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Solve for x and y if (x + y)/2 = 5 and sqrt(xy) = 3.

 May 31, 2020
 #1
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(x + y)/2 = 5

 

 

sqrt(xy) = 3

sqrt(xy)2=32                     Square both sides to eliminate the square root on the left side.

xy = 9

(xy)/x = 9/x                       Divide x on both sides.

y = 9/x

 

Now that you know y, you can solve for x by using the first equation given.

 

(x + y)/2 = 5

(x+(9/x))/2 = 5                   Substitute 9/x into the value of y.

(x+(9/x))/ 2 ⋅ 2 = 5 ⋅ 2        Multiply 2 on both sides to get rid of the denominator on the left side.

x+(9/x) = 10

x+(9/x) - x = 10 - x            Subtract x on both sides.

9/x = 10 - x

(9/x) ⋅ x = (10 - x) ⋅ x         Multiply x on both sides to get rid of the denominator once again.

9 = 10x - x2                       

9 - 9 = 10x -x2 -9               Subtract 9 to move everything onto one side.

-x2 +10x - 9 = 0                  Rewrite.

-1 (-x2 +10x - 9 = 0)           From here, we must factor to get the values of x. But first we must mulitply the equation by -1.

x2 - 10x + 9 = 0                  Factor.

(x+10) (x-1) = 0                  Now set each of the parenthesis to equal 0.

x + 10 = 0

x = -10

 

x - 1 = 0

x = 1

 May 31, 2020
 #2
avatar+47 
+1

\({(x+y)\over 2} =5\)

\( \sqrt{xy} =3\)

\(xy = 9\)

\(x = {9\over y}\)

\({({{9\over y}+y)}\over 2} = 5\)

\({({{9\over y}+y)}} =10\)

\({({{9\over y}+{y^2\over y})}} =10\)

\({9+y^2} = 10y\)

\((y-9)(y-1) = 0\)

 

 

\(y = 1\)

\(x = 9\)

 

 

\(y = 9\)

\(x = 1\)

.
 May 31, 2020

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