The sum of 2 number is 8 .the sum of its resiprocal is 1/4 find the numbers

5 and 20

Step-by-step explanation:

Let one of the numbers be x, and let the other number be y.

Since the sum of the two numbers is 25, we can write the following equation:  

x + y = 25

Since the sum of their multiplicative inverses (reciprocals) is 1/4, we can also write this equation:  

1/x + 1/y = 1/4

Since we have two equations in two unknowns, we can solve for x and y. We will use the substitution method to solve these two simultaneous equations.

Using the first equation, we’ll solve for y in terms of x:

y = 25 ‒ x.  

Now, substituting this result into the second equation, we get:

1/x + 1/(25 ‒ x) = 1/4. For this fractional equation, we can see by inspection that the lowest or least common denominator (LCD) is the product of all the different, individual denominators, i.e., 4(x)(25 ‒ x). Now, multiplying both sides of the fractional equation by the LCD in order to clear the equation of fractions, we get:

4(x)(25 ‒ x)[1/x + 1/(25 ‒ x)] = (1/4)[4(x)(25 ‒ x)]

4(x)(25 ‒ x)(1/x) + 4(x)(25 ‒ x)[1/(25 ‒ x)] = (1/4)[4(x)(25 ‒ x)]

4(x)(1/x)(25‒x) +4(x){(25 ‒ x)[1/(25 ‒ x)]}= x[(1/4)(4)](25‒x)]

4(1)(25‒x) + 4(x){1} = x[(1)](25‒x)]

4(25‒x) + 4x = x(25‒x)

100 ‒ 4x + 4x = 25x ‒ x²

100 + (4 ‒ 4)x = 25x ‒ x²

100 + (0)x = 25x ‒ x²

100 = 25x ‒ x²

Transposing everything to the left side of the equation, we get:

x² ‒ 25x + 100 = 0

Now, we conveniently have a quadratic equation in standard form, i.e., ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0.

By inspection, we see that we can easily solve this quadratic equation by factoring:

(x ‒ 5)(x ‒ 20) = 0

We know from a property of equality that the product ab = 0 if and only if a = 0 or b = 0; therefore, we can continue with our search for the solution to this problem by writing the following two equations and then solving for x for each:

x ‒ 5 = 0 or x ‒ 20 = 0

x ‒ 5 + 5 = 0 + 5 or x ‒ 20 + 20 = 0 + 20

x = 5 or x = 20

Now, solving for the other number y by substituting these two values for x as follows:

For x = 5:  

y = 25 ‒ x  

= 25 ‒ 5  

y = 20

For x = 20:

y = 25 ‒ x

= 25 ‒ 20

= 5

We see that y equals the same two numbers as x, but in reverse order from x! So, in terms of x and y, we have two solutions: x = 5, y = 20 and x = 20, y = 5.

CHECK:

For x = 5 and y = 20:

x + y = 25  

5 + 20 = 25  

25 = 25  

1/x + 1/y = 1/4

1/5 + 1/20 = 1/4

4/20 + 1/20 = 1/4

5/20 = 1/4

1/4 = 1/4

CHECK:

For x = 20 and y = 5:

x + y = 25  

20 + 5 = 25  

25 = 25  

1/x + 1/y = 1/4

1/20 + 1/5 = ¼

1/20 + 4/20 = 1/4

5/20 = 1/4

1/4 = 1/4

CONCLUSION: In reality, as we can see, there is really just one pair of numbers, regardless of with which variable, x or y, that each number in the pair is associated with, that satisfies the original conditions set forth by this problem; Those two numbers are, to answer the original question: “What are the two numbers?” : 5 and 20.

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