Calculating 3a-7b: Solving the Expression (2a-b-3)^2 + (3a+b-7)^2 = 0

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The equation (2a-b-3)² + (3a+b-7)² = 0 implies that both components must equal zero, leading to the system of equations 2a - b - 3 = 0 and 3a + b - 7 = 0. Solving these equations simultaneously reveals the values of a and b. Once a and b are determined, the expression 3a - 7b can be calculated. The discussion highlights the importance of recognizing that both squared terms must equal zero for real numbers. Ultimately, the solution hinges on solving the derived system of equations.
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If a and b are real numbers, and we know that (2a-b-3)^{2} + (3a+b-7)^{2}=0, how much is 3a-7b

Any ideas on this? I'm guessing the solution can go two ways: either i find a and b separately, or i calculate the expression above somehow and i'll be left with 3a-7b
 
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Chuckster said:
If a and b are real numbers, and we know that (2a-b-3)^{2} + (3a+b-7)^{2}=0, how much is 3a-7b

Any ideas on this? I'm guessing the solution can go two ways: either i find a and b separately, or i calculate the expression above somehow and i'll be left with 3a-7b

You're answering your own question. :smile:

Since you would have 1 equation and 2 unknowns, there can be many solutions depending on what cancels out. Your best bet is to multiply stuff out and see what happens.
 
gb7nash said:
You're answering your own question. :smile:

Since you would have 1 equation and 2 unknowns, there can be many solutions depending on what cancels out. Your best bet is to multiply stuff out and see what happens.

The key word would be somehow. If multiplication worked, i wouldn't be asking how :).

Anyway, i get a bunch of junk, and i can't seem to figure out what to do with it. How to create or find 3a-7b, that is...
 
Ahh, I see what's going on.

At first glance, I would go with multiplication. However, there's something special about this equation. I'll replace it with this equation:

c2 + d2 = 0, where c and d are real numbers.

What's the only way this could happen? What must c and d be equal to?
 
gb7nash said:
Ahh, I see what's going on.

At first glance, I would go with multiplication. However, there's something special about this equation. I'll replace it with this equation:

c2 + d2 = 0, where c and d are real numbers.

What's the only way this could happen? What must c and d be equal to?

It's possible if c and d are both equal zero!
Then i just solve the system of 2 equations with 2 unknowns.

THANKS gb7nash!

(hate it when i miss obvious catches)...
 
You got it. :smile:

It's ok, I missed it too.
 
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