5
$\begingroup$

How is the equation $x_1+5x_2-\sqrt{(2x_3)} = 1$ a linear equation? The answer given in the book is, "The Equation is linear".

How can an equation involving a square root like the above equation be a linear equation?

here is the cutting of the book, enter image description here

  • 2
    It isn't. But if you had $\sqrt 2 x_3$ instead of $\sqrt{2x_3}$, it would be... I suspect there was a typo.2012-12-26
  • 0
    The original post clearly did not have $2x_3$ nested in parentheses.2012-12-26
  • 0
    no this is not misprint you can see the cutting from the book, this is (a) part and at the bottom you can see the answer2012-12-26
  • 0
    @DavidMitra I only put parenthesis because LaTeX cut off the top of the square root sign, which sort of defeated the purpose of the question.2012-12-26
  • 2
    It's a typesetting error, clearly. The variables were not meant to be under the radicand (if the answer is to be correct).2012-12-26
  • 0
    @mathguy Ah. If I may suggest a TeX tip: If you want a bit of extra space, use "\," or "\thinspace" at the end (these are plain TeX commands). "\sqrt{2x_3}" gives $\sqrt{2x_3}$, while "\sqrt{2x_3\,}" gives $\sqrt{2x_3\,}$.2012-12-26
  • 0
    @DavidMitra Ohh...I'll definitely keep that in mind, thanks!2012-12-26
  • 0
    Let's hope *those* are ugly, big, annoying and confusing typesetting errors ( in (a) and (f) ) and not mathematical ones by that book's author...2012-12-26
  • 0
    Shouldn't they have written it as $x_3\sqrt2$ to avoid this mess in the first place?2012-12-26

3 Answers 3

4

Answer to title question: It's NOT!

Your question is legitimate:

$$x_1+5x_2-\sqrt{2x_3\;} = 1\tag{1}$$

$(1)$ is not a linear equation as you suggest.

Nor is $(f)$ linear, as typeset in the image.


I suspect there was a misprint in the problem set (book), or a careless typo that the author (and/or editor) over-looked, and which was intended to be:

$$x_1 + 5x_2 - \sqrt{2}\;\cdot x_3 = 1\tag{2}$$

NOW, $(2)$ is a linear equation.

  • 0
    this is not misprint you can see the cutting of the book in the question above2012-12-26
  • 0
    this is from elementary linear algebra by Howard anton and Chris rorres2012-12-26
  • 1
    @ZiaurRahman: Maybe that was mistyped in the book.2012-12-26
  • 2
    Zia ur: I meant a misprint in the BOOK, not by you!2012-12-26
  • 0
    i don't think so2012-12-26
  • 0
    @ZiaurRahman: May I ask you to write for me the exact definition which book did about linearity of an equation? Thanks.2012-12-26
  • 0
    yes you are right its typing error in the book2012-12-26
  • 0
    @ZiaurRahman, why did you change your opinion about typo now?2012-12-26
  • 0
    in linear equation all variables occurs only to the first power, linear equation does not involve any product or roots, and the variables must not appear as arguments to logrithmic, trignometric or exponential functions.2012-12-26
  • 1
    i changed my opinion because i saw the same question in the 9th edition of this book where the square root is only on 22012-12-26
  • 0
    No problem, Zia ur. I would have been perplexed too, had I encountered that question as written in the older edition!2012-12-26
13

Here is the real exercise found on Amazon...

enter image description here

  • 0
    You saved us here. ;-)2012-12-26
  • 1
    This picture is from the latest edition. It seems in some earlier edition the erroneous statements appear (see the question itself). Nowadays textbooks have on-line errata sites. In the olden days it was not so easy...2012-12-26
4

$$(x+5y-1)=\sqrt{2z}$$ so $$(x+5y-1)^2=2z$$ and this is not a linear equation because the order of variabes are 2.

  • 0
    @amWhy: thanks so much. $\ddot\smile$2013-02-07