Question

The denominator of a fraction is one more than twice the numerator.if the sum of fraction and its reciprocal is 58/21. Find the numbers.?

Answer 1

Hey!

Let the fraction be x/y.
As given in the question,
y=2x+1.

Fraction=x/y= x/2x+1.

x/y + y/x = 58/21.
(x^2+y^2)/xy = 58/21.
21(x^2+y^2)=58(xy).
21(x^2+(2x+1)^2=58(x(2x+1)).
21(x^2+4x^2+1+4x)=58(2x^2+x).
21(5x^2+4x+1)=116x^2+58x.
105x^2+84x+21= 116x^2+58x.
116x^2-105x^2+58x-84x-21=0.
11x^2 - 26x -21 =0.
11x^2 -33x+7x-21=0.
11x(x-3)+7(x-3)=0.
(11x+7)(x-3)=0.

x-3=0.
x=3.

So Fraction= x/y where y= 2x+1.
x=3, y= 2x+1=2×3+1=7.
So the fraction is 3/7.

Hope it helps.

Answer 2

Answer: 3/7

Step-by-step explanation:

Let the numerator  be  x and  denominator  be  y

According  to  first  condition, denominator of the fraction is one more than twice it's numerator.it  means,

y = 2x + 1

y - x = x +1  ................i)

According to second  condition,

sum of the fraction and it's reciprocal is $\frac{58}{21}$

it means,

$\frac{x}{y}+\frac{y}{x}=\frac{58}{21}\\\;\\\frac{x^2+y^2}{xy}=\frac{58}[12}\\\;\\\frac{x^2+y^2-2xy+2xy}{xy}=\frac{58}{12}\\\;\\\frac{(y-x)^2+2xy}{xy}=\frac{58}{12}\\\;\\\frac{(x+1)^2+2xy}{xy}=\frac{58}{12}\;\;\;(using\;eq\;i)\\\;$

Putting y = 2x+1 ,

$\frac{(x+1)^2+2x(2x+1)}{x(2x+1)}=\frac{58}{12}\\\;\\\frac{(x+1)^2}{x(2x+1)}+\frac{2x(2x+1)}{x(2x+1)}=\frac{58}{12}\\\;\\\frac{x^2+1+2x}{2x^2+x}+2=\frac{58}{21}\\\;\\\frac{x^2+1+2x}{2x^2+x}=\frac{58}{21}-2\\\;\\\frac{x^2+1+2x}{2x^2+x}=\frac{58-42}{21}\\\;\\\frac{x^2+1+2x}{2x^2+x}=\frac{16}{21}\\\;\\21x^2+21+42x=32x^2+16x\\\;\\0=32x^2-21x^2+16x-42x-21\\\;\\0=11x^2-26x-21\\\;\\11x^2-26x-21=0\\\;\\11x^2-33x+7x-21=0\\\;\\11x(x-3)+7(x-3)=0\\\;\\(x-3)(11x+7)=0$

$x=3\;\;or\;\;x=-\frac{7}{11}(Negligible)\\\;\\when\;x=3\;\;from\;eq.\;i)\\\;\\y=2\times3+1\\\;\\y=6+1\\\;\\y=7\\\;\\Hence\\\;\\\text{fraction is}=\frac{3}{7}$