I need help understanding, how to graph the inequality, for the following story problem. On average, it takes an accountant 1 hour to complete a simple tax return and 3 hours to complete a complicated return. If the accountant wants to work no more than 9 hours per day, use the illustration to graph an inequality that shows the possible ways that simple returns (x) and complicated returns (y) can be completed each day. I got the inequality written write, which is x + 3y is equal to or less than 9..I just don't understand how to graph it..see attachment for solution from answer key..
I don't understand how the solutions manual came up with what they did...
you probably have doubt with the ordered pairs
It's been a couple years since I've touched algebra at all...I'm in college now....I've forgot what the ordered pairs were all about, and how to find them...
ordered pairs are pair of x, y values which satisfy an equation. You arbitrarily choose (preferably simple) values of x and put it in the equation to get corresponding values of y.
since you are provided with an inequality, you treat it as an equation in order to calculate y-values x+3y=9
So I randomly put numbers in for x, which will give me different values for y...which will give me the graph that the solutions manual came up with? I guess I'm confused as to how their ordered pairs even correspond with the line they graphed..
yes, that is how ordered pairs of (x,y) are calculated. Now plot these point on graph. draw a line passing through these points. It is the required graph
I don't see a dot for any of the ordered pairs that they listed...their points appear to be within the shaded area to me...
Because the ordered pairs they have provided is the part of final solution. They are not on the line.
The shaded area are all the points that satisfy the inequality. So it's ok if they points they give are in it. That's what you would expect.
Any (x,y) value that is in the shaded region can be plugged in and you should get a true statement.
For example if we take the (1,1) that they have: \[1 + 3(1) \le 9 \implies 4 \le 9\] Which is true. 4 is less than or equal to 9.
Once you draw the line with the help of equation, then you see the condition of inequality, which states that x+3y be less than or equal to 9. So, any point below this line would satisfy the inequality. Hence, the solution is the shaded-area below the line.
So how do I figure out the shaded area and line? this picture is from the answer key...
The line is what you get when you consider the equality version of what you have. Instead of : \[x + 3y \le 9\] You look at this: \[x + 3y = 9\]
That last equation is a line. If you draw that line then you know where the edge of your inequality is.
Since you want to consider only the points less than that line you shade the region beneath it.
Any point in the shaded region will satisfy (make true) the original inequality.
Would I have to put the x + 3y = 9 into y= x + ect. form?
It probably makes it easier to graph that way yes.
so does y = -(1/3) + 3 look right?
It's missing an x, but other than that, yes. It's good.
oops the -1/3 was for the x..
I figured =)
But how can I graph this? The graph can't involve negative numbers...
Welcome back to college btw =)
Well like other graphs, you start with the y intercept. (3) then move according to the slope (down 1, right 3)
Then draw a line between those two points.
lightbulb finally came on..and I'm feeling pretty dumb now...
You're doing great.
I forgot that you start at the y intercept, not zero, with the 1/3 slope...
One of the better students we've had around here in fact.
um,I've hardly answered any questions on here, how can you say that?
I said student, not tutor ;)
funny...and plan to apply for a tutoring position at my college this fall...in math and science, lower levels anyway..:P
That's awsome!
So how long has it been since highschool?
2 years.
Oh, that's not bad at all. I waited 15 before I actually got around to going back to college.
actually, this is my first time at college....I'm all caught up on doing good with my online algebra course...because grades mean everything, when trying for nursing..
Yep. It's a good field to get into.
well...I should get back to my math, thanks polpalk...I'll probably be back on here off and on all day today....I'm trying to get a lot of math done, so I can crash for my Anatomy and Physiology 1 test this Monday tomorrow...:P
fun fun. =) Good luck.
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