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A positive integer is 1 less than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is frac(7,6), then find the two integers

User BigChief
by
2.9k points

2 Answers

1 vote

Answer:

The two integers are 2 and 3.

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;


(1)/(x-1)+(2)/(x)=\frac76

Now we will make the denominator common using LCM.


(x)/(x(x-1))+(2(x-1))/(x(x-1))=\frac76

Now denominator are common so we will solve the numerator;


(x+2x-2)/(x(x-1))=\frac76\\\\(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=(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.

User N Fard
by
3.0k points
1 vote

Answer:

2 and 3

Explanation:

A positive integer is 1 less than another.

If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is
(7)/(6)

Question asked:

Then find the two integers.

Solution:

Let a positive integer is
x

Then another positive integer will be =
x-1

According to question,

Sum of the reciprocal of the smaller and twice the reciprocal of the larger is
(7)/(6)

Reciprocal of the smaller =
(1)/(x-1)

Twice the reciprocal of the larger =
2*(1)/(x)=(2)/(x)

Now, equation will be:-


(1)/(x-1) +(2)/(x) =(7)/(6) \\ \\


Taking\ LCM\ of\ x\ and\ x-1,\ we\ get,\ x(x-1)


(x+2(x-1))/(x(x-1)) =(7)/(6) \\\\ (x+2x-2)/(x^(2) -x) =(7)/(6) \\\\ (3x-2)/(x^(2)-x ) =(7)/(6)

By cross multiplication:


6(3x-2)=7(x^(2) -x)\\18x-12=7x^(2) -7x

Adding both sides by
7x


25x-12=7x^(2) \ or\\7x^(2) -25x+12=0\\


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


Taking\ common\\7x(x-3)-4(x-3)=0\\7x-4=0,x-3=0\\7x=4,x=3\\x=(4)/(7) ,x=3

As we know that integer can never be a fraction but a whole number, hence
x=3, the 1 integer and the other will be
x-1 = 3 - 1 = 2

Therefore, two integers are 2 and 3.

User Anthony Roberts
by
3.7k points