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Find a solution to an equation with 2 variables

  1. Dec 9, 2011 #1
    Hi,

    I've this one equation.

    Is it possible to find a solution for this equation? How?

    5x+3y=29


    Is this a linear equation?
     
  2. jcsd
  3. Dec 9, 2011 #2
    Yes it is a linear equation.

    Given any set of values for x one can plot a graph showing the corresponding values of y which, together with the values of x, satisfy the equation.
     
  4. Dec 9, 2011 #3
    I'm going to replace 29 to 30, because it's easier to solve.
    Now the equation become 5x+3y=30.

    To solve this equation, I did the following:

    x=[itex]\frac{30-3y}{5} = 6-\frac{3}{5}y[/itex]

    y=[itex]\frac{30-5x}{3} = 10-\frac{5}{3}x[/itex]


    replacing the x variable in the y equation, I ended in the result:

    y=[itex]10-\frac{5}{3}(6-\frac{3}{5}y) = 10-10+y[/itex]

    Now, I'm stuck.

    y=y???? Really????

    I must be doing something wrong and I don't know where. Any help?
     
    Last edited: Dec 9, 2011
  5. Dec 9, 2011 #4

    hotvette

    User Avatar
    Homework Helper

    Where did you get 30? It should be 29 based on your original equation.
     
  6. Dec 9, 2011 #5
    Yes, it should be 29 on the original equation, but doesn't matter if it's 29 or 30. I just put 30 now, because because I think it's easier to solve it. I just want to know what am I doing wrong, to my solution end in the stuck stage.
     
  7. Dec 9, 2011 #6
    Getting y = y shows that the working was OK.

    But getting expressions of x and for y from the given equation and substituting these expressions back into the equation will lead nowhere except to tell us that y = y!
     
  8. Dec 9, 2011 #7
    So, how do I solve this equation?
     
  9. Dec 9, 2011 #8

    Mark44

    Staff: Mentor

    The equation 5x + 3y = 30 (your revised equation) represents a line in the plane. As such, there is no single solution. Every pair of points (x, y) that is on the line is a solution to this equation.

    Some points on the line are (6, 0), (0, 10), and an infinite number of others.
     
  10. Dec 9, 2011 #9
    1 - So, you mean that exists many solutions. This is because of this specific equation, or all these type of equations have several solutions, and what I did is enough?

    2 - If I draw a line in a plan between (6,0) and (0,10), it means that all the values that are inside the area of the line ( (6,0) -- (0,0) -- (0,10); points AOB) are solutions?
     
  11. Dec 9, 2011 #10

    Mark44

    Staff: Mentor

    Yes, there are many (an infinite number of) solutions to your equation. Any equation of the form Ax + By = C, where A and B are not both zero, represents a straight line. Any point (x, y) on the line is a solution to the equation, and any solution to the equation is a point on the line.

    As to what you did, all you did was write an equivalent equation y = 10 - (5/3)x. This is just another form for the same equation - you didn't find any solutions. You also wrote another equation in which you solved for x. That also is not finding a solution.

    Toward the end of your 2nd post, you did some more work in which you concluded that y = y. This is true, but not at all useful. A variable is obviously equal to itself.

    Any of the three equations can be used to find as many solutions as you want. For example, using the equation y = 10 - (5/3)x, if you let x = 0, it's easy to see that y = 10. So the point (0, 10) is a solution.

    If you let x = 3, then y = 5, so (3, 5) is a solution. And so on for as many values of x as you want to put into the equation.


    No. The only solutions to the equation 5x + 3y = 30 (or y = (-5/3)x + 10) are the points on the line - not under it or over it.

    The points (6, 0) and (0, 10) are solutions to the equation (and are points on the line) because each of these pairs of numbers makes 5x + 3y = 30 a true statement. The point (0, 0) is not a solution (and so is not on the line) because 5*0 + 3*0 [itex]\neq[/itex] 30.
     
  12. Dec 9, 2011 #11
    If you want a single solution for all the variables you'll have to have at least as many equations as variables. Here you have 2 variables and 1 equation, so a single solution isn't possible. Note that having as many equations as variables doesn't mean there will be a solution.

    In the case of your equation:
    y = -(5/3)x + 10
    This equation will give a single solution:
    y = x + 2

    In this case, the solution is where the two lines intersect, x=3, y=5.

    However this equation:
    y = -(5/3)x + 9
    will produce a parallel line that will never intersect. Thus, there is no solution.

    Here's a good online graphing calculator to play around with this stuff:
    http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html
     
  13. Dec 9, 2011 #12
    Yeah you need at least two equations in this case to find a single solution ;]
     
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