SOLUTION: this question has a few parts. does this count as my four question limit? a) find the slope of the line 2x -3y = 7 by placing the equation on slope intercept form. show work

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Question 256217: this question has a few parts. does this count as my four question limit?
a) find the slope of the line 2x -3y = 7 by placing the equation on slope intercept form. show work
b) johnny is to work the following problem. let x be the number of small doghouses to be bulit under the following constraints. let y be the number of large doghouses to be built under the following constraints. x < y; y < or = to 10. give a verbal statement about what y < or = 10 means about the doghouses being built. johnny also adds two other constraints x > or = to 0 and y > or = to 0. why would theses two constraints be logical and needed? johnny decides to graph his related x, y values (x,y) as points. he ended up with all of his graph being in one of the four quandrants. explain why he is correct using only one quandrant.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


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



Add -2x to both sides:



Multiply both sides by



b)



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