Question

$$\left. \begin{array} { l } { x = \frac { 7 x - 1 } { 2 } ; \frac { x - 4 } { 7 } = \frac { x - 24 } { 4 } } \\ { \text { S o \right.$$

Answer

$$-(So*e^2*tan(e)*l^2*v*s^2*m*u^3*o^2*q*a*t*n*(x-y))/742$$

Solution


Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{Solves\imath mul(\tan{e})ousequat\imath on(x-y)}{742}\]
Regroup terms.
\[\frac{llvssmuuuooqatnSoe\imath (\tan{e})e\imath (x-y)}{742}\]
Simplify  \(llvssmuuuooqatnSoe\imath (\tan{e})e\imath (x-y)\)  to  \({l}^{2}v{s}^{2}m{u}^{3}{o}^{2}qatnSoe\imath (\tan{e})e\imath (x-y)\).
\[\frac{{l}^{2}v{s}^{2}m{u}^{3}{o}^{2}qatnSoe\imath (\tan{e})e\imath (x-y)}{742}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{{l}^{2}v{s}^{2}m{u}^{3}{o}^{2}qatnSo{e}^{2}{\imath }^{2}\tan{e}(x-y)}{742}\]
Use Square Rule: \({i}^{2}=-1\).
\[\frac{{l}^{2}v{s}^{2}m{u}^{3}{o}^{2}qatnSo{e}^{2}\times -1\times \tan{e}(x-y)}{742}\]
Simplify  \({l}^{2}v{s}^{2}m{u}^{3}{o}^{2}qatnSo{e}^{2}\times -1\times \tan{e}(x-y)\)  to  \({l}^{2}v{s}^{2}m{u}^{3}{o}^{2}qatnSo{e}^{2}\times -\tan{e}(x-y)\).
\[\frac{{l}^{2}v{s}^{2}m{u}^{3}{o}^{2}qatnSo{e}^{2}\times -\tan{e}(x-y)}{742}\]
Regroup terms.
\[\frac{-So{e}^{2}\tan{e}{l}^{2}v{s}^{2}m{u}^{3}{o}^{2}qatn(x-y)}{742}\]
Move the negative sign to the left.
\[-\frac{So{e}^{2}\tan{e}{l}^{2}v{s}^{2}m{u}^{3}{o}^{2}qatn(x-y)}{742}\]