BECE Mathematics 2019 Paper

Same as $10-\frac{x+3}{2}=8$Same as $10-\frac{1(x+3)}{2}=8$ $10-\frac{1(x+3)}{2}=8$Get rid of the fraction by multiplying each side of the equation by the LCM of the denominators The denominators are 1,2 and 1 and their LCM is 2 NOTE:10 = 10/1 and $8=\frac{8}{1}$, thus every number has one dividing it but its not written$\Rightarrow 10-\frac{1(x+3)}{2}=8$$2 \times 10 - 1 \times \frac{2(x+3)}{2} = 8 \times 2$$20-1(x+3)=16$Applying $a(b+c)=a \times b + a \times c$$\Rightarrow -1(x+3) = -1 x x - 1 \times 3 = -x-3$$20-1(x+3)=16$$\Rightarrow 20-x-3=16$Making sure $x$ is at one side of the equation,$-x=16+3-20$$-x=-1$Divide both sides by -1$\Rightarrow x=1$Divide both sides by -1$\Rightarrow x=1$NOTE: You can as well move $x$ to the right size of the equation and 16 to the left and the answer will still be the same.Thus,$20-x-3=16$$20-3-16=x$$20-19=x$$1=x$