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Question-93428




Question Number 93428 by mhmd last updated on 13/May/20
Answered by john santu last updated on 13/May/20
10−x=2∣x∣ ⇒100−20x+x^2 =4x^2   3x^2 +20x−100=0  (3x−10)(x+10)=0   { ((x=((10)/3)⇒y=((20)/3))),((x=−10⇒y=20)) :}
$$\mathrm{10}−\mathrm{x}=\mathrm{2}\mid\mathrm{x}\mid\:\Rightarrow\mathrm{100}−\mathrm{20x}+\mathrm{x}^{\mathrm{2}} =\mathrm{4x}^{\mathrm{2}} \\ $$$$\mathrm{3x}^{\mathrm{2}} +\mathrm{20x}−\mathrm{100}=\mathrm{0} \\ $$$$\left(\mathrm{3x}−\mathrm{10}\right)\left(\mathrm{x}+\mathrm{10}\right)=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{x}=\frac{\mathrm{10}}{\mathrm{3}}\Rightarrow\mathrm{y}=\frac{\mathrm{20}}{\mathrm{3}}}\\{\mathrm{x}=−\mathrm{10}\Rightarrow\mathrm{y}=\mathrm{20}}\end{cases} \\ $$$$ \\ $$
Commented by i jagooll last updated on 13/May/20
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Commented by mhmd last updated on 13/May/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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