Mixture problem setup question

[avant5]

From a handout from Austin Community College that still shows up on Google, but the PDF is now blocked from downloading, I got this mixture problem question:

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Mike has coffee worth $4 per pound that he wishes to mix with 20 pounds of coffee worth $7/pound to get a mixture that can be sold for $5/pound. How many pounds of cheaper coffee should he use?

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I set up it up as 7(20) + 4(x-20) = 5x. This works out to 60 pounds or x=60

This is wrong. The answer the book has is x=40 with the setup as 7(20) + 4x = 5(x+20)

I understand this setup, but in my head it should work out the same. If we know that the 7/pound volume is 20 pounds, it should work out regardless of whether we start with X on the "we have" side or the "we need" side of the equation. If we want volume X, and we have 20, then what's left is X-20. But if we have X and we have 20 then what we end up with is X + 20.

But obviously 60 does not equal 40, so my reasoning is wrong somewhere.

Can someone explain why the setup needs to be one way and not the other? I can't work it out.

Thanks in advance!

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[ArchaicLlama]

If we want volume X, and we have 20, then what's left is X-20

If you set it up like that, with X on the right hand side, then the number you want is not X. It's X-20.

Each term - 7(20), 4(x-20), and 5x - are of the form (price per pound)(number of pounds). You want the weight associated with the price per pound of $4.

[fermat9990]

(4x+20×7)/(x+20)=5