How do you solve $1/y + 1/(y+2) = 1/3$ ?

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Answer 1 · 2015-11-05T00:28:55

You have two solutions: $y=2pmsqrt(10)$ .

Sum the two fractions, obtaining only one denominator:

$1/y+1/(y+2) = ((y+2)+y)/(y(y+2))=(2y+2)/(y(y+2))$

So, the expression becomes

$(2y+2)/(y(y+2)) = 1/3$

Multiply both sides by $y(y+2)$ and then by $3$ :

$3cancel(y(y+2))(2y+2)/cancel(y(y+2)) = 1/cancel(3) cancel(3) y(y+2)$

The equation becomes

$3(2y+2) = y(y+2)$

Expand both terms:

$6y+6 = y^2+2y$

Bring everything to one side:

$y^2+2y-6y-6=0 iff y^2 -4y -6=0$

Complete the square:

$y^2 -4y -6 = (y^2-4y +4) -10 = (y-2)^2 -10$

So, the equation can be rewritten as

$(y-2)^2 -10 = 0 iff (y-2)^2 =10 \iff y-2 = pmsqrt(10)$

And thus $y=2pmsqrt(10)$ .

How do you solve 1/y + 1/(y+2) = 1/31/y + 1/(y+2) = 1/3?