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Question-209672




Question Number 209672 by Ansu last updated on 18/Jul/24
Answered by som(math1967) last updated on 18/Jul/24
 ((3x−1)/4)=((9−x)/x)  ⇒3x^2 −x=36−4x  ⇒3x^2 +3x−36=0  ⇒x^2 +x−12=0  ⇒(x+4)(x−3)=0   x=3
$$\:\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{9}−{x}}{{x}} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} −{x}=\mathrm{36}−\mathrm{4}{x} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{36}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}−\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{4}\right)\left({x}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\:{x}=\mathrm{3} \\ $$
Answered by Sutrisno last updated on 18/Jul/24
((ad)/(ab))=((ae)/(ac))  ((3x−1)/(3x+3))=((9−x)/9)  ((3x−1)/(x+1))=((9−x)/3)  9x−3=−x^2 +8x+9  x^2 +x−12=0  (x−3)(x+4)=0  x=3
$$\frac{{ad}}{{ab}}=\frac{{ae}}{{ac}} \\ $$$$\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{3}{x}+\mathrm{3}}=\frac{\mathrm{9}−{x}}{\mathrm{9}} \\ $$$$\frac{\mathrm{3}{x}−\mathrm{1}}{{x}+\mathrm{1}}=\frac{\mathrm{9}−{x}}{\mathrm{3}} \\ $$$$\mathrm{9}{x}−\mathrm{3}=−{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{9} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{12}=\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)=\mathrm{0} \\ $$$${x}=\mathrm{3}\: \\ $$

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