How did you do it?
What I did:
If the relation is linear,
so
and to find ,
plug in one of the coordinates,
therefore,
I might be wrong, but this is a method for finding the equation of the line.
-Masoug
Okay, I came up with a wild answer for this problem:
Find the slope-intercept form of the line passing through the points (5,-2) and (-3,5)
I came up with this fraction-ridden answer that I know is wrong...
y = -7/8x - 1 and -43/8
I'm trying to figure out what I did wrong and how to properly solve this. Thanks!
Which did you use as and ? Either way works but you have to be consistent.
With and , .
Taking and [tex](x_2, y_2)= (5, -2) just changes the sign in both numerator and denominator and so gives the same slope: .
Taking , becomes so . .
Taking , becomes so . , exactly the same answer.
Of course, whichever way you choose, you can simplify by multiplying the whole equation, either here or, better, in the orginal form, by 8.
Multiplying both sides of by 8 gives or 8y- 40= -7x- 21 so 7x+ 8y= -21+ 40= 19.
Multiplying both sides of by 8 gives or 8y+ 16= -7x+ 35 so 7x+ 8y= 40- 16= 19 again.
Masoug said
and then said "b= 2.375"
He got that, of course, by adding to both sides of the equation:
He then divided: 8 goes into 19 twice with remainder 3 and 3/8= 0.375:
Personally, I would have left it as an improper fraction, or, as I have already said, multiply the entire equation by 8 to get rid of the fractions.
and, multiplying through by 8,Not to mention your final solution?
8y= -7x+ 19 or 7x+ 8y= 19.