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Question Number 173620 by eiverzurita last updated on 14/Jul/22

Answered by FelipeLz last updated on 14/Jul/22

2b^2 −2bx+2a^2 x+abx+ab^2 = 2abx+2a^2 b 2b^2 −2bx+2a^2 x+ab^2 = abx+2a^2 b 2b(b−x)−2a^2 (b−x)+ab(b−x) = 0 (b−x)(2b−2a^2 +ab) = 0 x = b

$$\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{bx}+\mathrm{2}{a}^{\mathrm{2}} {x}+{abx}+{ab}^{\mathrm{2}} \:=\:\mathrm{2}{abx}+\mathrm{2}{a}^{\mathrm{2}} {b} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{bx}+\mathrm{2}{a}^{\mathrm{2}} {x}+{ab}^{\mathrm{2}} \:=\:{abx}+\mathrm{2}{a}^{\mathrm{2}} {b} \\ $$$$\mathrm{2}{b}\left({b}−{x}\right)−\mathrm{2}{a}^{\mathrm{2}} \left({b}−{x}\right)+{ab}\left({b}−{x}\right)\:=\:\mathrm{0} \\ $$$$\left({b}−{x}\right)\left(\mathrm{2}{b}−\mathrm{2}{a}^{\mathrm{2}} +{ab}\right)\:=\:\mathrm{0} \\ $$$${x}\:=\:{b} \\ $$

Commented by Tawa11 last updated on 15/Jul/22

Great sirs

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

Answered by Rasheed.Sindhi last updated on 15/Jul/22

AnOther Way ((b−x)/a)+((ax)/b)+((x+b)/2)=x+a; x=? (b/a)−(x/a)+((ax)/b)+(x/2)+(b/2)−x=a −(x/a)+((ax)/b)+(x/2)−x=a−(b/a)−(b/2) x(−(1/a)+(a/b)+(1/2)−1)=((2a^2 −2b−ab)/(2a)) x(((−2b+2a^2 +ab−2ab)/(2ab)))=((2a^2 −2b−ab)/(2a)) x=((2a^2 −2b−ab)/(2a))×((2ab)/(−2b+2a^2 −ab)) =((b(2a^2 −2b−ab))/(−2b+2a^2 −ab))=b

$$\mathrm{AnOther}\:\mathrm{Way} \\ $$$$\frac{{b}−{x}}{{a}}+\frac{{ax}}{{b}}+\frac{{x}+{b}}{\mathrm{2}}={x}+{a};\:{x}=? \\ $$$$\frac{{b}}{{a}}−\frac{{x}}{{a}}+\frac{{ax}}{{b}}+\frac{{x}}{\mathrm{2}}+\frac{{b}}{\mathrm{2}}−{x}={a} \\ $$$$−\frac{{x}}{{a}}+\frac{{ax}}{{b}}+\frac{{x}}{\mathrm{2}}−{x}={a}−\frac{{b}}{{a}}−\frac{{b}}{\mathrm{2}} \\ $$$${x}\left(−\frac{\mathrm{1}}{{a}}+\frac{{a}}{{b}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)=\frac{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{b}−{ab}}{\mathrm{2}{a}} \\ $$$${x}\left(\frac{−\mathrm{2}{b}+\mathrm{2}{a}^{\mathrm{2}} +{ab}−\mathrm{2}{ab}}{\mathrm{2}{ab}}\right)=\frac{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{b}−{ab}}{\mathrm{2}{a}} \\ $$$${x}=\frac{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{b}−{ab}}{\mathrm{2}{a}}×\frac{\mathrm{2}{ab}}{−\mathrm{2}{b}+\mathrm{2}{a}^{\mathrm{2}} −{ab}} \\ $$$$\:\:\:\:=\frac{{b}\left(\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{b}−{ab}\right)}{−\mathrm{2}{b}+\mathrm{2}{a}^{\mathrm{2}} −{ab}}={b} \\ $$

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