Question Number 62609 by hovea cw last updated on 23/Jun/19
$$\mathrm{If}\:\:\mathrm{5}\mid{x}\mid\:+\:\mathrm{4}\mid{y}\mid\:=\:\mathrm{4}\:\mathrm{and}\:\mathrm{2}\mid{x}\mid\:−\:\mathrm{4}\mid{y}\mid\:=\:\mathrm{10}, \\ $$$$\mathrm{then}\:\mathrm{find}\:{x}\:\mathrm{and}\:{y}. \\ $$
Answered by $@ty@m last updated on 23/Jun/19
$${Let}\:\mid{x}\mid={a}\:\&\mid{y}\mid={b} \\ $$$$\mathrm{5}{a}+\mathrm{4}{b}=\mathrm{4}\:\: \\ $$$$\mathrm{2}{a}−\mathrm{4}{b}=\mathrm{10}\: \\ $$$$−−−−−− \\ $$$$\mathrm{7}{a}\:\:\:\:\:\:\:\:\:\:=\mathrm{14} \\ $$$$\Rightarrow{a}=\mathrm{2}\:\Rightarrow\mid{x}\mid=\mathrm{2}\Rightarrow{x}=\pm\mathrm{2} \\ $$$$\Rightarrow{b}=\frac{\mathrm{4}−\mathrm{5}×\mathrm{2}}{\mathrm{4}}\:=\:\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mid{y}\mid=\frac{−\mathrm{3}}{\mathrm{2}}\:{which}\:{is}\:{impossible}. \\ $$$${pl}.\:{check}\:{question}… \\ $$$$ \\ $$