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Linear equations

Few students have trouble solving linear equations as simple as

\begin{displaymath}3x+4=10\end{displaymath}

just subtract 4 from both sides and then divide by 3.

More complicated linear equations, like those arising in implicit differentiation, for instance, may cause more trouble. For example, suppose we want to solve the equation

\begin{displaymath}3xy^2 y' + y^3=\sin(x+y)(1+y')\end{displaymath}

for $y'$ in terms of $x$ and $y$. Start by multiplying out the right hand side:

\begin{displaymath}3xy^2 y' + y^3 = \sin(x+y) + \sin(x+y)  y'\end{displaymath}

Then move all terms with $y'$ to the left side of the equation and all other terms to the right:

\begin{displaymath}\left(3xy^2-\sin(x+y)\right) y' = \sin(x+y)-y^3\end{displaymath}

Then divide by the coefficient of $y'$:

\begin{displaymath}y'=\frac{\sin(x+y)-y^3}{\left(3xy^2-\sin(x+y)\right)}\end{displaymath}



Larry Stout 2003-01-09