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Thread: I hate word problems

  1. #1 I hate word problems 
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    Ok

    You know, any help would be greatly appreciated, because I'm really getting pissed off.

    I'm having word problems, u know, based on solving linear systems by graphing.

    I'm having a problem that is kind of making me confused and lost.

    So, I'll write down a system that I can't solve, instead of copying the whole word problem. Here it goes.

    1) d = -55t + 3000 2)d= -40t + 2400

    which is therefore,

    1) y = -55x + 3000 2) y = -40x + 2400

    I believe I need to find out the point of intersection, right? How do I do that? First, I'd need to use calculator, and type in both of these systems in "y=" button.

    After I did that, I'd need to set the window right. And that's where my big problem is, that where I'm stuck, setting the window right. How the hell do I set the window right before finding out the point of intersection? That's where it's pissing me off........

    Any input is greatly appreciated, because you know I will love you forever if you help me!


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  3. #2  
    Forum Bachelors Degree Demen Tolden's Avatar
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    If you knew that a = b, and b = c, then wouldn't stand to reason that a = c?

    If you knew that a = 2b, and b = c, then wouldn't a = 2c?

    If you knew that a = 2b + 3, and 2a = 6b + 8, then what is the numberical value of b? If you can find what the value of b is, then just plug that number in and find a.


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  4. #3  
    Forum Bachelors Degree Demen Tolden's Avatar
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    Actually I'll make the second line easier for you.
    2a = 6b + 8

    is the same as:
    a = 3b + 4

    If you notice, you could make both
    a = 2b

    and
    a = 3b + 4

    into terms of x and y, and have yourself a pair of lines with different slopes, and therefore they would intersect at some point. These values of a and b at your intersection point are the very same values you should get if you solved both equations. Meaning there is exactly one value of a for which
    a = 2b

    and
    a = 3b + 4

    and one value of b. If you wrote these two values like this (a,b), that would be the point where the two lines would intersect.
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  5. #4  
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    Is it required that you use a calculator?
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  6. #5  
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    Thanks for the replies.

    Demen, I'm not sure if you got my point but I'm talking about finding the point of intersection (the solution) from word problems.

    Ok, here I will copy the steps from notes I have I wrote from class...so to see if you have an idea of what I'm talking about.

    Steps -

    1. Put on calculator "y="

    2. "Windows" to fit the graph

    3. Point of intersection

    4. Pick 3 points to create graph

    Harold, I probably wouldn't say it's required, but it's also meant to practice graphing by hand. (slopes on grid paper) But, based on finding point of intersection, I would say yes...

    Let me try to make it more clear -

    In my first post in this thread, where I pointed out d's and t's - d represents the metres and t represents the time in seconds (out of the word problem) therefore making it a y=mx+b formula...

    Thanks again for the replies.
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  7. #6  
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    Well, I never had a graphing calculator but the usual way to solve two simultaneous equations like that would be to take one equation and subtract the other, eliminating the x, then solve for y. Then you take y and substitute it back into one of the equations to get x. You should get 800, 40 as the point where the two lines intersect.
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  8. #7  
    Forum Professor serpicojr's Avatar
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    Faron, Demen and Harold (and now myself) are really encouraging you not to use your calculator. You don't need it. Even if you were told to do so in class, you really should try to do it by hand.

    You're correct that a solution to that system of equations is a point of intersection of those two lines. So it's good enough to find that. There are a couple of ways. One uses graphs (geometry), and one uses algebra.

    If you're going to do this by graphing things by hand, then let's think about how to graph lines. One way is to recall that a line is determined by two points--i.e., given two points, there is a unique line connecting them. So if you just plug in a couple of values for x, find the corresponding values for y, and plot them, you can easily draw a graph for a line.

    Or there's a way of drawing lines of the form y = mx+b. m means something geometric, as does b, and you can use these to draw your line. You just have to recall what these guys mean.

    The problem with this approach is you're using some big values for m and b, so off the bat I know you can't fit these graphs on a piece of graphing paper containing the origin and with each square representing one square unit. You either have to be crafty about how you graph things, or you have to look for another way of doing things.

    One way is by using your calculator. Put it down now.

    The other way is by algebra, which is what Demen Tolden was suggesting. If we have two lines and we want to find their point of intersection, we know we're looking for a point (x,y) which satisfies both equations. So, for example, if your case, if I have the point of intersection (x,y), then I know:

    y = -55x + 3000

    and

    y = -40x + 2400

    This is the same x and the same y in these equations. So y is equal to -55x+3000 and is equal to -40x+2400. y is -55x+3000 is -40x+2400. A rose is a rose is a rose.
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  9. #8  
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    Hmmmm.........

    Ok, well yeah that's what I'm trying to do, to graph by hand, with the support of calculator.

    I believe I need to find y-intercept and x-intercept from both equations. And if I do, that would help me to set the window right and then I'd find the point of intersection.

    So the main issue is finding y-intercept and x-intercept....

    I just don't wanna be confused..
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  10. #9  
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    Oh for speaking of y-intercept, I also think I got this:

    y=-55x+3000

    y-int=3000

    x-int= ?

    - - -

    y=-40x+2400

    y-int=2400

    x-int = ?

