what's wrong here?

[allegansveritatem]

Here is the exercise:
The first thing I wanted to do was to find the intersection point of the first two equations. the last two equations are not intersecting, they are more like limiters. Anyway, here is my dilemma: When I try to solve these two--I switched the inequality signs to equal signs--for the intersection point I first cancel the x's. and I get y=5 and x=1/3. then I try solving for the intersection point by cancelling out the y's first and get y=3, x=1. This second solution is the right one, according to my graphing calculator. Now, I went over the first ty again and again and couldn't find any error in my work, but still I don't get the right answer so, what is wrong here. I am not asking how to deal with the exercise as a system of inequalities right now, but I am asking why, when I cancel x first I get one solutiona and when I cancel y first I get another? Here is what I did:

 

[pka]

Have a look at this plot.

 

[Harry_the_cat]

At the end of your first line you have 5y=15 then y=5. Should be y=3.

 

[Steven G]

the last two equations are not intersecting. Are you sure about that??????

 

[allegansveritatem]

At the end of your first line you have 5y=15 then y=5. Should be y=3.

I knew I was doing something like that but I just couldn't see it. It's enough to make a feller blush. Thanks.

 

[HallsofIvy]

Here is the exercise:View attachment 21711
The first thing I wanted to do was to find the intersection point of the first two equations. the last two equations are not intersecting, they are more like limiters.[/quore]
On the contrary, the lies bounding the last two inequalities are x= -2, a vertical line, and y= 4, a horizontal line, and they intersect at (-2, 4).

The lines bounding the first two inequalities are 3x+ y= 6 and y- 2x= 1. From the second y= 2x+ 1. Putting that into the first, 3x+ 2x+ 1= 5x+ 1= 6 so 5x= 5 and x= 1. Then y= 2+ 1= 3. The first two lines intersect at (1, 3). The lines y- 2x= 1 and y= 4 intersect where 4- 2x= 1 so x= 3/2, y= 4. The lines y- 2x= 1 and x= -2 intersect where y- 2(-2)= 1, so y= 1- 4= -3. The lines intersect at (-2, -3).

The region defined by the inequalities is the irregular quadrilateral having vertices at (-2, 4), (3/2, 4), (1, 3), and (-2, -3).