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: Β
x + y = 25
Since the sum of their multiplicative inverses (reciprocals) is 1/4, we can also write this equation: Β
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, weβll solve for y in terms of x:
y = 25 β x. Β
Now, substituting this result into the second equation, we get:
1/x + 1/(25 β 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 β 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 β x)[1/x + 1/(25 β x)] = (1/4)[4(x)(25 β x)]
4(x)(25 β x)(1/x) + 4(x)(25 β x)[1/(25 β x)] = (1/4)[4(x)(25 β x)]
4(x)(1/x)(25βx) +4(x){(25 β x)[1/(25 β x)]}= x[(1/4)(4)](25βx)]
4(1)(25βx) + 4(x){1} = x[(1)](25βx)]
4(25βx) + 4x = x(25βx)
100 β 4x + 4x = 25x β xΒ²
100 + (4 β 4)x = 25x β xΒ²
100 + (0)x = 25x β xΒ²
100 = 25x β xΒ²
Transposing everything to the left side of the equation, we get:
xΒ² β 25x + 100 = 0
Now, we conveniently have a quadratic equation in standard form, i.e., axΒ² + bx + c = 0, where a, b, and c are real numbers and a β 0.
By inspection, we see that we can easily solve this quadratic equation by factoring:
(x β 5)(x β 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 β 5 = 0 or x β 20 = 0
x β 5 + 5 = 0 + 5 or x β 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: Β
y = 25 β x Β
= 25 β 5 Β
y = 20
For x = 20:
y = 25 β x
= 25 β 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 Β
5 + 20 = 25 Β
25 = 25 Β
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 Β
20 + 5 = 25 Β
25 = 25 Β
1/x + 1/y = 1/4
1/20 + 1/5 = ΒΌ
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: βWhat are the two numbers?β : 5 and 20.