How to solve sqrt(20-x)+6=sqrt(9-x)+9
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How to solve sqrt(20-x)+6=sqrt(9-x)+9

[From: ] [author: ] [Date: 12-11-29] [Hit: ]
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I know the answer is 80/9 (teacher), but I don't know how to do it. Please show steps and explain! Thank you so much!

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Isolate one of the square roots:

√(20 - x) = √(9 - x) + 3

Square both sides to cancel off the square root:

20 - x = [√(9 - x) + 3] [√(9 - x) + 3]

20 - x = 9 - x + 6√(9 - x) + 9

Combine the like terms:

20 - x = 18 - x + 6√(9 - x)

Add x to both sides:

20 = 18 + 6√(9 - x)

Subtract 18 from both sides:

2 = 6√(9 - x)

Divide both sides by 6:

1/3 = √(9 - x)

Square both sides to cancel off the square root:

1/9 = 9 - x

Subtract 9 from both sides:

-80/9 = -x

Divide both sides by -1:

x = 80/9

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sqrt(20-x)+6=sqrt(9-x)+9
sqrt(20-x) = sqrt(9-x) - 3 (square both side)
(sqrt(20-x))^2 = (sqrt(9-x) - 3)^2
20 - x = 9-x + 2 . 3 sqrt(9-x) + 9
20 - x = 18 - x + 6 sqrt(9-x)
2 = 6 sqrt(9-x)
sqrt(9-x) = 1/3 (square both side again)
9 - x = 1/9
x = 9 - 1/9 = 81/9 - 1/9 = 80/9
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