Question

Q1.The denominator of a fraction is one more than twice it's numerator.if the sum of the fraction and it's reciprocal is 2 16/21,find the fraction.??c

Answer 1

Step-by-step explanation:

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

$2\frac{16}{21}$

it means,

$\begin{gathered}\frac{x}{y}+\frac{y}{x}=2\frac{16}{21}\\\;\\\frac{x}{y}+\frac{y}{x}=\frac{42+16}{21}\\\;\\\frac{x}{y}+\frac{y}{x}=\frac{58}{21}\\\;\\\frac{x^2+y^2}{xy}=\frac{58}[12}\\\;\\\frac{x^2+y^2-2xy+2xy}{xy}=\frac{58}{12}\\\;\\\frac{(y-x)^2+2xy}{xy}=\frac{58}{12}\\\;\\\frac{(x+1)^2+2xy}{xy}=\frac{58}{12}\;\;\;(using\;eq\;i)\\\;\\\end{gathered}$

Putting y = 2x+1 ,

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

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

Answer 2

Answer:

the person answered before me has given the right answer