Find the two numbers with the difference of 3 between them and the difference of their squares is 51?

All the answers below are correct

so 7 and 10 or -7 and -10

Let numbers be x and x + 3

So (x + 3)^2 - x^2 = 51

x^2 + 6x + 9 - x^2 = 51

6x = 42

x = 7

x + 3 = 7 + 3 = 10

Numbers are 7 and 10

they are 10 and 7

because their difference is 3

the difference of there squares is= 100^2-7^2=100-49=51

7 -49

10 - 100

difference is 51

x - y = 3

x = 3 + y

x^2 - y^2 = 51

(3 + y)^2 - y^2 = 51

9 + 6y + y^2 - y^2 = 51

6y = 42

y = 7

x = 10

i) Let the two numbers be a & b, with a > b

ii) So by given data: a - b = 3 ----- (1) and

a² - b² = 51 ----- (2)

iii) Factoring the second one, (a + b)(a - b) = 51

==> a + b = 17 [Since from (1) a - b = 3] ---- (3)

iv) Solving (1) & (3): a = 10 and b = 7

x and x+3

so (x+3)^2 - x^2= 51

==> x^2 + 6x + 9 - x^2 = 51

==> 6x = 42

==> x = 7

so 7 and 10

... or....

-10 and -7

x -y =3

x^2 -2xy +y^2 = 9

x^2 - y^2 = 51

2 x^2 - 2xy = 60

2 x^2 - 2x(x-3) = 60

2 x^2 - 2 x^2 + 6x = 60

6 x = 60

x= 10 and y= 7

let the numbers be a and b

a-b=3

a*a-b*b=51

by solving we will get 7 and 10

simple simultaneous equations