Solve Algebra: 16+(y-6)^2=9+(y-5)^2

[DB]

When

$$\sqrt{16 + (y-6)^2}=\sqrt{9+(y-5)^2}$$

is simplified to:

$$16+(y-6)^2=9+(y-5)^2$$

How come the 16 and 9 arent square rooted?

Thanks

 

[Muzza]

Do you agree that $\sqrt{a} = \sqrt{b}$ implies a = b?

If so, what happens when you set a = 16 + (y - 6)^2 and b = 9 + (y - 5)^2?

 

[DB]

ahhh, o you its because they are equal, i got mixed up thanks.
But $$\sqrt{16 + (y-6)^2}$$
doesn't simplify to $$16+(y-6)^2$$ right?

 

[Muzza]

Correct. It would have if the expression was $\sqrt{ (16 + (y - 6)^2)^2 }$ though ;)

 

[DB]

ya thanks muzza

 

[HallsofIvy]

If you square both sides of the equation,
$$\sqrt{16- (y-6)^2}= \sqrt{9- (y-5)^2}$$
you eliminate the two square roots and you get
16- (y-6)²= 9- (y-5)².