(gelijknamig maken)
\(y = {1 \over x + 4} - 1\)
\(\text{} = {1 \over x + 4} - {1 \over 1}\)
\(\text{} = {1 \over x + 4} - {x + 4 \over x + 4}\)

(haakjes uitwerken)
\(y = {1 - (x + 4) \over x + 4}\)
\(\text{} = {1 - x - 4 \over x + 4}\)
\(\text{} = {-x - 3 \over x + 4}\)

(gelijknamig maken)
\(y = {1 \over x} - {5 x \over 1}\)
\(\text{} = {1 \over x} - {5 x ⋅ x \over x}\)

(herleiden)
\(y = {1 \over x} - {5 x^{2} \over x}\)
\(y = {1 - 5 x^{2} \over x}\)

(breuk vermenigvuldigen)
\(y = {x + 1 \over 1} ⋅ \frac{4}{5} ⋅ {5 x \over x - 3}\)
\(\text{} = {(x + 1) ⋅ 4 ⋅ 5 x \over 1 ⋅ 5 ⋅ (x - 3)}\)

(vereenvoudigen)
\(y = {20 x (x + 1) \over 5 (x - 3)}\)
\(y = {4 x (x + 1) \over x - 3}\)

(haakjes wegwerken)
\(y = {4 x^{2} + 4 x \over x - 3}\)

(balansmethode)
\(y - 8 = {3 \over x}\)

(wisseleigenschap, ofwel \({A \over B} = C\) geeft \({A \over C} = B \text{)}\)
\(x = {3 \over y - 8}\)

(\({A \over B} ⋅ C = {C \over B} ⋅ A\) geeft)
\(y = {32 \over 8} ⋅ (5 x - 3) - 9\)

(breuk vereenvoudigen)
\(y = 4 ⋅ (5 x - 3) - 9\)

(haakjes wegwerken)
\(y = 20 x - 12 - 9\)
\(y = 20 x - 21\)

(balansmethode)
\({y \over x + 8} = -5\)

(kruislings vermenigvuldigen)
\({y \over x + 8} = {-5 \over 1}\)
\(y = -5 ⋅ (x + 8)\)

(haakjes wegwerken)
\(y = -5 x - 40\)

(kruislings vermenigvuldigen)
\({y \over 1} = {3 x - 18 \over 15}\)
\(3 x - 18 = 15 y\)

(balansmethode)
\(3 x = 15 y + 18\)
\(x = 5 y + 6\)

(uitdelen)
\(y = {2{,}5 x^{2} \over x} - {9 x \over x} + {3 \over x} + 8 x + 7{,}5\)
\(\text{} = 2{,}5 x - 9 + {3 \over x} + 8 x + 7{,}5\)

(herleiden)
\(y = 10{,}5 x - 1{,}5 + {3 \over x}\)