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Solve for y

√y- 8 + y=4

Feb 22, 2018

#1
+2297
0

Let's start off by Subtracting -8+y from each side which would now make this: √y^2 - 8 + y=4 into this: √y^2=-y+12

(I'm horrible at explaining things, sorry!)

Now we square both sides, (√y^2)^2=(-y+12)^2

First let's do (√y^2)^2

√y=y^1/2 so let's replace the square root with 1/2 now we have (y^2 and 1/2)^2

Let's apply the exponent rule: (a^b)^c=a^bc which makes us have y^2 and 1/2*2 which makes it A LOT easier to solve this problem. 1/2*2=1

1*2/2=2/2=1, we end up with y^2

Now we have to solve (-y+12)^2

Apply the distributing rule: (a+b)^2=a^2+2ab+b^2 (a=-y and b=12)

We now have (-y)^2 +2(-y)*12+12^2 (SIMPLIFY)

Remove the parenthesis. y^2 -2y*12+12^2

(-a)^n=a^n (only if n is even)

(-y^2)=y^2

y^2-2*12y+12^2

Refine: y^2-24y+144

Now we have y^2=y^2-24y+144, so we have to solve that now.

Subtract y^2 from both sides and we get: 0=-24y+144

Let's subtract 144 from both sides which= -24y=-144

Divide both sides by -24: -24y/-24=y and -144/-24=6

So, y=6

I just made it really confusing xD but yeah that's it.

Feb 22, 2018
#2
+3729
+2

Hey, Manuel!

We can first subtract $$-8+y$$ from both sides, to get $$\sqrt{y^2}-8+y-(-8+y)=4-(-8+y), \sqrt{y^2}=-y+12$$. After that, we can square both sides, expand, and expand again, to get $$y^2=y^2-24y+144$$ . Solve, to get $$y=6$$

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Feb 22, 2018
#3
+1397
+1

Thanks All of you

Feb 22, 2018