From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	If we subtract <math>2z</math> from both sides,
-	<center> [[Image:3_1_4c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>iz+2-2z=-3</math>}}
		+
		+	and then subtract <math>2</math> from both sides, we have <math>z</math> terms left only on the left-hand side,
		+
		+	{{Displayed math||<math>iz-2z=-3-2\,\textrm{.}</math>}}
		+
		+	After taking out a factor <math>z</math> from the left-hand side,
		+
		+	{{Displayed math||<math>(i-2)z=-5\,,</math>}}
		+
		+	we obtain, after dividing by <math>-2+i</math>,
		+
		+	{{Displayed math||<math>\begin{align}
		+	z &= \frac{-5}{-2+i}
		+	= \frac{-5(-2-i)}{(-2+i)(-2-i)}
		+	= \frac{(-5)\cdot(-2)-5\cdot(-i)}{(-2)^2-i^2}\\[5pt]
		+	&= \frac{10+5i}{4+1}
		+	= \frac{10+5i}{5}
		+	= 2+i\,\textrm{.}\end{align}</math>}}
		+
		+	A quick check shows that <math>z=2+i</math> satisfies the original equation,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\text{LHS} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt]
		+	\text{RHS} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

If we subtract \displaystyle 2z from both sides,

\displaystyle iz+2-2z=-3

and then subtract \displaystyle 2 from both sides, we have \displaystyle z terms left only on the left-hand side,

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

After taking out a factor \displaystyle z from the left-hand side,

\displaystyle (i-2)z=-5\,,

we obtain, after dividing by \displaystyle -2+i,

\displaystyle \begin{align} z &= \frac{-5}{-2+i} = \frac{-5(-2-i)}{(-2+i)(-2-i)} = \frac{(-5)\cdot(-2)-5\cdot(-i)}{(-2)^2-i^2}\\[5pt] &= \frac{10+5i}{4+1} = \frac{10+5i}{5} = 2+i\,\textrm{.}\end{align}

A quick check shows that \displaystyle z=2+i satisfies the original equation,

\displaystyle \begin{align} \text{LHS} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt] \text{RHS} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.} \end{align}