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# Slope of the line

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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.

tertre  Mar 31, 2018

#1
+7266
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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 .

hectictar  Mar 31, 2018
edited by hectictar  Mar 31, 2018
#1
+7266
+4

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 .

hectictar  Mar 31, 2018
edited by hectictar  Mar 31, 2018
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Thanks so much, hectictar!

tertre  Apr 2, 2018
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Nice, hectictar  !!!!

CPhill  Mar 31, 2018