# simplify fractio f(x,y)=(yx^2-2xy+7/4x^2-7/2x+y+7/4)/(yx+7/4)^2-(7/4x+y)^2

sciencesolve | Certified Educator

You need to form groups of terms to numerator such that:

`f(x,y)=((yx^2+7/4x^2) + (-2xy-7/2x) + (y+7/4))/((yx+7/4)^2-(7/4x+y)^2)`

You should factor out `x^2`  in the first group and -2x in the second group such that:

`f(x,y)=(x^2(y + 7/4) -2x(y + 7/4) + (y + 7/4))/((yx+7/4)^2-(7/4x+y)^2)`

Notice that you may factor out `(y+7/4)`  to numerator such that:

f`(x,y)=((y + 7/4)(x^2 -2x + 1))/((yx+7/4)^2-(7/4x+y)^2)`

Notice that you may substitute `(x-1)^2`  for `x^2 -2x + 1`  such that:

`f(x,y)=((y + 7/4)(x-1)^2)/((yx+7/4)^2-(7/4x+y)^2)`

You need to substitute the special product for difference of squares to denominator such that:

`f(x,y)=((y + 7/4)(x-1)^2)/((yx+7/4-7/4x-y)(yx+7/4+7/4x+y))`

You may group the terms in the factor `yx+7/4-7/4x-y`  such that:

`yx+7/4-7/4x-y = (yx-y) + (7/4-7/4x)`

Factoring out y in the first group and `7/4`  in the second group yields:

`yx+7/4-7/4x-y = y(x-1) + 7/4(x-1)`

Factoring out (x-1) yields:

`yx+7/4-7/4x-y = (x-1)(y+7/4)`

You may group the terms in the factor `yx+7/4+7/4x+y`  such that:

`yx+7/4+7/4x+y = (x+1)(y+7/4)`

You need to substitute `(x-1)(y+7/4)`  for `yx+7/4-7/4x-y`   and `(x+1)(y+7/4)`  for `yx+7/4+7/4x+y`  such that:

`f(x,y)=((y + 7/4)(x-1)^2)/((x-1)(y+7/4)(x+1)(y+7/4))`

You need to reduce the like terms such that:

`f(x,y)=(x-1)/((y+7/4)(x+1))`

Hence, reducing the fraction to its lowest terms yields `f(x,y)=(x-1)/((y+7/4)(x+1)).`