If three times the multiplicative inverse of a number is added to the number, the result is 13/2. Find the number?

x + 3/x = 13/2
2x + 6/x = 13
2x^2 + 6 = 13x
2x^2 - 13x + 6 = 0
x = (13 +/- sqrt(169 - 48)) / 4
x = (13 +/- sqrt(121)) / 4
x = (13 +/- 11) / 4
x = 24/4 , 2/4
x = 6 , 1/2

6 + 3/6 = 6 + 1/2 = 13/2

1/2 + 3/(1/2) = 1/2 + 3 * 2 = 1/2 + 6 = 13/2

6 and 1/2 both fit the bill

translation:
3 * (1/x) + x = 13/2

multiply by x
3 + x^2 = 13/2 x

rearrange
x^2 -13/2 x +3 = 0

use QF or complete the square:
x = -b/2a ±√(b² -4ac)/2a
(written as the vertex ± intercept offsets)

x=1/2, x=6

Let this number be x.

The multiplicative inverse is 1/x since (1/x)(x) = 1

Three times the multiplicative inverse is 3/x

We are given that 3/x + x = 13/2
We need to solve this equation.

Multiply throughout by x;
3/x(x) + x(x) = 13/2(x)
=> 3 + x^2 = 13x/2

Multiply throughout by 2;\
6 + 2x^2 = 13x

Rearrange so that the equation is in the form ax^2 + bx + c = 0
2x^2 - 13x + 6 = 0

Factorize;
2x^2 -12x - x +6 = 0
2x(x -6) -1(x - 6) = 0
(2x - 1) (x - 6) = 0

=> x = 1/2 or x = 6

The number is therefore either 1/2 or 6

3 * 1/x + x = 13/2
3/x + x = 13/2
3 + x^2 = 13x/2
x^2 - 13x/2 + 3 = 0
2x^2 - 13x + 6 = 0
(2x - 1)(x - 6) = 0
x = 1/2 or x = 6