Question

please tell me accordingly cause I am in 6th
thankss​

Answer 1

Answer:

let x be the first no.

and the second no. differs by 6

so second no. is x+6

now ATQ

x×(x+6) = 27

x^2 +6x = 27

x^2 + 6x+9= 27+9 (adding 9 to both sides)

(x+3)^2 = 36 ( using identity (a+b)^2 = a^2+b^2+2ab)

x+3 = +-6

the value of x is 6-3 and -6-3

x = 3

x= -9

the value of first and second no.s are

when x is 3

first no. is 3 and second no. is 9

when x is -9

first no. is -9 and second no. is -3

verifying we get 3×9 = 27

and -9×-3 = 27

wait have you learned quadratic equations yet no I don't they teach you that in 6th omg did I just waste my time and you did not get it at all

Answer 2

Step-by-step explanation:

Let’s assign variables to the two numbers: x and y.

Now, we can create equations: y=x+6 and xy = 27.

We can use substitution to get x(x+6)=27. Expanding, we get x^2 + 6x - 27 = 0.

We can factor this quadratic to get (x+9)(x-3)=0. x=3 and x=-9.

Plugging x=3 back into xy=27, we get y=9. We must check our answer with the equation y=x+6, and find that x=3 y=9 is indeed a valid solution.

We can do the same for x=-9. Plugging x=-9 back into xy=27, we get y=-3. Plugging this back into y=x+6, we find that -3 does indeed equal -9+6.

There are two sets of solutions: x=-9, y=-3 and x=3, y=9.