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Question Number 110722 by ZiYangLee last updated on 30/Aug/20

Answered by ajfour last updated on 30/Aug/20

(a−b)x+(b−a)=0 ⇒ x=1 ⇒ a+b=−1 .

$$\left({a}−{b}\right){x}+\left({b}−{a}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{x}=\mathrm{1} \\ $$$$\Rightarrow\:\:{a}+{b}=−\mathrm{1}\:. \\ $$

Answered by bobhans last updated on 30/Aug/20

⇔ x^2 = x^2 ⇔ −b−ax = −bx−a ⇔ (b−a)x = b−a ; b ≠ a ⇔ x = 1 ⇒1^2 +a+b = 0; a+b = −1

$$\Leftrightarrow\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{x}^{\mathrm{2}} \\ $$$$\Leftrightarrow\:−\mathrm{b}−\mathrm{ax}\:=\:−\mathrm{bx}−\mathrm{a} \\ $$$$\Leftrightarrow\:\left(\mathrm{b}−\mathrm{a}\right)\mathrm{x}\:=\:\mathrm{b}−\mathrm{a}\:;\:\mathrm{b}\:\neq\:\mathrm{a}\: \\ $$$$\Leftrightarrow\:\mathrm{x}\:=\:\mathrm{1}\:\Rightarrow\mathrm{1}^{\mathrm{2}} +\mathrm{a}+\mathrm{b}\:=\:\mathrm{0};\:\mathrm{a}+\mathrm{b}\:=\:−\mathrm{1} \\ $$

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