Question

$$2ut+\frac{1}{9}at^{2}=ut+\frac{1at^{2}}{9}at^{2}$$

Answer

$$u=(a*t*(a*t^2-1))/9$$

Solution


Simplify  \(1\times a{t}^{2}\)  to  \(a{t}^{2}\).
\[2ut+\frac{1}{9}a{t}^{2}=ut+\frac{a{t}^{2}}{9}a{t}^{2}\]
Simplify  \(\frac{1}{9}a{t}^{2}\)  to  \(\frac{a{t}^{2}}{9}\).
\[2ut+\frac{a{t}^{2}}{9}=ut+\frac{a{t}^{2}}{9}a{t}^{2}\]
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[2ut+\frac{a{t}^{2}}{9}=ut+\frac{a{t}^{2}a{t}^{2}}{9}\]
Regroup terms.
\[2ut+\frac{a{t}^{2}}{9}=ut+\frac{aa{t}^{2}{t}^{2}}{9}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[2ut+\frac{a{t}^{2}}{9}=ut+\frac{{a}^{1+1}{t}^{2+2}}{9}\]
Simplify  \(1+1\)  to  \(2\).
\[2ut+\frac{a{t}^{2}}{9}=ut+\frac{{a}^{2}{t}^{2+2}}{9}\]
Simplify  \(2+2\)  to  \(4\).
\[2ut+\frac{a{t}^{2}}{9}=ut+\frac{{a}^{2}{t}^{4}}{9}\]
Subtract \(\frac{a{t}^{2}}{9}\) from both sides.
\[2ut=ut+\frac{{a}^{2}{t}^{4}}{9}-\frac{a{t}^{2}}{9}\]
Subtract \(ut\) from both sides.
\[2ut-ut=\frac{{a}^{2}{t}^{4}}{9}-\frac{a{t}^{2}}{9}\]
Simplify  \(2ut-ut\)  to  \(ut\).
\[ut=\frac{{a}^{2}{t}^{4}}{9}-\frac{a{t}^{2}}{9}\]
Join the denominators.
\[ut=\frac{{a}^{2}{t}^{4}-a{t}^{2}}{9}\]
Factor out the common term \(a{t}^{2}\).
\[ut=\frac{a{t}^{2}(a{t}^{2}-1)}{9}\]
Divide both sides by \(t\).
\[u=\frac{\frac{a{t}^{2}(a{t}^{2}-1)}{9}}{t}\]
Simplify  \(\frac{\frac{a{t}^{2}(a{t}^{2}-1)}{9}}{t}\)  to  \(\frac{a{t}^{2}(a{t}^{2}-1)}{9t}\).
\[u=\frac{a{t}^{2}(a{t}^{2}-1)}{9t}\]
Simplify  \(\frac{a{t}^{2}(a{t}^{2}-1)}{9t}\)  to  \(\frac{at(a{t}^{2}-1)}{9}\).
\[u=\frac{at(a{t}^{2}-1)}{9}\]