to the equation [tex]\( 25x^2y^3 = 25xy \)[/tex] includes the following possibilities:
- [tex]\( x = 0 \)[/tex]
- [tex]\( y = 0 \)[/tex] (not providing a valid x)
- [tex]\( x = \frac{1}{y^2} \)[/tex]
Thus, the final solutions are:
[tex]\[ x = 0 \][/tex]
[tex]\[ x = y^{-2} \][/tex]
Since we were asked to solve for [tex]\( x \)[/tex], the solutions are:
[tex]\[ x = 0 \text{ or } x = y^{-2} \][/tex]
This is consistent with what we found!
