Question

$$\frac { } { x = \frac { 3 y - 9 } { 2 } }$$

Answer

x=(3*(y-3))/(2*Su*e*b*j*c*t)

Solution


Factor out the common term \(3\).
\[Subjectx=\frac{3(y-3)}{2}\]
Divide both sides by \(Su\).
\[bjectx=\frac{\frac{3(y-3)}{2}}{Su}\]
Simplify  \(\frac{\frac{3(y-3)}{2}}{Su}\)  to  \(\frac{3(y-3)}{2Su}\).
\[bjectx=\frac{3(y-3)}{2Su}\]
Divide both sides by \(b\).
\[jectx=\frac{\frac{3(y-3)}{2Su}}{b}\]
Simplify  \(\frac{\frac{3(y-3)}{2Su}}{b}\)  to  \(\frac{3(y-3)}{2Sub}\).
\[jectx=\frac{3(y-3)}{2Sub}\]
Divide both sides by \(j\).
\[ectx=\frac{\frac{3(y-3)}{2Sub}}{j}\]
Simplify  \(\frac{\frac{3(y-3)}{2Sub}}{j}\)  to  \(\frac{3(y-3)}{2Subj}\).
\[ectx=\frac{3(y-3)}{2Subj}\]
Divide both sides by \(e\).
\[ctx=\frac{\frac{3(y-3)}{2Subj}}{e}\]
Simplify  \(\frac{\frac{3(y-3)}{2Subj}}{e}\)  to  \(\frac{3(y-3)}{2Suebj}\).
\[ctx=\frac{3(y-3)}{2Suebj}\]
Divide both sides by \(c\).
\[tx=\frac{\frac{3(y-3)}{2Suebj}}{c}\]
Simplify  \(\frac{\frac{3(y-3)}{2Suebj}}{c}\)  to  \(\frac{3(y-3)}{2Suebjc}\).
\[tx=\frac{3(y-3)}{2Suebjc}\]
Divide both sides by \(t\).
\[x=\frac{\frac{3(y-3)}{2Suebjc}}{t}\]
Simplify  \(\frac{\frac{3(y-3)}{2Suebjc}}{t}\)  to  \(\frac{3(y-3)}{2Suebjct}\).
\[x=\frac{3(y-3)}{2Suebjct}\]