From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	A general strategy when solving equations is to try to get the unknown variable by itself on one side.
-	<center> [[Bild:3_1_4a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	In this case, we start by subtracting <math>z</math> from both sides,
		+
		+	{{Displayed math||<math>z+3i-z=2z-2-z\,\textrm{.}</math>}}
		+
		+	Then we have a <math>z</math> left on the right-hand side,
		+
		+	{{Displayed math||<math>3i=z-2\,\textrm{.}</math>}}
		+
		+	We add <math>2</math> to both sides to remove the <math>-2</math> from the right hand side,
		+
		+	{{Displayed math||<math>3i+2=z-2+2\,,</math>}}
		+
		+	and after that we can just read off the solution,
		+
		+	{{Displayed math||<math>2+3i=z\,\textrm{.}</math>}}
		+
		+	To check that we have calculated correctly, we substitute <math>z=2+3i</math> into the original equation and see that it is satisfied,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\text{LHS} &= z +3i = 2+3i+3i=2+6i\,,\\[5pt]
		+	\text{RHS} &= 2z-2 = 2(2+3i)-2 = 4 + 6i -2 = 2+6i\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

A general strategy when solving equations is to try to get the unknown variable by itself on one side.

In this case, we start by subtracting \displaystyle z from both sides,

\displaystyle z+3i-z=2z-2-z\,\textrm{.}

Then we have a \displaystyle z left on the right-hand side,

\displaystyle 3i=z-2\,\textrm{.}

We add \displaystyle 2 to both sides to remove the \displaystyle -2 from the right hand side,

\displaystyle 3i+2=z-2+2\,,

and after that we can just read off the solution,

\displaystyle 2+3i=z\,\textrm{.}

To check that we have calculated correctly, we substitute \displaystyle z=2+3i into the original equation and see that it is satisfied,

\displaystyle \begin{align} \text{LHS} &= z +3i = 2+3i+3i=2+6i\,,\\[5pt] \text{RHS} &= 2z-2 = 2(2+3i)-2 = 4 + 6i -2 = 2+6i\,\textrm{.} \end{align}