BECE Mathematics 2020 Paper

Get rid of the fraction by multiplying each side of the equation by the LCM of the denominators $(3,1$ and 1 )Thus $3(w-1)=\frac{3(w-1)}{1}$ (every number is being divided by 1 which has no effect on the number)$-1=\frac{-1}{1}$Hence the denominators are 3,1 and 1 and the LCM is 3Multiply both sides of the equation by 3$\;3 \times \frac{w}{3}=3 \times 3(w-1)-3 \times 1$$\;3 \;\text{cancels}\; 3 \;\text{in}\; 3 \times \frac{w}{3}=w$$\;3 \times \frac{w}{3}=3 \times 3(w-1)-3 \times 1$$\; \Rightarrow w=9(w-1)-3$Applying the concept, $a(b+c)=a \times b+a x c, a(b-c)=a \times b-a x c$$\Rightarrow 9(w-1)=9 \times w-9 \times 1=9 w-9$$\Rightarrow w=9 w-9-3$Grouping the letters $(w)$ at one side and the numbers at the other side of the equation Concept, when negative moves to the other side of the equation, becomes positive and positive becomes negativeOption 1, grouping the $w$ at the left side of equation$\;w=9 w-9-3$$\; \Rightarrow w-9 w=-9-3$$\;-8 w=-12$Divide both sides by -8$\frac{-8w}{-8}=\frac{-12}{-8}$NOTE: The negative cancels each other$w=\frac{12}{8}=\frac{3}{2}$, thus 4 goes into 12,3 times and into 8,2 times Option 2, grouping the $w$ at the right side of equation$\;w=9 w-9-3$$\; \Rightarrow 9+3=9 w-w$$\; 12=8 w$Divide both sides by 8$\frac{12}{8}=\frac{8w}{8}$$w=\frac{12}{8}=\frac{3}{2}$, thus 4 goes into 12,3 times and into 8,2 times