Question 256217
<font face="Garamond" size="+2">


To put your equation into slope-intercept form, just solve it for *[tex \Large y].  That is, perform whatever manipulations are required to get *[tex \Large y] all by itself on the LHS and everything else in the RHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ -\ 3y\ =\ 7]


Add -2x to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3y\ =\ -2x\ +\ 7]


Multiply both sides by *[tex \Large -\frac{1}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{2}{3}x\ -\ \frac{7}{3}]


b)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ \leq\ 10]


means that for some reason no more than 10 large doghouses can be built in a given time period.  It might mean that Market Research says you can't sell more than 10 large ones in that time period, or the factory never has orders for more than 10 in that time period.


Both of the variables have to be greater than or equal to zero.  The factory certainly can produce 0 or any positive number (as long as it isn't more than 10 large ones) of either size doghouse.  But a negative number of doghouses of either sized doesn't make any sense -- what are they going to do, disassemble one they built yesterday?


Since both of the variables must always be positive and Quadrant 1 is the only quadrant where both variables are positive, it makes perfectly good sense that the usable part of the graph is in Quadrant 1 only.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>