Question

The sum of the reciprocals of two consecutive even integers is 9/40. This can be represented by the equation shown.1/x + 1/x+2 = 9/40

Answer 1

Step-by-step explanation

`(1)/(x)+(1)/(x+2)=(9)/(40)`

Adding the two fractions:

`(x+2+x)/(x(x+2))=(9)/(40)`
`(2x+2)/(x^2+2x)=(9)/(40)`

We subtract 9/40 to equal zero:

`(2x+2)/(x^(^2)+2x)-(9)/(40)=0`

Subtracting the two fractions:

`(40(2x+2)-9(x^2+2x))/(40(x^2+2x))=0`
`(80x+80-9x^2-18x)/(40x^2+80x)=0`
`(-9x^2+62x+80)/(40x^2+80x)=0`

Factoring:

`(-(9x+10)(x-8))/(40x(x+2))=0`

This equation is zero when:

`\begin{gathered} 9x+10=0 \\ 9x=-10 \\ x=-(10)/(9) \end{gathered}`

or

`\begin{gathered} x-8=0 \\ x=8 \end{gathered}`

The only integer solution is x=8.

`\begin{gathered} (1)/(8)+(1)/(10)=(10+8)/(80)=(18)/(80)=(9)/(40) \\ \\ \frac{}{} \\ \end{gathered}`

Answer

x=8