+4609 1. What is the slope of the line determined by any two solutions to the equation \(\frac{2}{x}+\frac{3}{y} = 0\)?Express your answer as a common fraction.
+9466 Let's find two solutions by choosing a value for x and solving for y .
To find a first solution I choose x = 1 .
\(\frac21+\frac3y=0\\ 2+\frac3y=0\\ \frac3y=-2\\ 3=-2y\\ -\frac{3}{2}=y\)
When x = 1 , y = -3/2 . So one solution is (1, -3/2) .
To find a second solution I choose x = 2 .
\(\frac22+\frac3y=0\\ 1+\frac3y=0\\ \frac3y=-1\\ 3=-1y\\ -3=y\)
When x = 2 , y = -3 . So another solution is (2, -3)
slope between (1, -3/2) and (2, -3) \(=\,\frac{(-3)-(-\frac32)}{2-1}\\ =\,\frac{(-3)-(-\frac32)}{1}\\ =\,(-3)-(-\frac32)\\ =\,-3+\frac32\\ =\,-\frac62+\frac32\\ =\,-\frac32\)
Another way to find the slope is to get the original equation into slope intercept form.
\(\frac2x+\frac3y=0\\ 2+\frac{3x}{y}=0\qquad\text{and }x\neq0\\ 2y+3x=0\qquad\text{and }y\neq0\\ 2y=-3x\\ y=-\frac32x\)
The slope between any two points is -3/2 .
+9466 Let's find two solutions by choosing a value for x and solving for y .
To find a first solution I choose x = 1 .
\(\frac21+\frac3y=0\\ 2+\frac3y=0\\ \frac3y=-2\\ 3=-2y\\ -\frac{3}{2}=y\)
When x = 1 , y = -3/2 . So one solution is (1, -3/2) .
To find a second solution I choose x = 2 .
\(\frac22+\frac3y=0\\ 1+\frac3y=0\\ \frac3y=-1\\ 3=-1y\\ -3=y\)
When x = 2 , y = -3 . So another solution is (2, -3)
slope between (1, -3/2) and (2, -3) \(=\,\frac{(-3)-(-\frac32)}{2-1}\\ =\,\frac{(-3)-(-\frac32)}{1}\\ =\,(-3)-(-\frac32)\\ =\,-3+\frac32\\ =\,-\frac62+\frac32\\ =\,-\frac32\)
Another way to find the slope is to get the original equation into slope intercept form.
\(\frac2x+\frac3y=0\\ 2+\frac{3x}{y}=0\qquad\text{and }x\neq0\\ 2y+3x=0\qquad\text{and }y\neq0\\ 2y=-3x\\ y=-\frac32x\)
The slope between any two points is -3/2 .
