Question

$$\left. \begin{array} { l } { 2 } \\ { \frac { 4 x } { 5 } - x \frac { 1 } { 7 } } \\ { \Rightarrow } \end{array} \right.$$

Answer

$$-(248*e*t^2*h*f^2*o^3*l^3*w*n^2*g*r*a^2*c)/35$$

Solution


Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{2thefollow\imath ngfract\imath on\times 4\times 31al}{5\times 7}\]
Take out the constants.
\[\frac{(2\times 4\times 31)tthffooolllwnngraace\imath \imath }{5\times 7}\]
Simplify  \(2\times 4\)  to  \(8\).
\[\frac{(8\times 31)tthffooolllwnngraace\imath \imath }{5\times 7}\]
Simplify  \(8\times 31\)  to  \(248\).
\[\frac{248tthffooolllwnngraace\imath \imath }{5\times 7}\]
Simplify  \(248tthffooolllwnngraace\imath \imath \)  to  \(248{t}^{2}h{f}^{2}{o}^{3}{l}^{3}w{n}^{2}gr{a}^{2}ce\imath \imath \).
\[\frac{248{t}^{2}h{f}^{2}{o}^{3}{l}^{3}w{n}^{2}gr{a}^{2}ce\imath \imath }{5\times 7}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{248{t}^{2}h{f}^{2}{o}^{3}{l}^{3}w{n}^{2}gr{a}^{2}ce{\imath }^{2}}{5\times 7}\]
Use Square Rule: \({i}^{2}=-1\).
\[\frac{248{t}^{2}h{f}^{2}{o}^{3}{l}^{3}w{n}^{2}gr{a}^{2}ce\times -1}{5\times 7}\]
Simplify  \(248{t}^{2}h{f}^{2}{o}^{3}{l}^{3}w{n}^{2}gr{a}^{2}ce\times -1\)  to  \(-248{t}^{2}h{f}^{2}{o}^{3}{l}^{3}w{n}^{2}gr{a}^{2}ce\).
\[\frac{-248{t}^{2}h{f}^{2}{o}^{3}{l}^{3}w{n}^{2}gr{a}^{2}ce}{5\times 7}\]
Regroup terms.
\[\frac{-248e{t}^{2}h{f}^{2}{o}^{3}{l}^{3}w{n}^{2}gr{a}^{2}c}{5\times 7}\]
Simplify  \(5\times 7\)  to  \(35\).
\[\frac{-248e{t}^{2}h{f}^{2}{o}^{3}{l}^{3}w{n}^{2}gr{a}^{2}c}{35}\]
Move the negative sign to the left.
\[-\frac{248e{t}^{2}h{f}^{2}{o}^{3}{l}^{3}w{n}^{2}gr{a}^{2}c}{35}\]