# how this equation is a linear equation?

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,

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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. –  David Mitra Dec 26 '12 at 17:16
The original post clearly did not have $2x_3$ nested in parentheses. –  David Mitra Dec 26 '12 at 17:20
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 answer –  Zia ur Rahman Dec 26 '12 at 17:22
@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. –  mathguy Dec 26 '12 at 17:23
It's a typesetting error, clearly. The variables were not meant to be under the radicand (if the answer is to be correct). –  David Mitra Dec 26 '12 at 17:25

## 3 Answers

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.

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this is not misprint you can see the cutting of the book in the question above –  Zia ur Rahman Dec 26 '12 at 17:23
this is from elementary linear algebra by Howard anton and Chris rorres –  Zia ur Rahman Dec 26 '12 at 17:24
@ZiaurRahman: Maybe that was mistyped in the book. –  Babak S. Dec 26 '12 at 17:24
Zia ur: I meant a misprint in the BOOK, not by you! –  amWhy Dec 26 '12 at 17:26
i changed my opinion because i saw the same question in the 9th edition of this book where the square root is only on 2 –  Zia ur Rahman Dec 26 '12 at 17:48

Here is the real exercise found on Amazon...

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You saved us here. ;-) –  Babak S. Dec 26 '12 at 18:01
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... –  GEdgar Dec 26 '12 at 22:32

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

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+1 Short and simple! (It's a new day now! + 2:45) –  amWhy Feb 7 '13 at 2:45
@amWhy: thanks so much. $\ddot\smile$ –  Babak S. Feb 7 '13 at 2:47