Hi, can you please explain how to solve this? I dont understand why the answer cannot be 21 and 25?
What values of A and B would get you 4A + 7B = 25?
If B=3 and a=1. But that would mean that 2a+3b=10 is not satisfied. I get that. But is trial and error the only way to do this question? it isn’t 18 because 4a+6b is 20. So 4a+7b has to be more than 20. But the question does not even specify that a and be are integers. So how do we go about solving it?
The solution is not given… that’s the reason i’m asking…
4A + 7B = \underbrace{4A + 6B}_{20} + B = 20 + B
Can you find the range of B and hence the range of 4A + 7B ? As a hint, you should look at the first equation to figure out the range of B.
Yes, because they need not be integers.
Got it. B should be less than 3 as per the first equation… so it cannot be 5. Thank you!
B must be less than \frac {10}{3} because, as we mentioned, the prices need not be integers. But yes, besides that, your subsequent idea of showing that B = 5 lies outside the interval of possible B’s is correct
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