Question Number 109004 by Don08q last updated on 20/Aug/20
$$\overset{} {\:}\:\:{Solve}\:\:\:\:\mid\mathrm{3}{x}+\mathrm{5}\mid\:=\:\mid\mathrm{4}{x}−\mathrm{3}\mid \\ $$$$\:\:\:{where}\:{x}\:\in\:\mathbb{R} \\ $$$$ \\ $$
Answered by bemath last updated on 20/Aug/20
$$\Rightarrow\:\left(\mathrm{3}{x}+\mathrm{5}+\mathrm{4}{x}−\mathrm{3}\right)\left(\mathrm{3}{x}+\mathrm{5}−\mathrm{4}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\mathrm{7}{x}+\mathrm{2}\right)\left(\mathrm{8}−{x}\right)=\mathrm{0}\:\rightarrow\begin{cases}{{x}=\mathrm{8}}\\{{x}=−\frac{\mathrm{2}}{\mathrm{7}}}\end{cases} \\ $$$${cheking}\:\rightarrow\begin{cases}{\mid\mathrm{24}+\mathrm{5}\mid=\mid\mathrm{32}−\mathrm{3}\mid\:\langle{true}\rangle}\\{\mid\frac{−\mathrm{6}+\mathrm{35}}{\mathrm{7}}\mid\:=\:\mid\frac{−\mathrm{8}−\mathrm{21}}{\mathrm{7}}\mid\:\langle{true}\rangle}\end{cases} \\ $$
Commented by john santu last updated on 20/Aug/20
$${short}\:{cut}\: \\ $$
Answered by Aziztisffola last updated on 20/Aug/20
$$\mid\mathrm{3}{x}+\mathrm{5}\mid\:=\:\mid\mathrm{4}{x}−\mathrm{3}\mid\Rightarrow\begin{cases}{\mathrm{3}{x}+\mathrm{5}=\mathrm{4}{x}−\mathrm{3}}\\{\mathrm{3}{x}+\mathrm{5}=−\mathrm{4}{x}+\mathrm{3}}\end{cases} \\ $$$$\:\begin{cases}{{x}=\mathrm{8}}\\{{x}=−\frac{\mathrm{2}}{\mathrm{7}}}\end{cases} \\ $$
Commented by Rasheed.Sindhi last updated on 21/Aug/20
$${Standard}\:{way}!\: \\ $$
Answered by floor(10²Eta[1]) last updated on 20/Aug/20
$$\mathrm{squaring}\:\mathrm{both}\:\mathrm{sides}: \\ $$$$\mathrm{9x}^{\mathrm{2}} +\mathrm{30x}+\mathrm{25}=\mathrm{16x}^{\mathrm{2}} −\mathrm{24x}+\mathrm{9} \\ $$$$\mathrm{7x}^{\mathrm{2}} −\mathrm{54x}−\mathrm{16}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{54}\pm\sqrt{\mathrm{54}^{\mathrm{2}} +\mathrm{4}.\mathrm{7}.\mathrm{16}}}{\mathrm{2}.\mathrm{7}}=\frac{\mathrm{27}\pm\mathrm{29}}{\mathrm{7}} \\ $$$$\Rightarrow\mathrm{x}=\frac{−\mathrm{2}}{\mathrm{7}},\:\mathrm{x}=\mathrm{8} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Aug/20
$$\mid\mathrm{3}{x}+\mathrm{5}\mid\:=\:\mid\mathrm{4}{x}−\mathrm{3}\mid\Rightarrow\frac{\mid\mathrm{3}{x}+\mathrm{5}\mid}{\mid\mathrm{4}{x}−\mathrm{3}\mid}=\mathrm{1} \\ $$$$\Rightarrow\mid\frac{\mathrm{3}{x}+\mathrm{5}}{\mathrm{4}{x}−\mathrm{3}}\mid=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{3}{x}+\mathrm{5}}{\mathrm{4}{x}−\mathrm{3}}=\mathrm{1}\:\vee\:\frac{\mathrm{3}{x}+\mathrm{5}}{\mathrm{4}{x}−\mathrm{3}}=−\mathrm{1} \\ $$$$\:\:\:\:\mathrm{4}{x}−\mathrm{3}=\mathrm{3}{x}+\mathrm{5}\:\vee\:\mathrm{3}{x}+\mathrm{5}=−\mathrm{4}{x}+\mathrm{3} \\ $$$$\:\:\:\:\:\:{x}=\mathrm{8}\:\vee\:{x}=−\mathrm{2}/\mathrm{7} \\ $$