Question 178979
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To put an equation into slope-intercept form, solve for y:


*[tex \Large 3x + 2y = 6]


Add -3x to both sides:


*[tex \Large 3x + (-3x) + 2y = -3x + 6]


*[tex \Large 2y = -3x + 6]


Multiply by *[tex \Large \frac {1}{2}]


*[tex \Large \left(\frac {1}{2} \right) 2y = \left(\frac {1}{2} \right)(-3x + 6)]


*[tex \Large y = \frac {-3}{2} x + 3]


The slope is now the coefficient on x, namely *[tex \Large \frac {-3}{2}]


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Here's another way to look at it, if perhaps you remain unconvinced:


Start with your original equation.


*[tex \Large 3x + 2y = 6]


Select a value for <i>x</i>, anything will do so pick something that will make the arithmetic easy, like 1.  Substitute this value into the original equation and calculate the value of <i>y</i>:


*[tex \Large 3(1) + 2y = 6]


*[tex \Large 2y = 3]


*[tex \Large y = \frac {3}{2}]


So now we know that when *[tex \Large x = 1], *[tex \Large y = \frac {3}{2}], meaning that one of the points on the line is described by the ordered pair *[tex \Large \left( 1, \frac {3}{2} \right)].


Repeat the process with another value of <i>x</i>, say 2.


*[tex \Large 3(2) + 2y = 6]


*[tex \Large 2y = 0]


*[tex \Large y = 0]


And we have another point on the line: *[tex \Large \left( 2, 0 \right)]. 


Given two points *[tex \Large \left( x_1,y_1 \right)] and *[tex \Large \left( x_2, y_2 \right)], the slope can be calculated using:


*[tex \Large m = \frac {y_1 - y_2}{x_1 - x_2}]


Substituting our derived coordinates:


*[tex \Large m = \frac {{3 \over 2} - 0}{1 - 2}=\frac {{3 \over 2}}{-1}= -\frac {3}{2}]


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One more idea to carry around with you in your back pocket.  Any time you have a 2-variable linear equation <i><b>written in standard form</b></i> where the signs on the two variable terms are the same, you will have a negative slope.

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