Question Number 57175 by Tinkutara last updated on 31/Mar/19
Commented by Tinkutara last updated on 31/Mar/19
Please help in all the 3 questions!
Commented by mr W last updated on 31/Mar/19
$${Q}\mathrm{11}: \\ $$$$\mid{a}+{b}\frac{{y}}{{x}}\mid\mid{x}\mid+\mid{b}\frac{{x}}{{y}}+{a}\mid\mid{y}\mid=\mid{x}\mid+\mid{y}\mid \\ $$$${or} \\ $$$$\mid{a}\frac{{x}}{{y}}+{b}\mid\mid{y}\mid+\mid{b}+{a}\frac{{y}}{{x}}\mid\mid{x}\mid=\mid{x}\mid+\mid{y}\mid \\ $$$$ \\ $$$${b}=\mathrm{0} \\ $$$${a}=\pm\mathrm{1} \\ $$$${or} \\ $$$${a}=\mathrm{0} \\ $$$${b}=\pm\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\mid{a}\mid+\mid{b}\mid=\mathrm{1} \\ $$
Commented by Tinkutara last updated on 31/Mar/19
But Sir how to solve if a and b are both other than 0?
Commented by mr W last updated on 31/Mar/19
$${there}\:{is}\:{no}\:{solution}\:{with}\:{a}\neq\mathrm{0}\:{and}\:{b}\neq\mathrm{0}. \\ $$$${since}\:{x}\:{and}\:{y}\:{can}\:{be}\:{of}\:{any}\:{values}, \\ $$$${we}\:{will}\:{get}\:{a}+{b}=\pm\mathrm{1}\:{and}\:{a}−{b}=\pm\mathrm{1}, \\ $$$${therefore}\:\left({a}=\pm\mathrm{1}\:{and}\:{b}=\mathrm{0}\right)\:{or}\:\left({a}=\mathrm{0}\:{and}\right. \\ $$$$\left.{b}=\pm\mathrm{1}\right).\:{no}\:{other}\:{possibility}. \\ $$
Commented by Tinkutara last updated on 02/Apr/19
Thank you Sir!