Determine whether the equation is linear in x and y
- Thread starter frctl
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Tough call on this one. I'll give you one for free, just because it's hard to type what is actually meant.
1) [math]2^{1/3}x + \sqrt{3}y = 1[/math]
or
2) [math]2^{1/3}x + \sqrt{3y} = 1[/math]
It's not perfectly clear which it is in the inline version that you wrote. You can use parentheses to clarify intent. (√3)y or √(3y). Or, like I showed, you can learn just a little LaTeX and make it right!
The first is linear in x an y. The second is not linear. Do you see the difference?
1) [math]2^{1/3}x + \sqrt{3}y = 1[/math]
or
2) [math]2^{1/3}x + \sqrt{3y} = 1[/math]
It's not perfectly clear which it is in the inline version that you wrote. You can use parentheses to clarify intent. (√3)y or √(3y). Or, like I showed, you can learn just a little LaTeX and make it right!
The first is linear in x an y. The second is not linear. Do you see the difference?
Suppose y = 4. Then
[MATH]\sqrt{3}y = \sqrt{3} * 4 = 4\sqrt{3} \ne 2\sqrt{3} = \sqrt{4} * \sqrt{3} = \sqrt{4 * 3} = \sqrt{3y}.[/MATH]
So the answer depends on whether you meant sqrt(3) * y or sqrt((3y), which are quite different.
The problem neither asks nor requires you to isolate either variable. Moreover you can never isolate multiple variables simultaneously.
A linear equation is one that can be expressed with an exponent of exactly 1 on each variable.
I do not recommend algebra students to use LaTeX, which is a very fussy code and therefore very time consuming. However, an easy way to start learning is to hit reply to a post containing LaTeX. You will then see the exact code used in the copy of what you are replying to.
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- Apr 12, 2005
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Agreed. I did say "just a little". Go ahead and use more parentheses than you think are necessary. Make sure what you write is clear.
You are NOT asked to solve for x and y if the title of your post reflects what the question asked is, and in fact you cannot find a unique answer from the information given
You seem to just be being asked whether the equation is linear. It is if it can be expressed solely in terms of [MATH]x^1[/MATH] and [MATH]y^1.[/MATH]
So what's your answer?
Yes.