Question

$$\overline{11x}+y=9$$

Answer

$$x=(9-y)/(11*e^2*IM*o*v*r*l*n)$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[ov{e}^{2}rl\imath n\times 11x+y=9\]
Regroup terms.
\[11{e}^{2}\imath ovrlnx+y=9\]
Regroup terms.
\[y+11{e}^{2}\imath ovrlnx=9\]
Subtract \(y\) from both sides.
\[11{e}^{2}\imath ovrlnx=9-y\]
Divide both sides by \(11\).
\[{e}^{2}\imath ovrlnx=\frac{9-y}{11}\]
Divide both sides by \({e}^{2}\).
\[\imath ovrlnx=\frac{\frac{9-y}{11}}{{e}^{2}}\]
Simplify  \(\frac{\frac{9-y}{11}}{{e}^{2}}\)  to  \(\frac{9-y}{11{e}^{2}}\).
\[\imath ovrlnx=\frac{9-y}{11{e}^{2}}\]
Divide both sides by \(\imath \).
\[ovrlnx=\frac{\frac{9-y}{11{e}^{2}}}{\imath }\]
Simplify  \(\frac{\frac{9-y}{11{e}^{2}}}{\imath }\)  to  \(\frac{9-y}{11{e}^{2}\imath }\).
\[ovrlnx=\frac{9-y}{11{e}^{2}\imath }\]
Divide both sides by \(o\).
\[vrlnx=\frac{\frac{9-y}{11{e}^{2}\imath }}{o}\]
Simplify  \(\frac{\frac{9-y}{11{e}^{2}\imath }}{o}\)  to  \(\frac{9-y}{11{e}^{2}\imath o}\).
\[vrlnx=\frac{9-y}{11{e}^{2}\imath o}\]
Divide both sides by \(v\).
\[rlnx=\frac{\frac{9-y}{11{e}^{2}\imath o}}{v}\]
Simplify  \(\frac{\frac{9-y}{11{e}^{2}\imath o}}{v}\)  to  \(\frac{9-y}{11{e}^{2}\imath ov}\).
\[rlnx=\frac{9-y}{11{e}^{2}\imath ov}\]
Divide both sides by \(r\).
\[lnx=\frac{\frac{9-y}{11{e}^{2}\imath ov}}{r}\]
Simplify  \(\frac{\frac{9-y}{11{e}^{2}\imath ov}}{r}\)  to  \(\frac{9-y}{11{e}^{2}\imath ovr}\).
\[lnx=\frac{9-y}{11{e}^{2}\imath ovr}\]
Divide both sides by \(l\).
\[nx=\frac{\frac{9-y}{11{e}^{2}\imath ovr}}{l}\]
Simplify  \(\frac{\frac{9-y}{11{e}^{2}\imath ovr}}{l}\)  to  \(\frac{9-y}{11{e}^{2}\imath ovrl}\).
\[nx=\frac{9-y}{11{e}^{2}\imath ovrl}\]
Divide both sides by \(n\).
\[x=\frac{\frac{9-y}{11{e}^{2}\imath ovrl}}{n}\]
Simplify  \(\frac{\frac{9-y}{11{e}^{2}\imath ovrl}}{n}\)  to  \(\frac{9-y}{11{e}^{2}\imath ovrln}\).
\[x=\frac{9-y}{11{e}^{2}\imath ovrln}\]