Calvinism

$Let\quad the\quad total\quad number\quad of\quad coins\quad be\quad m.\\ Let\quad it\quad be\quad divided\quad into\quad 2\quad unequal\quad parts,\quad x\quad and\quad y,\\ where\quad x>y\\ Given,\\ 48(x-y)\quad =\quad { x }^{ 2 }-y^{ 2 }\\ \boxed { By\quad identity\quad { a }^{ 2 }-{ b }^{ 2 }\quad =\quad (a+b)(a-b) } \\ 48(x-y)\quad =\quad (x+y)(x-y)\\ \quad \quad \quad 48\quad =\quad (x+y)\\ Thus\quad m\quad =\quad 48\quad coins.$

-Let the number of coins be x. x is divided into two different numbers so let them be y and z.

We know that 48 times(X) the difference between the two numbers (y and z) is equal to the difference between the square of the two numbers (y and z). Which then you can come up with this equation:

48(y - z) = y square - z square

You should know a2 - b2 = (a+b) (a-b)

Then you can change y2 - z2 to (y+z) (y-z), which you can cancel out the (y-x) on both side, which leaves you:

48 = y + z

We know that y + z is the sum of the total number of coins, y + z = x

Therefore x = total number of coins = 48

let the unequal division be x & y; So, 48(x-y)=x^2-y^2; 48(x-y)=(x+y)(x-y); x+y=48

let no. of gold coins------>x when, divided into 2 unequal parts------>n,x-n(assume) by given, 48(x-n-n)=(x-n)^2-n^2 48(x-2n) =x(x-2n) (use a^2-b^2=(a-b)(a+b) ) dividing both the sides by (x-2n), 48 =x therefore, the coins he has is 48.

let x and y be the numbers and x+y is the total number of coins, obviously
48(x-y) = x2-y2
48(x-y) = (x-y) (x+y)
x+y=48

The wording is a bit misleading but it refers to the difference of two perfect squares i.e. x^2 - y^2 = (x+y)(x-y) = 48 (x-y) which gives >> (x+y)=48

Let the two unequal parts be x & y. As per the question, 48(x-y)=( x^2 - y^2 ) . This implies that 48(x-y)=(x+y)(x-y) Now by cancelling out the like terms i.e. (x-y), we get x+y=48, which is the total no. of coins.

48(x-y)=x^2-y^2 x=y=48 and x=y is the total number of coin