# Thread: If this = 0 then that = ?

1. ## 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.

2. Originally Posted by ikislav
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.
note that $\displaystyle (anything)^2 \ge 0$

so ... the only way that equation will equal zero is if

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

... take it from here?

3. Take $\displaystyle (2a-b-3)^2$ from both sides, square root both sides and then isolate $\displaystyle 3a-7b$ to one side.

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

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

$\displaystyle b=2a-3$

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

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

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

$\displaystyle a=2$

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

$\displaystyle 4-b-3=0$

$\displaystyle -b=-4+3$

$\displaystyle -b=-1$

$\displaystyle b=1$

$\displaystyle a=2$

$\displaystyle b=1$

$\displaystyle 3a-7b=?$

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

$\displaystyle 6-7=-1$

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

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

5. solving the system using substitution is fine ... elimination would also work

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

$\displaystyle 3a+b-7 = 0$
------------------
$\displaystyle 5a - 10 = 0$

$\displaystyle a = 2$ , and so on

6. Originally Posted by pickslides
Take $\displaystyle (2a-b-3)^2$ from both sides, square root both sides and then isolate $\displaystyle 3a-7b$ to one side.
Can you show me this method please? I'm very interrested.

7. Originally Posted by ikislav
Can you show me this method please? I'm very interrested.
You do not need to that.
Look carefully at replies #2 & #5. Follow those.

8. Yes, I already solved it, but I am stil interested in method mentioned by mr. pickslides.

9. Originally Posted by pickslides
Take $\displaystyle (2a-b-3)^2$ from both sides, square root both sides and then isolate $\displaystyle 3a-7b$ to one side.
$\displaystyle \displaystyle (2a - b - 3)^2 + (3a + b - 7)^2 = 0$

$\displaystyle \displaystyle (3a + b - 7)^2 = -(2a - b - 3)^2$
And you can't take the square root. (Pesky little negative sign!)

-Dan