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Question Number 109004 by Don08q last updated on 20/Aug/20

^ Solve ∣3x+5∣ = ∣4x−3∣ where x ∈ R

$$\overset{} {\:}\:\:{Solve}\:\:\:\:\mid\mathrm{3}{x}+\mathrm{5}\mid\:=\:\mid\mathrm{4}{x}−\mathrm{3}\mid \\ $$$$\:\:\:{where}\:{x}\:\in\:\mathbb{R} \\ $$$$ \\ $$

Answered by bemath last updated on 20/Aug/20

⇒ (3x+5+4x−3)(3x+5−4x+3)=0 (7x+2)(8−x)=0 → { ((x=8)),((x=−(2/7))) :} cheking → { ((∣24+5∣=∣32−3∣ ⟨true⟩)),((∣((−6+35)/7)∣ = ∣((−8−21)/7)∣ ⟨true⟩)) :}

$$\Rightarrow\:\left(\mathrm{3}{x}+\mathrm{5}+\mathrm{4}{x}−\mathrm{3}\right)\left(\mathrm{3}{x}+\mathrm{5}−\mathrm{4}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\mathrm{7}{x}+\mathrm{2}\right)\left(\mathrm{8}−{x}\right)=\mathrm{0}\:\rightarrow\begin{cases}{{x}=\mathrm{8}}\\{{x}=−\frac{\mathrm{2}}{\mathrm{7}}}\end{cases} \\ $$$${cheking}\:\rightarrow\begin{cases}{\mid\mathrm{24}+\mathrm{5}\mid=\mid\mathrm{32}−\mathrm{3}\mid\:\langle{true}\rangle}\\{\mid\frac{−\mathrm{6}+\mathrm{35}}{\mathrm{7}}\mid\:=\:\mid\frac{−\mathrm{8}−\mathrm{21}}{\mathrm{7}}\mid\:\langle{true}\rangle}\end{cases} \\ $$

Commented by john santu last updated on 20/Aug/20

short cut

$${short}\:{cut}\: \\ $$

Answered by Aziztisffola last updated on 20/Aug/20

∣3x+5∣ = ∣4x−3∣⇒ { ((3x+5=4x−3)),((3x+5=−4x+3)) :} { ((x=8)),((x=−(2/7))) :}

$$\mid\mathrm{3}{x}+\mathrm{5}\mid\:=\:\mid\mathrm{4}{x}−\mathrm{3}\mid\Rightarrow\begin{cases}{\mathrm{3}{x}+\mathrm{5}=\mathrm{4}{x}−\mathrm{3}}\\{\mathrm{3}{x}+\mathrm{5}=−\mathrm{4}{x}+\mathrm{3}}\end{cases} \\ $$$$\:\begin{cases}{{x}=\mathrm{8}}\\{{x}=−\frac{\mathrm{2}}{\mathrm{7}}}\end{cases} \\ $$

Commented by Rasheed.Sindhi last updated on 21/Aug/20

Standard way!

$${Standard}\:{way}!\: \\ $$

Answered by floor(10²Eta[1]) last updated on 20/Aug/20

squaring both sides: 9x^2 +30x+25=16x^2 −24x+9 7x^2 −54x−16=0 x=((54±(√(54^2 +4.7.16)))/(2.7))=((27±29)/7) ⇒x=((−2)/7), x=8

$$\mathrm{squaring}\:\mathrm{both}\:\mathrm{sides}: \\ $$$$\mathrm{9x}^{\mathrm{2}} +\mathrm{30x}+\mathrm{25}=\mathrm{16x}^{\mathrm{2}} −\mathrm{24x}+\mathrm{9} \\ $$$$\mathrm{7x}^{\mathrm{2}} −\mathrm{54x}−\mathrm{16}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{54}\pm\sqrt{\mathrm{54}^{\mathrm{2}} +\mathrm{4}.\mathrm{7}.\mathrm{16}}}{\mathrm{2}.\mathrm{7}}=\frac{\mathrm{27}\pm\mathrm{29}}{\mathrm{7}} \\ $$$$\Rightarrow\mathrm{x}=\frac{−\mathrm{2}}{\mathrm{7}},\:\mathrm{x}=\mathrm{8} \\ $$

Answered by Rasheed.Sindhi last updated on 21/Aug/20

∣3x+5∣ = ∣4x−3∣⇒((∣3x+5∣)/(∣4x−3∣))=1 ⇒∣((3x+5)/(4x−3))∣=1 ⇒((3x+5)/(4x−3))=1 ∨ ((3x+5)/(4x−3))=−1 4x−3=3x+5 ∨ 3x+5=−4x+3 x=8 ∨ x=−2/7

$$\mid\mathrm{3}{x}+\mathrm{5}\mid\:=\:\mid\mathrm{4}{x}−\mathrm{3}\mid\Rightarrow\frac{\mid\mathrm{3}{x}+\mathrm{5}\mid}{\mid\mathrm{4}{x}−\mathrm{3}\mid}=\mathrm{1} \\ $$$$\Rightarrow\mid\frac{\mathrm{3}{x}+\mathrm{5}}{\mathrm{4}{x}−\mathrm{3}}\mid=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{3}{x}+\mathrm{5}}{\mathrm{4}{x}−\mathrm{3}}=\mathrm{1}\:\vee\:\frac{\mathrm{3}{x}+\mathrm{5}}{\mathrm{4}{x}−\mathrm{3}}=−\mathrm{1} \\ $$$$\:\:\:\:\mathrm{4}{x}−\mathrm{3}=\mathrm{3}{x}+\mathrm{5}\:\vee\:\mathrm{3}{x}+\mathrm{5}=−\mathrm{4}{x}+\mathrm{3} \\ $$$$\:\:\:\:\:\:{x}=\mathrm{8}\:\vee\:{x}=−\mathrm{2}/\mathrm{7} \\ $$

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