Respuesta :
Answer:
2 and 3
Step-by-step 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 [tex]\frac{7}{6}[/tex]
Question asked:
Then find the two integers.
Solution:
Let a positive integer is [tex]x[/tex]
Then another positive integer will be = [tex]x-1[/tex]
According to question,
Sum of the reciprocal of the smaller and twice the reciprocal of the larger is [tex]\frac{7}{6}[/tex]
Reciprocal of the smaller = [tex]\frac{1}{x-1}[/tex]
Twice the reciprocal of the larger = [tex]2\times\frac{1}{x}=\frac{2}{x}[/tex]
Now, equation will be:-
[tex]\frac{1}{x-1} +\frac{2}{x} =\frac{7}{6} \\ \\[/tex]
[tex]Taking\ LCM\ of\ x\ and\ x-1,\ we\ get,\ x(x-1)[/tex]
[tex]\frac{x+2(x-1)}{x(x-1)} =\frac{7}{6} \\\\ \frac{x+2x-2}{x^{2} -x} =\frac{7}{6} \\\\ \frac{3x-2}{x^{2}-x } =\frac{7}{6}[/tex]
By cross multiplication:
[tex]6(3x-2)=7(x^{2} -x)\\18x-12=7x^{2} -7x[/tex]
Adding both sides by [tex]7x[/tex]
[tex]25x-12=7x^{2} \ or\\7x^{2} -25x+12=0\\[/tex]
[tex]7x^{2} -21x-4x+12=0\\[/tex]
[tex]Taking\ common\\7x(x-3)-4(x-3)=0\\7x-4=0,x-3=0\\7x=4,x=3\\x=\frac{4}{7} ,x=3[/tex]
As we know that integer can never be a fraction but a whole number, hence [tex]x=3[/tex], the 1 integer and the other will be [tex]x-1[/tex] = 3 - 1 = 2
Therefore, two integers are 2 and 3.
Answer:
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 = [tex]x-1[/tex]
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 [tex]\frac76[/tex]
So we can say that;
[tex]\frac{1}{x-1}+\frac{2}{x}=\frac76[/tex]
Now we will make the denominator common using LCM.
[tex]\frac{x}{x(x-1)}+\frac{2(x-1)}{x(x-1)}=\frac76[/tex]
Now denominator are common so we will solve the numerator;
[tex]\frac{x+2x-2}{x(x-1)}=\frac76\\\\\frac{3x-2}{x(x-1)}=\frac76[/tex]
Using cross multiplication we get;
[tex]6(3x-2)=7x(x-1)\\\\18x-12=7x^2-7x\\\\7x^2-7x-18x+12=0\\\\7x^2-25x+12=0[/tex]
Now we will find the roots of the equation we get;
[tex]7x^2-21x-4x+12=0\\\\7x(x-4)-4(x-4)=0\\\\(7x-4)(x-3)=0[/tex]
Now we will solve separately to find the value of x we get;
[tex]7x-4=0 \ \ \ Or \ \ \ \ x-3=0\\\\7x=4 \ \ \ \ \ Or \ \ \ \ x=3\\\\x=\frac{4}{7}\ \ \ \ \ \ Or \ \ \ \ x=3[/tex]
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 = [tex]x=3[/tex]
Smaller number = [tex]x-1=3-1=2[/tex]
Hence the two integers are 2 and 3.
