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xy = 4 => y = 4/x => 1/y = x/4

1/x + 1/y = 65/56

1/x + x/4 = 65/56

Multiply both sides by 56x:

56 + 14x^2 = 65x

14x^2 - 65x + 56 = 0

[I chose to find Vertex Form and solve that.]

h = -b/2a = - -65/(2*14) = 65/28

k = c - ah^2 = 56 - 14(65/28)^2

= (56^2 - 65^2)/56 = (56-65)(56+65)/56 = -9(121)/56

x = h ± √(-k/a) = 65/28 ± √(- -9(121)/(56*14))

= 65/28 ± √(3^2(11^2)/(4*14^2))

= 65/28 ± 3*11/(2*14) = (65 ± 33)/28

x = 7/2 or 8/7

y = 4/x = 8/7 or 7/2

So the two numbers are 8/7 and 7/2.

check:

xy =? 4

(8/7)(7/2) =? 4

4 = 4 √

1/x + 1/y =? 65/56

1/(8/7) + 1/(7/2) =? 65/56

7/8 + 2/7 =? 65/56

7*7/(8*7) + 2*8/(7*8) =? 65/56

(49 + 16)/56 =? 65/56

65/56 = 65/56 √

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

XY = 4

1/ X + 1/ Y = 65/ 56

X = 4 / Y

Substitute in 2nd equation:

Y/ 4 + 1 / Y = 65 / 56

Y^2 + 4 = 65/ 56

4 Y

56 Y ^2 + 224 = 260 Y

14 Y ^2 - 65 Y - 56 = 0

14Y^2 - 16 Y - 49 Y - 56 =0

2 Y ( 7 Y - 8 ) - 7 ( 7Y - 8 ) =0

( 7Y - 8 ) ( 2Y - 7 ) = 0

7y - 8 = 0 Y = 8/7 X = 4 / ( 8/7) = 7/2

Test:

7/8 + 2 /7 = ( 49 + 16) /56 = 65/ 56

True:

2Y - 7 = 0 Y = 7/2 X = 8/7

This will work too, only X, y change their position.

Hi Willie,

Let twonumbers are x & y.

x*y=4...................Eq-1

1/x+1/y=65/56.....Eq-2

Now from Eq-1, y=4/x

substituting value of y in Eq-2,

we get,1/x+1/4/x=65/56

1/x+x/4=65/56

4+x^2=65/56*4x

x^2+4=65/14 *x

x^2-65x/14+4=0

14x^2-65x+56=0

14x^2-49x-16x+56=0

7x(2x-7)-8(2x-7)=0

(2x-7)(7x-8)=0

2x-7=0 or 7x-8=0

x=7/2 or x=8/7

y=4/x=4/7/2=8/7 or y=4/8/7=7/2.

Ans: two numbers are 7/2 & 8/7.

Meena from Strongsville,OH

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