for solving equation which one we use ⇒ and =, i mean where we use ⇒ and where we use, = and where we use one of them to consider wrong.


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Question Number 157272 by zakirullah last updated on 21/Oct/21

$f o r s o l v i n g e q u a t i o n w h i c h o n e w e u s e$ $\Rightarrow a n d =, i m e a n w h e r e w e u s e \Rightarrow a n d w h e r e w e u s e, =$ $a n d w h e r e w e u s e o n e o f t h e m t o$ $c o n s i d e r w r o n g .$
	Answered by TheHoneyCat last updated on 21/Oct/21

	$When you write a sentence S_{1} that can be$ $either true or false,$ $f o r e x a m p l e ‘ ‘ I t i s {r a i n i n g}^{″} o r ‘ ‘ I a m {w e t}^{″}$ $o r ‘ ‘ x i s a s o l u t i o n t o t h e {e q u a t i o n}^{″} o r ‘ ‘ t h e$ $f u n c t i o n f i s {l i n e a r}^{″}$ $It is usulay assumed you belive that S_{1} is true$ $So if I write$ $x = {b l a b l a}_{1}$ $= {b l a b l a}_{2}$ $= {b l a b l a}_{3}$ $I, in fact, mean that$ $‘ ‘ x = {b l a b l a}_{1}^{″}$ $and that ‘ ‘ x = {b l a b l a}_{2}^{″}$ $and that ‘ ‘ x = {b l a b l a}_{3}^{″}$ $This is the most common use of the sign ‘ ‘ =^{″}$ $‘ ‘ \Leftrightarrow^{″} is very different, it is not simply used in$ $sentences, like ‘ ‘ =^{″}, it is used to make$ $sentences about other sentences .$ $‘ ‘ S_{1} \Leftrightarrow S_{2}^{″} means that the sentence S_{1} is true$ $if S_{2} is and wrong if S_{2} is .$ $So during a math proof (or any speech about$ $some math), you may use it to this purpose .$ $For example$ $i f x^{2} = 2$ $y o u {d o n}^{'} t k n o w i f ‘ ‘ x = - {\sqrt{2}}^{″} o r n o t$ $i t c o u l d b e t r u e, b e c a u s e ‘ ‘ {(- \sqrt{2})}^{2} = 2^{″} i s t r u e$ $b u t x c o u l d a l s o b e + \sqrt{2}$ $s o y o u m i g t h w r i t e s o m e t i n g a s$ $‘ ‘ x ⩽ 0 \Leftrightarrow x = - {\sqrt{2}}^{″} a n d ‘ ‘ x ⩾ 0 \Leftrightarrow x = {\sqrt{2}}^{″}$ $N o w t h a t m i g t h s e e q u i t e u s e l e s s, b u t i t i s$ $v e r y u s e f u l l w h e n y o u w a n t t o s o l v e a$ $q u e s t i o n l i k e :$ $D o e s x = y ?$ $y o u m i g h t t h e n w r i t e :$ $x = y \Leftrightarrow x - y = 0$ $\Leftrightarrow {b l a b l a}_{1}$ $\Leftrightarrow {b l a b l a}_{2}$ $\Leftrightarrow {b l a b l a}_{3}$ $a n d t h e n y o u c h e c k i f b l a b l a 3 i s t r u e o r n o t$
	Commented by TheHoneyCat last updated on 21/Oct/21
	$Here is an example: (my answere in blue) Find all the positive solutions x of the following equation: 3x^3 +4x^2 −5x=2 3x^3 +4x^2 −5x=2 ⇔3x^3 +4x^2 −5x−2=0 But, I also know that: (3x+1)(x−1)(x+2) =(3x+1)(x^2 +2x−x−2) =(3x+1)(x^2 +x−2) =3x^3 +3x^2 −6x+x^2 +x−2 =3x^3 +4x^2 −5x−2 So, 3x^3 +4x^2 −5x=2 ⇔(3x+1)(x−1)(x+2)=0 ⇔3x+1=0 or x−1=0 or x+2=0 because if a product is 0, one of its numbers is too 3x+1=0⇔x=((−1)/3) x−1=0⇔x=1 x+2=0⇔x=−2 So: 3x^3 +4x^2 −5x=2 ⇔ x∈{1,−2,−1/3} And since x must be pisitive, the solution is : x=1 See the difference?$
	$H e r e i s a n e x a m p l e :$ $(m y a n s w e r e i n b l u e)$ $Find all the positive solutions x of the$ $following equation :$ $3 x^{3} + 4 x^{2} - 5 x = 2$ $3 x^{3} + 4 x^{2} - 5 x = 2$ $\Leftrightarrow 3 x^{3} + 4 x^{2} - 5 x - 2 = 0$ $But, I also know that :$ $(3 x + 1) (x - 1) (x + 2)$ $= (3 x + 1) (x^{2} + 2 x - x - 2)$ $= (3 x + 1) (x^{2} + x - 2)$ $= 3 x^{3} + 3 x^{2} - 6 x + x^{2} + x - 2$ $= 3 x^{3} + 4 x^{2} - 5 x - 2$ $So,$ $3 x^{3} + 4 x^{2} - 5 x = 2$ $\Leftrightarrow (3 x + 1) (x - 1) (x + 2) = 0$ $\Leftrightarrow 3 x + 1 = 0 o r x - 1 = 0 o r x + 2 = 0$ $because if a product is 0, one of its numbers is$ $too$ $3 x + 1 = 0 \Leftrightarrow x = \frac{- 1}{3}$ $x - 1 = 0 \Leftrightarrow x = 1$ $x + 2 = 0 \Leftrightarrow x = - 2$ $So :$ $3 x^{3} + 4 x^{2} - 5 x = 2 \Leftrightarrow x \in {1, - 2, - 1 / 3}$ $And since x must be pisitive, the solution$ $is : x = 1$ $See the difference ?$
	Commented by zakirullah last updated on 21/Oct/21

	$g r e a t$
	Answered by talminator2856791 last updated on 22/Oct/21

	$i think this is one interpretation :$ $a = b$ $same mathematical object$ $a \Rightarrow b$ $b \Rightarrow a$ $different mathematical objects$ $dependant on each other .$
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