A positive integer is 1 less than another. If the

Ответ:

jrassicworld4ever

22.01.2022

The two integers are 2 and 3.

Step-by-step explanation:

Let the larger positive integer be 'x'.

Now given:

A positive integer is 1 less than another.

So Smaller integer will be = x-1

We need to find the two integers.

Solution:

Now given:

If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is $\frac76$

So we can say that;

$\frac{1}{x-1}+\frac{2}{x}=\frac76$

Now we will make the denominator common using LCM.

$\frac{x}{x(x-1)}+\frac{2(x-1)}{x(x-1)}=\frac76$

Now denominator are common so we will solve the numerator;

$\frac{x+2x-2}{x(x-1)}=\frac76\\\\\frac{3x-2}{x(x-1)}=\frac76$

Using cross multiplication we get;

$6(3x-2)=7x(x-1)\\\\18x-12=7x^2-7x\\\\7x^2-7x-18x+12=0\\\\7x^2-25x+12=0$

Now we will find the roots of the equation we get;

$7x^2-21x-4x+12=0\\\\7x(x-4)-4(x-4)=0\\\\(7x-4)(x-3)=0$

Now we will solve separately to find the value of x we get;

$7x-4=0 \ \ \ Or \ \ \ \ x-3=0\\\\7x=4 \ \ \ \ \ Or \ \ \ \ x=3\\\\x=\frac{4}{7}\ \ \ \ \ \ Or \ \ \ \ x=3$

Now we get 2 values 1 in fraction and 1 integer but given the two numbers are positive integer hence we will discard the fraction value we get;

Larger number = x=3

Smaller number = x-1=3-1=2

Hence the two integers are 2 and 3.

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