(a) make x the subject of the formular $${\frac{{\mathtt{1}}}{{\mathtt{z}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{x}}}{{\mathtt{y}}}} = {\frac{{\mathtt{5}}}{{\mathtt{y}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{y}}\right)}}$$ ,
(b) next,if x:y:z=2:5:3 , find the value(s) of x satisfyinh the formular in (a)
I found the answer in (a) is $${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}{{\mathtt{\,-\,}}\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}}$$ , how to solve (b)?
next,if x:y:z=2:5:3 , find the value(s) of x satisfyinh the formular in (a)
$$\\y=\frac{x}{2}*5=\frac{5x}{2}, \qquad z=\frac{x}{2}*3=\frac{3x}{2}\\\\\\
x=\frac{10z-2y}{5z}\\\\
x=\frac{10*\frac{3x}{2}-2*\frac{5x}{2}}{5*\frac{3x}{2}}\\\\
x=\frac{15x-5x}{\frac{15x}{2}}\\\\
x=10x*\frac{2}{15x}\qquad \mbox{Note x cannot equal 0}\\\\
x=\frac{4}{3}$$
I think that your part a question can be simplified further to
$$x=\dfrac{10z-2y}{5z}$$
I found the answer in (a) is $${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}{{\mathtt{\,-\,}}\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}}$$ , how to solve (b)?
$$x=\dfrac{10zy-2y^2}{5zy} = 2-\dfrac{2}{5}*\dfrac{y}{z} \quad | \quad \dfrac{y}{z} = \dfrac{5}{3}\\\\
x= 2 - \frac{2}{3} = 1\frac{1}{3}$$
next,if x:y:z=2:5:3 , find the value(s) of x satisfyinh the formular in (a)
$$\\y=\frac{x}{2}*5=\frac{5x}{2}, \qquad z=\frac{x}{2}*3=\frac{3x}{2}\\\\\\
x=\frac{10z-2y}{5z}\\\\
x=\frac{10*\frac{3x}{2}-2*\frac{5x}{2}}{5*\frac{3x}{2}}\\\\
x=\frac{15x-5x}{\frac{15x}{2}}\\\\
x=10x*\frac{2}{15x}\qquad \mbox{Note x cannot equal 0}\\\\
x=\frac{4}{3}$$