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algebra

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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
+47
+1

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

$$\sqrt{xy} =3$$

$$xy = 9$$

$$x = {9\over y}$$

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

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

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

$${9+y^2} = 10y$$

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

$$y = 1$$

$$x = 9$$

$$y = 9$$

$$x = 1$$

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May 31, 2020