\[x^2y^3=1125\]
(1)\[x^3y^2-675\]
(2)Can most easily be solved by raising each equation to some power so that the powers of
\[x\]
( or \[y\]
are equalised, then dividing the equationsRaise (1) to the power of 3 and (2) to the power of 2 giving
\[x^6y^9=1125^3\]
(3)\[x^6y^4=675^2\]
(4)(3) divided by (4) gives
\[\frac{x^6y^9}{x^6y^4}=\frac{1125^3}{675^2} \rightarrow y^5=3125 \rightarrow y=\sqrt[5]{3125}=5\]
Then from (1)
\[x^25^3=1125 \rightarrow x^2=\frac{1125}{5^3}=9 \rightarrow x=\sqrt{9}=3\]
Only one square root is taken since
\[x=-3\]
does not fit equation (2)