Question Number 200139 by hardmath last updated on 14/Nov/23
$$\mathrm{If} \\ $$$$\mathrm{x}\::\:\mathrm{y}\::\:\mathrm{z}\:=\:\frac{\mathrm{1}}{\mathrm{7}}\::\:\frac{\mathrm{1}}{\mathrm{3}}\::\:\frac{\mathrm{1}}{\mathrm{21}} \\ $$$$\mathrm{5x}\:−\:\mathrm{2y}\:+\:\mathrm{z}\:=\:\mathrm{16} \\ $$$$ \\ $$$$\mathrm{Find}:\:\:\:\mathrm{y}\:=\:? \\ $$
Answered by ajfour last updated on 14/Nov/23
$$\mathrm{5}\left(\frac{{x}}{{y}}\right){y}−\mathrm{2}{y}+\left(\frac{{z}}{{y}}\right){y}=\mathrm{16} \\ $$$${y}=\frac{\mathrm{16}}{\mathrm{5}\left(\frac{\mathrm{3}}{\mathrm{7}}\right)−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{7}}}=\frac{\mathrm{16}×\mathrm{7}}{\mathrm{15}−\mathrm{14}+\mathrm{1}}=\mathrm{56} \\ $$
Answered by mr W last updated on 15/Nov/23
$${say}\:{x}=\mathrm{3}{k},\:{y}=\mathrm{7}{k},\:{z}={k} \\ $$$$\mathrm{5}×\mathrm{3}{k}−\mathrm{2}×\mathrm{7}{k}+{k}=\mathrm{16} \\ $$$$\Rightarrow{k}=\mathrm{8}\:\Rightarrow{y}=\mathrm{7}×\mathrm{8}=\mathrm{56} \\ $$
Commented by hardmath last updated on 15/Nov/23
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}… \\ $$$$\mathrm{sorry},\:\mathrm{but}\:\left(\mathrm{x}=\mathrm{3k},\:\mathrm{y}=\mathrm{7k},\:\mathrm{z}=\mathrm{k}\right)\:\mathrm{where}\:\mathrm{did}\:\mathrm{it} \\ $$$$\mathrm{come}\:\mathrm{from} \\ $$
Commented by mr W last updated on 15/Nov/23
$${x}:{y}:{z}=\frac{\mathrm{1}}{\mathrm{7}}:\frac{\mathrm{1}}{\mathrm{3}}:\frac{\mathrm{1}}{\mathrm{21}}\:{is}\:{the}\:{same}\:{as} \\ $$$${x}:{y}:{z}=\mathrm{3}:\mathrm{7}:\mathrm{1} \\ $$$${so}\:{you}\:{can}\:{take} \\ $$$${x}=\mathrm{3}{k},\:{y}=\mathrm{7}{k},\:{z}={k} \\ $$
Commented by hardmath last updated on 15/Nov/23
$$\mathrm{dear}\:\mathrm{professor},\:\mathrm{for}\:\mathrm{example},\:\mathrm{why}\:\mathrm{x}=\mathrm{3k} \\ $$$$\mathrm{and}\:\mathrm{not}\:\mathrm{7k}… \\ $$
Commented by mr W last updated on 15/Nov/23
$${because}\:{you}\:{have}\:{stated}\:{that} \\ $$$${x}:{y}:{z}=\frac{\mathrm{1}}{\mathrm{7}}:\frac{\mathrm{1}}{\mathrm{3}}:\frac{\mathrm{1}}{\mathrm{21}} \\ $$$${that}\:{means} \\ $$$${if}\:{z}=\frac{\mathrm{1}}{\mathrm{21}},\:{then}\:{y}=\frac{\mathrm{1}}{\mathrm{3}},\:{i}.{e}.\:{y}=\mathrm{7}{z},\:{and} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{7}},\:{i}.{e}.\:{x}=\mathrm{3}{z},\:{so}\:{we}\:{can}\:{generally} \\ $$$${say} \\ $$$${z}={k},\:{y}=\mathrm{7}{k},\:{x}=\mathrm{3}{k}. \\ $$
Commented by hardmath last updated on 15/Nov/23
$$\mathrm{super}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{tbank}\:\mathrm{you} \\ $$