Hey,
I have this question thats been bugging me for a few days now.
$\displaystyle 3y-15=2x-8$
Whats the answer if we put it in the eqN of a line?
$\displaystyle 2x-3y+7=0$ OR $\displaystyle 2x-3y-7=0$?
Thanks for the help.
-NZF
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Hey,
I have this question thats been bugging me for a few days now.
$\displaystyle 3y-15=2x-8$
Whats the answer if we put it in the eqN of a line?
$\displaystyle 2x-3y+7=0$ OR $\displaystyle 2x-3y-7=0$?
Thanks for the help.
-NZF
are you asking the solution of the line (when y equals 0) or are you asking for us to put the equation in standard/y-intercept/point-slope form?Quote:
Originally Posted by 905
Put the equation in standard form. :pQuote:
Originally Posted by Quick
EDIT: Actually it's just part of a question, the only part I don't fully understand.
-NZF
alright, standard form is in the form of $\displaystyle ax+by=c$ therefore we start with your equation...Quote:
Originally Posted by NineZeroFive
$\displaystyle 3y-15=2x-8$
add 15 to both sides $\displaystyle 3y=2x-8+15$
subtract 2x from both sides: $\displaystyle 3y-2x=7$
voila!
~ $\displaystyle Q\!u\!i\!c\!k$
Algebraically it would have to be $\displaystyle 2x-3y+7=0$. Other standard forms:Quote:
Originally Posted by NineZeroFive
$\displaystyle y = \frac{2}{3}x + \frac{7}{3}$ (Slope - Intercept form)
and
$\displaystyle (y - 3) = \frac{2}{3}(x - 1)$ (An example of point-slope form)
and
$\displaystyle \frac{x}{3} - \frac{y}{2} = - \frac{7}{6}$ (I forget what this one is called.)
-Dan
$\displaystyle 3y-2x-7=0$? Because it the textbook it says its $\displaystyle 2x-3y+7=0$?Quote:
Originally Posted by Quick
:confused:
multiply both sides of $\displaystyle 3y-2x-7=0$ by negative 1Quote:
Originally Posted by NineZeroFive
Sorry,
I think I made a typo somewhere.
3y-15=2x-8 is the equation you need to put in the standard form.
Is the answer 2x-3y+7=0? (textbook answer) Why? -.-
Why is it NOT 2x-3y-7=0?
-NZF
$\displaystyle 3y - 15 = 2x - 8$Quote:
Originally Posted by NineZeroFive
$\displaystyle 3y - 15 - 3y = 2x - 8 - 3y$
$\displaystyle -15 = 2x - 3y - 8$
$\displaystyle -15 + 15 = 2x - 3y - 8 + 15$
$\displaystyle 0 = 2x - 3y + 7$
-Dan
I assume you mean:Quote:
Originally Posted by NineZeroFive
but the answer, both equations are correct, your book is just picky.Quote:
Originally Posted by NineZeroFive
although I must say, I've never actually seen standard form equalling zero...
One standard way of writing a multinomial (the "most" standard I've seen, if there is such a thing) is to write the terms of highest degree first down to the lowest degree such that the expression equals zero. For example:Quote:
Originally Posted by Quick
$\displaystyle 3x^2y^3 - xy^2 + 12x - 3y + 5 = 0$
In the case for a linear equation the form simply becomes:
$\displaystyle ax + by + c = 0$
I'll admit I don't usually see this form for a line, but considering the expression as a multinomial it would be standard.
-Dan
I'm still kinda confused because the book says "The equation of the line through A and B is 2x-3y+7=0. :confused:Quote:
Originally Posted by Quick
Why is it 2x-3y+7=0 and NOT 2x-3y-7=0?
hmm..
-NZF
Unless you have a question about my derivation, please see post #9 under this thread. I derived the form there.Quote:
Originally Posted by NineZeroFive
-Dan
