SOLUTION: hi how are you doing today? well im stuck on this one promblem. the directions say, "Find the slope of each equation." and the equation is 2x+3y+32=0. now what i tried to do was ge

Question 7198: hi how are you doing today? well im stuck on this one promblem. the directions say, "Find the slope of each equation." and the equation is 2x+3y+32=0. now what i tried to do was get the y and x intercepts and some how put them in this form>>>(x,y) (x2,y2) so that i can use the slope formula m=y2-y1/(that means over)x2-x1. thank you so much for your help and i hope i didnt confuse you in anything and if i did then let me know and i'll try to clarify it. thanks again
Answer by prince_abubu(198) (Show Source): You can put this solution on YOUR website!
I think you attempted too hard on this problem. All you need to do is to get the equation in y= form. Once you've done that you can easily pick out the slope.

<---- start.

<---- since we want to get y by itself on one side of the equation (usually on the left), we need to first let the term 3y by itself on one side. We moved the 2x and the 32 to the right side. and they became -2x - 32. I know that many Algebra teachers don't like that way of thinking but it works. (That is, when you transfer terms to the other side of the equation, you flip the signs)

<----- To get the y by itself, we divided 3y by 3, AND also the -2x AND the -32 by 3. If you multiply or divide a term by a certain number, you MUST also multiply or divide ALL THE OTHER TERMS on BOTH SIDES of the equation by that same number.

Now that it's in y= form, you can pick out the slope, which is the constant number multiplied to the x, in this case, the -2/3.
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