Question

Find the value of the variable y, for which:

the difference between the fractions 6/y−4 and y/y+2 is equal to their product.

Answer 1

Answer: 6

Explanation:

`\text{Difference:}\\\\.\quad (6)/(y-4)-(y)/(y+2)\\\\\\=\bigg((y+2)/(y+2)\bigg)(6)/(y-4)-(y)/(y+2)\bigg((y-4)/(y-4)\bigg)\\\\\\=(6y+12)/((y+2)(y-4))-(y^2-4y)/((y+2)(y-4))\\\\\\=(6y+12-(y^2-4y))/((y+2)(y-4))\\\\\\=(-y^2+10y+12)/((y+2)(y-4))`

`\text{Product:}\\\\.\quad ((6)(y))/((y-4)(y+2))\\\\\\=(6y)/((y-4)(y+2))\\\\\\\text{Difference = Product:}\\\\(-y^2+10y+12)/((y+2)(y-4))=(6y)/((y+2)(y-4))\qquad Restriction: y\\eq -2, 4\\\\\\\rightarrow -y^2+10y+12=6y\\\\\rightarrow 0=y^2-4y-12\\\\\rightarrow 0=(y-6)(y+2)\\\\\rightarrow 0=y-6\quad and\quad 0=y+2\\\\\rightarrow 6=y\qquad and\quad -2=y\\\\\text{Since y = -2 is a restricted value, it is not a valid solution}\\\text{So, y = 6}`