    If I find x-intercept then I can be able to set the window and then point of intersection, right? So...
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  11. #10  
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    You knew what the y-intercept is because you memorized the fact that when you put the equation in the form y=mx+b, the y-intercept is b. But think about why that is true. It's because that is the point on the line where x=0, so if you set x=0, then y=b, the y-intercept. You can find the x intercept using the same kind of logic.
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  12. #11  
    Forum Professor serpicojr's Avatar
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    What does the x intercept mean? Thinking about the definition will help you figure out how to find it.

    What is your strategy for solving this? Literally, step by step, is your plan to graph your lines roughly on paper so that you can set the window correctly on your calculator and use, say, the cursor to find the intersection?

    And, just curious, what level class is this for?
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  13. #12  
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    I believe that the best strategy to solve this (out of word problems) is to use x-intercept and y-intercept (bigger numbers than x-int and y-int) on a window settings and then you can find the point of intersections.......here is what i got so far -

    Xmin = 0
    Xmax = ?
    Xscl = 1
    Ymin = 0
    Ymax = 5000
    yscl = 1
    xres = 1

    It's Math 20 applied. I'm just in need of finding the point of intersection and other points to make a graph on a graphing paper.

    I think there is also a "Let x & Let y" method involved, too.
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  14. #13  
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    Ok.........looks like I got it, I think. To my guess, I used this -

    y=-55x+3000
    0=-55x+3000
    -55x=3000
    /-55 /-55
    x = -54.5454545455

    y=-40x+2400
    0=-40x+2400
    -40x=2400
    /-40 /-40
    x=-60

    Therefore, the x-max is 100. Then I was able to find the point of intersection perfect.....
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  15. #14  
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    You made an arithmetic error in the third line.
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  16. #15  
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    Then the pt of int is (40, 800) and then to find other points...

    Let x = 100 (any number larger than 40)
    y = - 55(100) + 3000
    = -2500

    (100, -2500)

    let x = 100
    y = -40 (100) + 2400
    = -1600

    (100, -1600)

    Hmmm.........well, I kind of expected the y to be in positive, weird....

    edit:

    Oh nvm I got it......I used the number 50.

    let x = 50
    y = - 55 (50) + 3000
    = 250

    (50, 250)

    let x = 50
    y = -40 (50) + 2400
    = 400

    (50, 400)

    Yes, looks like I got it. Thnx guys! I'm going to try make the graph on graphing paper now.
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  17. #16  
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    Great, now I'm not so sure anymore.

    The purpose of this problem is to make the proper dots on the graph paper and then connect the dots.
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  18. #17  
    Forum Professor serpicojr's Avatar
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    Okay, so you found the point on intersection. You've got the y-intercepts of each line. So, for each line, you have two points. You should have all the information you need to graph the lines. What aren't you sure of?
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  19. #18  
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    Omg....

    Ok, what I've stated, in the first page in this thread, in the last post, is all wrong, I think.

    Yes, now I have I made dots and numbers. I put a dot of point of intersection. And i put two dots of the y-intercerpts.

    Now what is missing is the dots of x-intercepts. That's what I need to figure out.........
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  20. #19  
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    That's what I have on graphing/grid paper right now.

    5000|
    4500|
    4000|
    3500|
    3000|-
    2500|-
    2000|
    1500|
    1000|
    500 | *
    100 |
    0|_________________________________
    0 10 20 30 40 50 60 70 80 90 100

    ("-", "-", "*" are dots, and "*" is the point of intersection)


    Ok here is the screenshot since "*" dot wouldn't appear as i wanted it to be.



    I need to find dots on x-axis.......
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  21. #20  
    Forum Professor serpicojr's Avatar
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    The one explicit hint I will give you is that you only need two points to determine a line. So, unless you want to be able to label your graph, you don't need to find the x-intercepts--you're done. However, if you want or need to find them, think about what it means for a point to lie on the x axis. What can you say about its coordinates?
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  22. #21  
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    So, you don't think I need to find x's? Just make a line to connect to the point of intersection from y-axis dots? Well, wouldn't that make a bit inaccurate? I know the slope, but still.....

    I really think I need to find x's............

    y=-55x+3000

    y=-40x+2400

    Hmm......well, I did try to solve for x's, except the x's appeared to be negative but I need them to be positive....

    y=-55x+3000
    -55x=3000+0
    /-55 /-55
    x=-54.54545455

    y=-40x+2400
    -40x=2400+0
    /-40 /-40
    x=-60

    Hmm..........from this solution, the lines don't connect with the point of intersections, though?
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  23. #22  
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    Ok, I think I believe I understand now, but I still would like to find out the x's.
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  24. #23  
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    so what can you tell me about the points, A(1,0), B(2,0), C(3,0) and D(4,0)?
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  25. #24  
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    Quote Originally Posted by wallaby
    so what can you tell me about the points, A(1,0), B(2,0), C(3,0) and D(4,0)?
    What about them? That's x-axis/x-intersection, therefore is a horizontal slope?

    But I think i got it now. If i have a different system of equations, like 500x+y=125 and 800x+y=155, then i probably would need to find the x's and the point of intersection, everything.
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