Therefore,
\(\dfrac{1}{60} + \dfrac{1}{t}\)
\(=\dfrac{1}{45},\ t \gt 0\)
\(180t\left(\dfrac{1}{60}+\dfrac{1}{t}\right)\)
\(= \left(\dfrac{1}{45}\right)180t\)
\(\overset{\color{BrickRed}3}{\cancel{180}} t\left(\dfrac{1}{ \cancel{60}} \right)+ 180 \cancel{t} \left(\dfrac{1}{ \cancel{t}}\right)\)
\(= \left(\dfrac{1}{ \cancel{45}}\right) \overset{\color{BrickRed}4}{\cancel{180}}t\)
Thus \(t=180\), which satisfies the condition, \(t \gt 0\).
Therefore, it would take Greg \(3\) hours \(\left( 180 \right.\) minutes\(\left. \right)\) to cut the lawn on his own, using the push mower.