5 two natural numbers differ by 3 find the numbers if the sum of their reciprocals is 7

How about we believe the two normal numbers to be\[x\text{ }and\text{ }x\text{ }+\text{ }3\] . (As they vary by 3)

now,

\[20x\text{ }+\text{ }30\text{ }=\text{ }7×2\text{ }+\text{ }21x\]

\[7×2\text{ }+\text{ }x\text{ }\text{ }30\text{ }=\text{ }0\]

\[7×2\text{ }\text{ }14x\text{ }+\text{ }15x\text{ }\text{ }30\text{ }=\text{ }0\]

\[7x\left( x\text{ }\text{ }2 \right)\text{ }+\text{ }15\left( x\text{ }\text{ }2 \right)\text{ }=\text{ }0\]

\[\left( 7x\text{ }+\text{ }15 \right)\text{ }\left( x\text{ }\text{ }2 \right)\text{ }=\text{ }0\]

In this way, \[7x\text{ }+\text{ }15\text{ }=\text{ }0\text{ }or\text{ }x\text{ }\text{ }2\text{ }=\text{ }0\]

\[x\text{ }=\text{ }-\text{ }15/7\text{ }or\text{ }x\text{ }=\text{ }2\]

As, x is a characteristic number. Just \[x\text{ }=\text{ }2\] is a legitimate arrangement.

Subsequently, the two normal numbers are \[2\text{ }and\text{ }5.\]

To solve this system, let's use the method of substitution.
From the first equation we can express #x# in terms of #y# and substitute into the second equation.

From equation (1) we can derive:
(3) #x = y+3#

Substitute it into equation (2):
(4) #1/(y+3) + 1/y = 7/10# Incidentally this necessitates another restriction:

#y+3!=0#, that is #y!=-3#.

Using common denominator #10y(y+3)# and considering only numerators, we transform equation (4) into:
#10y+10(y+3)=7y(y+3)#

This is a quadratic equation that can be rewritten as:
#20y+30=7y^2+21y# or
#7y^2+y-30=0#

Two solutions to this equation are:
#y_(1,2)=(-1+-sqrt(1+840))/14# or

#y_(1,2)=(-1+-29)/14#

So, we have two solutions for #y#:
#y_1=2# and #y_2=-30/14=-15/7#

Correspondingly, using #x=y+3#, we conclude that there are two solutions to a system:
#(x_1, y_1)=(5,2)#
#(x_2, y_2)=(6/7,-15/7)#

In both cases #x# is greater than #y# by #3#, so the first condition of a problem is satisfied. Let's check the second condition:

(a) for a solution #(x_1, y_1)=(5,2)#:

Both solutions are correct.

Page 2

Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is 7/10.

Let the numbers be x and x + 3

From the given information

`1/x + 1/(x + 3) = 7/10`

`(x + 3  + x)/(x(x + 3)) = 7/10`

`(2x + 30)/(x^2 + 3x) = 7/10`

`20x + 30 = 7x^2 + 21x`

`7x^2 + x - 30 = 0`

`7x^2 - 14x + 15x - 30 = 0`

7x(x - 2)  + 15(x - 2) = 0

(x - 2)(7x + 15) = 0

`x = 2, (-15)/7`

Since x is a natural number so x= 2

Thus the number are 2 and 5

Concept: Quadratic Equations

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