Hello.

I'll get right to the point:

IF $\displaystyle (2a-b-3)^2+(3a+b-7)^2=0$ THEN $\displaystyle 3a-7b=?$

The solution is $\displaystyle -1$, but I need to know how they got that so I can tackle problems like this in the future by myself.

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- Mar 13th 2011, 12:33 PM #1

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## If this = 0 then that = ?

Hello.

I'll get right to the point:

IF $\displaystyle (2a-b-3)^2+(3a+b-7)^2=0$ THEN $\displaystyle 3a-7b=?$

The solution is $\displaystyle -1$, but I need to know how they got that so I can tackle problems like this in the future by myself.

- Mar 13th 2011, 12:45 PM #2

- Mar 13th 2011, 12:46 PM #3

- Mar 13th 2011, 01:03 PM #4

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$\displaystyle 2a-b-3=0$

$\displaystyle -b=-2a+3$

$\displaystyle b=2a-3$

$\displaystyle 3a+b-7=0$

$\displaystyle 3a+2a-3-7=0$

$\displaystyle a=\frac{10}{5}$

$\displaystyle a=2$

$\displaystyle 2a-b-3=0$

$\displaystyle 4-b-3=0$

$\displaystyle -b=-4+3$

$\displaystyle -b=-1$

$\displaystyle b=1$

$\displaystyle a=2$

$\displaystyle b=1$

$\displaystyle 3a-7b=?$

$\displaystyle 3*2-7*1=?$

$\displaystyle 6-7=-1$

$\displaystyle 3a-7b=-1$

Yes, that is the correct answer. Is that the right path for solving this after the hint you gave me?

- Mar 13th 2011, 01:13 PM #5

- Mar 13th 2011, 01:17 PM #6

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- Mar 13th 2011, 01:23 PM #7

- Mar 13th 2011, 01:29 PM #8

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- Mar 13th 2011, 05:08 PM #9