Question

The difference between two integers is 5. If the multiplicative inverse of the smaller is added to twice the multiplicative inverse of the larger the result is 23/66. What are the two integers?

Answer 1

The multiplicative inverse is the reciprocal.
Let the smaller integer = x
Let the larger integer = y
y - x = 5
Therefore y = 5 - x ...............(1)
\[\frac{1}{x}+\frac{2}{y}=\frac{23}{66}\ ............(2)\]
To find the value of x, substitute the value of y given in equation (1) into equation (2) and solve.

Answer 2

I've gotten that far, but I am having trouble after this point.

Answer 3

Please post what you have so far.

Answer 4

(1/x-5)+(2/x)=23/66
then I try to find the common denominator which would be 66(x-5)?

Answer 5

\[\frac{1}{x}+\frac{2}{5-x}=\frac{23}{66}\]
The common denominator for the left hand side is x(5-x)
\[\frac{5-x+2x}{x(5-x)}=\frac{23}{66}\ ..........(3)\]
Now you can cross multiply equation (3) to get rid of the fractions.

Answer 6

So now I have 330=49x-23x^2 is this correct so far?

Answer 7

Good work! So now we have a quadratic to solve:
\[23x ^{2}-49x+330=0\ ...........(4)\]

Answer 8

Am I able to just plug this into the quadratic formula?

Answer 9

That's right :)

Answer 10

Is the question right. I get a negative disciminant when I try to solve (4).

Answer 11

*discriminant

Answer 12

I got: \[x=\frac{ -49\pm i \sqrt{27959} }{ -46 } \]

Answer 13

I just checked and the answer is actually 11 and 6

Answer 14

My result is not the same. However we both get imaginary solutions rather than an integer. Please check the question again. Do you have answer choices given or a given answer?

Answer 15

The given answers are 11 and 6

Answer 16

Sorry. My bad. Equation (1) should have been y = x + 5
y = x + 5 .................(1)
\[\frac{1}{x}+\frac{2}{y}=\frac{23}{66}\ ...........(2)\]
Sustituting (1) in (2) gives:
\[\frac{1}{x}+\frac{2}{5+x}=\frac{23}{66}\]
The quadratic then becomes:
\[23x ^{2}-83x-330=0\ ......(3)\]
Solving (3) using the quadratic solution formula gives the correct answer fo x (the smaller integer.

Answer 17

Ok thank you for your help!

Answer 18

\[\frac{1}{x}+\frac{2}{5+x}=\frac{5+x+2x}{x(5+x)}=\frac{5+3x}{5x+x ^{2}}=\frac{23}{66}\ .........(4)\]
Cross multiplying (4) gives:
\[330+198x=115x+23x ^{2}\ ..........(5)\]
Rearranging (5) gives the quadratic:
\[23x ^{2}-83x-330=0\ .........(6)\]

Answer 19

You're welcome :)