# Thread: If this = 0 then that = ?

1. ## If this = 0 then that = ?

Hello.

I'll get right to the point:

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

The solution is $-1$, 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 $(2a-b-3)^2+(3a+b-7)^2=0$ THEN $3a-7b=?$

The solution is $-1$, but I need to know how they got that so I can tackle problems like this in the future by myself.
note that $(anything)^2 \ge 0$

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

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

... take it from here?

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

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

$-b=-2a+3$

$b=2a-3$

$3a+b-7=0$

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

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

$a=2$

$2a-b-3=0$

$4-b-3=0$

$-b=-4+3$

$-b=-1$

$b=1$

$a=2$

$b=1$

$3a-7b=?$

$3*2-7*1=?$

$6-7=-1$

$3a-7b=-1$

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

$2a-b-3 = 0$

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

$a = 2$ , and so on

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

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

-Dan