Question

$$\frac { 21 } { 27 } \frac { 7 } { x } \frac { 2 x } { 3 } = \cdots \cdots$$

Answer

$$-(98*If*e^3*s*q*u*v*a*l*n^2*t^3*o*h)/27$$

Solution


Cancel \(x\).
\[If\times \frac{21}{27}\imath sequ\imath valentto\times 7then\times \frac{2}{3}\]
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{If\times 21\imath sequ\imath valentto\times 7then\times 2}{27\times 3}\]
Take out the constants.
\[\frac{(21\times 7\times 2)squvalnntttohIf\imath e\imath ee}{27\times 3}\]
Simplify  \(21\times 7\)  to  \(147\).
\[\frac{(147\times 2)squvalnntttohIf\imath e\imath ee}{27\times 3}\]
Simplify  \(147\times 2\)  to  \(294\).
\[\frac{294squvalnntttohIf\imath e\imath ee}{27\times 3}\]
Simplify  \(294squvalnntttohIf\imath e\imath ee\)  to  \(294squval{n}^{2}{t}^{3}ohIf\imath e\imath ee\).
\[\frac{294squval{n}^{2}{t}^{3}ohIf\imath e\imath ee}{27\times 3}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{294squval{n}^{2}{t}^{3}ohIf{\imath }^{2}{e}^{3}}{27\times 3}\]
Use Square Rule: \({i}^{2}=-1\).
\[\frac{294squval{n}^{2}{t}^{3}ohIf\times -1\times {e}^{3}}{27\times 3}\]
Simplify  \(294squval{n}^{2}{t}^{3}ohIf\times -1\times {e}^{3}\)  to  \(-294squval{n}^{2}{t}^{3}ohIf{e}^{3}\).
\[\frac{-294squval{n}^{2}{t}^{3}ohIf{e}^{3}}{27\times 3}\]
Regroup terms.
\[\frac{-294If{e}^{3}squval{n}^{2}{t}^{3}oh}{27\times 3}\]
Simplify  \(27\times 3\)  to  \(81\).
\[\frac{-294If{e}^{3}squval{n}^{2}{t}^{3}oh}{81}\]
Move the negative sign to the left.
\[-\frac{294If{e}^{3}squval{n}^{2}{t}^{3}oh}{81}\]
Simplify  \(\frac{294If{e}^{3}squval{n}^{2}{t}^{3}oh}{81}\)  to  \(\frac{98If{e}^{3}squval{n}^{2}{t}^{3}oh}{27}\).
\[-\frac{98If{e}^{3}squval{n}^{2}{t}^{3}oh}{27}\]