The sum of a number and its reciprocal is 10 over 3. What is the number?

Answer 1

Let x be the number.

x + 1/x = 10/3
3x^2 + 3 = 10x
3x^2 -10x + 3 = 0
(3x - 1)(x - 3) = 0
x = 1/3, 3

So there are two solutions, 1/3 and 3.

Answer 2

Whenever we do word problems involving algebra, we first determine the standard of comparison; what is everything being compared to? In this case, it's a number. Therefore, we assign x to the number.

Let x = the number in question. Then
1/x = the number's reciprocal.

Translating the world problem, the sum of them is equal to 10/3. Therefore

x + 1/x = 10/3. We have to solve for x.

First, let's bring the 10/3 to the left hand side. This will give us

x + 1/x - 10/3 = 0

Now, put the terms under a common denominator. In this case, our common denominator will be 3x.

3x^2/3x + 3/3x - 10x/3x = 0

Now, we can merge them as a single fraction.

(3x^2 + 3 - 10x) / 3x = 0

Let's change the numerator to descending powers.

(3x^2 - 10x + 3) / 3x = 0

Now, we factor the numerator.

(3x - 1) (x - 3) / [3x] = 0

To solve this, we need to equate each of the factors in the numerator to 0.

3x - 1 = 0, x - 3 = 0
This will yield the result x = {1/3, 3}

Therefore, we have *two* numbers: 1/3, and 3. Both are equally valid responses, and it's not surprising that they are reciprocals of each other (since reciprocals of reciprocals is an inverse process).

Answer 3

1. the sum of a number and its reciprocal: "sum" means to add a number, "x", and its reciprocal which is 1/x. "Over" 3 means divide by 3:

x + 1/x = 10/3

First: get rid of fractions > take the denominators and multiply them by everything:

3x(x) + 3x(1/x) = 3x(10/3)

Second: cancel "like" terms and combine remaining terms:

3x^2 + 3 = 10x

Third: set the equation to equal "0" > subtract 10x from both sides:

3x^2 - 10x + 3 = 0

Fourth: factor > take the first coefficient and multiply it by the last coefficient (3*3 = 9). Find two numbers that give you (9) when multiplied and the middle coefficient (-10) when added/subtracted. The numbers are: (-1, -9)

3x^2 - 1x - 9x + 3 = 0
(3x^2 - 1x) - (9x + 3) = 0
x(3x - 1) - 3(3x - 1) = 0
(3x - 1)(x - 3) = 0

3x - 1 = 0
3x - 1 +1 = 0 + 1
3x = 1
3x/3 = 1/3 > x = 1/3

x - 3 = 0
x - 3 + 3 = 0 + 3 > x = 3

Answer 4

Let x be the number

x + 1/x = 10/3

or (x^2 + 1)/x = 10/3

Simplfying, 3x^2 - 10x + 3 = 0

Solving, x = 1/3 or 3

The number can either be 1/3 or 3.
But if the qn. states an integer then its 3.

Answer 5

Let x denote the number sought.

x + 1/x = 10/3
3x^2 + 3 = 10x [Multiplied equation by the G.C.D., 3x.]
3x^2 - 10x + 3 = 0
(3x - 1)(x - 3) = 0
x = 3 or x = 1/3.

Answer 6

So...if the number is y:

y + 1/y = 10/3
(y^2 + 1)/y = 10/3

Cross multiplication:
3(y^2+1) = 10y
3y^2 + 3 = 10y
3y^2 - 10y + 3 = 0

(3y - 1)(y - 3) = 0

Therefore y = 1/3 or y = 3