Math, asked by Anonymous, 1 year ago

Hey Mates....
PLS ANSWER➡➡

The difference of two numbers is 5 and the difference of their reciprocal is 1/10. Find the numbers.....
.
.
NO SPAM ___❌❌
✔✔


vineet5364: let x=10,y=5then. check out x-y=5means. 10-5=5 So x=10 y=5
Anonymous: ok ok i got the ans u are late

Answers

Answered by Itsmagix123
142

Answer:

Let the two number be x and x+5.

We have, 1/x-1/x+5=1/10

x+5-x/(x)(x+5)=1/10

5/x^2+5x=1/10

5×10=x^2+5x

50=x^2+5x

x^2+5x-50=0

x^2+10x-5x-50=0

x(x+10)-5(x+10)=0

(x-5)(x+10)=0

Therefore x-5=0 Or x+10=0

Hence, x=5,-10

Now as per question, given that the two numbers are natural nos.

Hence x=5

Second no.=5+5=10

I hope the helps you!!


Anonymous: And pls do not perform it again
akshay200123: I don't believe a ....
Anonymous: I said no comments
Anonymous: u don't understand
Near954: hi
Near954: hi
Anonymous: who
RuhiKhurana: stop commenting ✋
Anonymous: u stop
Anonymous: its my question
Answered by fiercespartan
174

Answer:

x = 10, y = 5

Step-by-step explanation:

First, let's start with what a reciprocal is.

A reciprocal is simply the reverse of the number. The numerator switches with the denominator. In mathematical terms, a reciprocal is 1 divided by the number.

We are now given two cases. The difference of the numbers and the difference of the reciprocals.

Let's take the first number as 'x' and the second number as 'y'

First case → x - y = 5

Reciprocal of 'x' is 1/x and the reciprocal of 'y' is 1/y

Second case → ( 1 / y ) - ( 1 / x ) = 1 / 10

Because, in the first equation, we get a positive output which is 5. This means that x > y.

In the second equation too we get positive solution that means the greater number goes first. But remember that we are taking is reciprocals. Hence, in the second equation 1/y > 1/x.

So, 1/y goes first.

Now, let's solve these equations.

we know that x - y = 5

Then, x = y + 5

(1 / y) - (1 / x) = 1/10

(1 / y) - (1 / y+5) = 1/10

The LCM would be y² + 5y

This would become :

\frac{(y+5)-y}{y^{2}+5y}

\frac{5}{y^{2}+5y}=\frac{1}{10}

y² + 5y = 50

y² + 5y - 50 = 0

y - 5y + 10y - 50 = 0

y( y - 5 ) + 10( y - 5 ) = 0

( y + 10 )( y - 5 ) 0

y = -10 and y = 5

There will be two solutions.

Let's take y = -10

x - y = 5

x - (-10) = 5

x + 10 = 5

x = -5                      First solution : (-5,-10)

But if we substitute these values in the second equation, it wouldn't work. This is considered as an extraneous solution. THIS IS NOT THE SOLUTION.

Now, let's take y = 5

x - y = 5

x - 5 = 5

x = 10                          Second solution : (10,5)


pretam20: can you join in my side
devanshi1560: to whom u r asking bro
pretam20: to jeny34
jeny34: Yes ask
dhan67: fu..
intelligentlearners: hey
jeny34: What
Anonymous: All of u...Stop your RUBBISH comments
Anonymous: Don't comment in My question
Anonymous: Comment elsewhere or U will get reported..!!
Similar questions