Help understanding when an equation is linear

[Nemanjavuk69]

I am studying Linear Equations and need help understanding WHY the equation I have put a red square around IS NOT linear. I have a hard time figuring out when something is linear and when it is not.

I am using the book (linear algebra and its applications sixth global edition).

I understand that the first equation is linear, since we can write it as [imath]a_1x_1+a_2x_2+a_3x_3=b[/imath]
However, why is that not possible with the second equation? Why can't we just do this [math]x_2=2\sqrt{x_1}-6 <=> 6=2\sqrt{x_1}-x_2[/math]. so that way we can have all our variables on one side and the coefficient on the other side. Is the second equation not linear BECAUSE [math]4x_1-5x_2=x_1x_2[/math] consists only of variables? If so, why does the author not just write that instead of saying "and [imath]\sqrt{x_1}[/imath] in the second" hinting that there is something wrong with equation 2.2 also?

 

[topsquark]

I am studying Linear Equations and need help understanding WHY the equation I have put a red square around IS NOT linear. I have a hard time figuring out when something is linear and when it is not.

I am using the book (linear algebra and its applications sixth global edition).

I understand that the first equation is linear, since we can write it as [imath]a_1x_1+a_2x_2+a_3x_3=b[/imath]
However, why is that not possible with the second equation? Why can't we just do this [math]x_2=2\sqrt{x_1}-6 <=> 6=2\sqrt{x_1}-x_2[/math]. so that way we can have all our variables on one side and the coefficient on the other side. Is the second equation not linear BECAUSE [math]4x_1-5x_2=x_1x_2[/math] consists only of variables? If so, why does the author not just write that instead of saying "and [imath]\sqrt{x_1}[/imath] in the second" hinting that there is something wrong with equation 2.2 also?
View attachment 33988

In [imath]4x_1 - 5x_2 = x_1 x_2[/imath] you are multiplying the [imath]x_1[/imath] and [imath]x_2[/imath] variables. All the variables need to be only added to each other, not multiplied.

In [imath]x_2 = 2 \sqrt{x_1} - 6[/imath] you are taking the square root of one of the variables. All variables have to be to the first power.

-Dan

 

[Nemanjavuk69]

In [imath]4x_1 - 5x_2 = x_1 x_2[/imath] you are multiplying the [imath]x_1[/imath] and [imath]x_2[/imath] variables. All the variables need to be only added to each other, not multiplied.

In [imath]x_2 = 2 \sqrt{x_1} - 6[/imath] you are taking the square root of one of the variables. All variables have to be to the first power.

-Dan

Ohhh, that really makes sense. I have one more curious question I hope you have time to answer. Regarding your second line of comment saying "All variables have to be to the first power". I know that taking the square root of a variable is the same as saying [imath]x^{\frac{1}{2}}[/imath]. Is this what you mean by, that all variables have to be to the first power? I am curious, because, what if the variable was, let's say 4, the square root of 4 would be 2 which is the same as saying [imath]2^{1}[/imath]. Is it just a definition case where we agreed, that taking a square root of a variable is not fulfilling the condition of the system being linear?

Thank you for your help and answer!

-Nemanja

 

[Steven G]

The coefficients of the x_i can have powers other than 1. However, to be linear you have to have the variable, x_i, raised to the 1st power and nothing else. Just look carefully at the definition of a linear equation that you posted above.

 

[Nemanjavuk69]

The coefficients of the x_i can have powers other than 1. However, to be linear you have to have the variable, x_i, raised to the 1st power and nothing else. Just look carefully at the definition of a linear equation that you posted above.

Makes completely sense. I understand. Thank you very much for elaborating. Both you and Dan helped a lot. Have a great day/ night forward!

-Nemanja