Question 118246: 1- Find the roots- 1/x+4 - 1/x-7= 11/30, x is not equal to -4,7.
2- Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24m, find the sides of the two squares.
3-The sum of the reciprocals of Rehman's ages(in years) 3 years ago and 5 years from now is 1/3. Find his present age.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1- Find the roots- 1/(x+4) - 1/(x-7)= 11/30, x is not equal to -4,7.
Multiply thru by 30(x+4)(x-7) to get:
30(x-7)-30(x+4) = 11(x-7)(x+4)
-210-120 = 11(x^2-3x-28)
-30 = x^2-3x-28
x^2-3x+2=0
(x-2)(x-1)=0
x=2 or x=1
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2- Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24m, find the sides of the two squares.
Let the side of one be "s"; Let the side of the other be "S"
EQUATIONS:
S^2+s^2 = 468
4S-4s = 24
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S=s+6
Substitute to get:
(s+6)^2+s^2 = 468
s^2+12s+36+s^2 = 468
2s^2+12s-32=0
s^2+6s-16=0
(s+8)(s-2)=0
Positive solution:
s = 2 m (side of one of the squares)
S = s+6 = 2+6 = 8 m (side of the other square)
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3-The sum of the reciprocals of Rehman's ages(in years) 3 years ago and 5 years from now is 1/3. Find his present age.
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Let his present age be "x"
EQUATION:
1/(x-3) + 1/(x+5) = 1/3
Multiply thru by 3(x-3)(x+5)
3(x+5) + 3(x-3) = (x-3)(x+5)
6x+6 = x^2+2x-15
x^2-4x-21 = 0
(x-7)(x+3)=0
Positive answer:
x = 7 (his current age)
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Cheers,
Stan H.
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