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Is this solvable? I think not, I think something is missing.

Am i wrong?

If 4a + b = 3 then 20a + 5 = ?
a. 15
b. 1/4
c. 4/3
d. 3/4
  • 6
    That problem must have a typo: it should be asking about $20a+5b$, which of course is $15$.2012-04-17
  • 1
    perhaps the edit which corrects the typo is not so helpful...2012-04-17
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    @Ronald: I think I see what you're saying. Brian's comment looks as though it's correcting an error that isn't there...2012-04-17
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    I find this a bit strange to get tagged "linear algebra" instead of "precalculus", maybe that's just me. I mean, in linear algebra, don't you try and find all solutions to equations like ((4*a)+b)=3? Digressing, ((4*a)+b)=3 has an infinity of solutions which you can describe by a line, or equivalently an infinite set of pairs on the reals. All members of the set of pairs have form (a, (3-(4*a))). So, (1, -1), (2, -5), (3, -9), etc. all work as particular solutions for ((4*a)+b)=3, though of course this isn't what this question is asking. The tag "linear-algebra" just trips me up here.2012-04-17
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    @Doug: Feel free to change it!2012-04-17

2 Answers 2

3

You are quite correct: something has to be missing. No matter what value you give $a$, there is a value of $b$ that makes the first equation true, so $20a+5$ is indeterminate.

From the answers given, it’s clear that the second expression was supposed to be $20a+5b$; since this is $5(4a+b)$, the intended correct answer must be $15$.

  • 0
    Ok thanks, That sounds about right.2012-04-17
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You got 1 equation with 2 unknowns and you're trying to find the first one (finding 20a+5 requires a). Therefore it's unsolvable.

  • 0
    Thanks!, thats what I thought.2012-04-17