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

Question

Julia Adams · Accepted Answer

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)#.

Eva Abramson · Answer

undefined To solve the equation 1/y + 1/(y+2) = 1/3, we can start by finding a common denominator. Multiplying every term by 3y(y+2) will eliminate the denominators. This leads to the equation 3(y+2) + 3y = y(y+2). Simplifying further, we get 3y + 6 + 3y = y^2 + 2y. Combining like terms, we have 6y + 6 = y^2 + 2y. Rearranging the equation, we get y^2 - 4y - 6 = 0. This is a quadratic equation, which can be solved using factoring, completing the square, or the quadratic formula. Factoring the equation, we have (y - 3)(y + 2) = 0. Therefore, the solutions are y = 3 and y = -2.