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				$A·B=0$ geeft $A=0$ of $B=0$ | 
			
				 
					$x^2-6x-16=0$ 
					$(x-8)(x+2)=0$ 
					$x-8=0$ of $x+2=0$ 
					$x=8$ of $x=-2$ 
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				$A^2=B^2$ geeft $A=B$ of $A=-B$ | 
			
				 
					$(2x+4)^2=(x-3)^2$ 
					$2x+4=x-3$ of $2x+4=-(x-3)$ 
					$x=-7$ of $2x+4=-x+3$ 
					$x=-7$ of $3x=-1$ 
					$x=-7$ of $x=-\frac{1}{3}$ 
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				$A·B=A·C$ geeft $A=0 \vee B=C$ | 
			
				$(x+4)\sqrt{2x-4}=(x-3)(x+4)$ 
				$x+4=0$ of $\sqrt{2x-4}=x-3$ 
				$x=-4$ of $2x-4=(x-3)^2$ 
				$x=-4$ of $2x-4=x^2-6x+9$ 
				$x=-4$ of $x^2-8x+13=0$ 
				$x=-4$ of $(x-4)^2-16+13=0$ 
				$x=-4$ of $(x-4)^2-3=0$ 
				$x=-4$ of $(x-4)^2=3$ 
				$x=-4$ of $x=4-\sqrt{3}$ (v.n.) of $x=4+\sqrt{3}$ 
				$x=-4$ of $x=4+\sqrt{3}$ | 
		
		
			| 
				$A·B=A$ geeft $A=0 \vee B=1$ | 
			
				$2x(x^3-4x+1)=2x$ 
				$2x=0$ of $x^3-4x+1=1$ 
				$x=0$ of $x^3-4x=0$ 
				$x=0$ of $x(x^2-4)=0$ 
				$x=0$ of $x=-2$ of $x=2$ | 
		
		
			| 
				$\eqalign{\frac{A}{B}=0}$ geeft $A=0$ | 
			
				$\eqalign{\frac{2x-3}{x^4-3x^3+2x^2-x+1}=0}$ 
				$2x-3=0$ 
				$2x=3$ 
				$x=1\frac{1}{2}$ | 
		
		
			| 
				$\eqalign{\frac{A}{B}=C}$ geeft $A=BC$ | 
			
				$\eqalign{\frac{2}{x-1}=x+3}$ 
				$(x-1)(x+3)=2$ 
				$x^2+2x-3=2$ 
				$x^2+2x-5=0$ 
				$(x+1)^2-1-5=0$ 
				$x=-1-\sqrt{6}$ of $x=-1+\sqrt{6}$ | 
		
		
			| 
				$\eqalign{\frac{A}{B}=\frac{A}{C}}$ geeft $A=0$ of $B=C$ | 
			
				$\eqalign{\frac{2x-5}{x^2-1}=\frac{2x-5}{4x+1}}$ 
				$2x-5=0$ of $x^2-1=4x+1$ 
				$2x=5$ of $x^2-4x-2=0$ 
				$x=2\frac{1}{2}$ of $(x-2)^2-6=0$ 
				$x=2\frac{1}{2}$ of $x=2-\sqrt{6}$ of $x=2+\sqrt{6}$ | 
		
		
			| 
				$\eqalign{\frac{A}{B}=\frac{C}{D}}$ geeft $AD=BC$ | 
			
				$\eqalign{ 
				&\frac{{x - 4}}{{x - 1}} = \frac{{x + 2}}{{x - 3}} \cr 
				&\left( {x - 4} \right)\left( {x - 3} \right) = \left( {x - 1} \right)\left( {x + 2} \right) \cr 
				&{x^2} - 7x + 12 = {x^2} + x - 2 \cr 
				&-8x = - 14 \cr 
				&x = 1\frac{3}{4} \cr} $ | 
		
		
			| 
				$\eqalign{\frac{A}{B}=\frac{C}{B}}$ geeft $A=C$ | 
			
				$\eqalign{ 
				& \frac{{4x + 1}}{{{x^4} - 6{x^3} + 2{x^2}}} = \frac{{x - 1}}{{{x^4} - 6{x^3} + 2{x^2}}} \cr 
				& 4x + 1 = x - 1 \cr 
				& 3x = - 2 \cr 
				& x = - \frac{2}{3} \cr} $ |