Question 171501: My niece was given an algebra problem for extra credit. I tried to solve it for her. But there were 2 Var. and only one equation. I tried to simplify the question:
5.5x -1.5y= 3.5x
soln.
5.5x-3.5x-1.5y=3.5x-3.5x
2x-1.5y=0
She said te answers were x=3 & y=4
How did they get that?
Found 3 solutions by Mathtut, solver91311, stanbon:
Answer by Mathtut(3670)   (Show Source):
You can put this solution on YOUR website! 5.5x-1.5y=3.5x
:
2x-1.5y=0....yes x=3 and y =4 works but so does x=6 and y=8 and x=1 and y=4/3......The answers are actually infinite. In order to get a one solution answer there has to be something to compare it to.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website! (3,4) is certainly A solution to your equation because
However, it is not the only solution. In fact, any two numbers that are in proportion 3:4 will satisfy the equation and that means that the solution set has infinite elements. What I suspect is that there is another constraint on the problem that you haven't mentioned (and perhaps your daughter has failed to mention to you) that will uniquely determine a single ordered pair as the solution set.
If this equation is the result of interpretation of a word problem, share the word problem as stated and perhaps there is a clue to that other constraint that you are missing.
Hope this helps.
John
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! My niece was given an algebra problem for extra credit. I tried to solve it for her. But there were 2 Var. and only one equation. I tried to simplify the question:
5.5x -1.5y= 3.5x
soln.
5.5x-3.5x-1.5y=3.5x-3.5x
2x-1.5y=0
She said the answers were x=3 & y=4
How did they get that?
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That is one of an infinite number of solutions for the equation.
If 2x - 1.5y = 0
then y = (2/1.5)x = (4/3)x
or the same statement is
3y = 4x
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Obviously x = 3 and y = 4 fits that equation (makes it true).
So does x = 6 and y = 8, or x = 9 and y = 12, etc.
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I believe that is what the instructor is hoping the students
will see, viz. that the equation has a numberless number of
solutions.
That would be true of most equations with two variables; I'm
allowing for an exception that does come to mind at the moment.
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Cheers,
Stan H.
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