Question 1123485

solve variable in equation

{{{ (2)/(y+2)-(6)/(y^2+5y+6)=(5)/((y+2)^2)}}}
<pre>{{{ (2)/(y+2)-(6)/(y^2+5y+6)=(5)/((y+2)^2)}}}
{{{matrix(1,3, 2/(y + 2) - 6/(y + 3)(y + 2), "=", 5/(y + 2)^2)}}} ------ Factoring denominator
From this, we can see that {{{matrix(1,3, x <> - 2, and, x <> - 3)}}}
{{{matrix(1,3, 2(y + 3)(y + 2) - 6(y + 2), "=", 5(y + 3))}}} -------- Multiplying by LCD, (y + 3)(y + 2)<sup>2</sup>
{{{matrix(1,3, 2(y^2 + 5y + 6) - 6y - 12, "=", 5y + 15)}}}
{{{matrix(1,3, 2y^2 + 10y + 12 - 6y - 12, "=", 5y + 15)}}}
{{{matrix(1,3, 2y^2 + 10y - 6y - 5y + 12 - 12 - 15, "=", 0)}}}
{{{matrix(1,3, 2y^2 - y - 15, "=", 0)}}}
(2y + 5)(y - 3) = 0 ------ Factoring TRINOMIAL
2y + 5 = 0       OR         y - 3 = 0
Solving for y above gives you: {{{highlight_green(matrix(1,8, y, "=", - 5/2, ",", or, y, "=", 3))}}} <======= These values are NOT part of the restricted values of x.