# Solve for y (y+3)/(y+3)=1/(y+2)

y+3y+3=1y+2
Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.
(y+3)⋅(y+2)=(y+3)⋅1
Solve the equation for y.
Simplify (y+3)⋅(y+2).
Expand (y+3)(y+2) using the FOIL Method.
Apply the distributive property.
y(y+2)+3(y+2)=(y+3)⋅1
Apply the distributive property.
y⋅y+y⋅2+3(y+2)=(y+3)⋅1
Apply the distributive property.
y⋅y+y⋅2+3y+3⋅2=(y+3)⋅1
y⋅y+y⋅2+3y+3⋅2=(y+3)⋅1
Simplify and combine like terms.
Simplify each term.
Multiply y by y.
y2+y⋅2+3y+3⋅2=(y+3)⋅1
Move 2 to the left of y.
y2+2⋅y+3y+3⋅2=(y+3)⋅1
Multiply 3 by 2.
y2+2y+3y+6=(y+3)⋅1
y2+2y+3y+6=(y+3)⋅1
y2+5y+6=(y+3)⋅1
y2+5y+6=(y+3)⋅1
y2+5y+6=(y+3)⋅1
Multiply y+3 by 1.
y2+5y+6=y+3
Move all terms containing y to the left side of the equation.
Subtract y from both sides of the equation.
y2+5y+6-y=3
Subtract y from 5y.
y2+4y+6=3
y2+4y+6=3
Move 3 to the left side of the equation by subtracting it from both sides.
y2+4y+6-3=0
Subtract 3 from 6.
y2+4y+3=0
Factor y2+4y+3 using the AC method.
Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is 3 and whose sum is 4.
1,3
Write the factored form using these integers.
(y+1)(y+3)=0
(y+1)(y+3)=0
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
y+1=0
y+3=0
Set the first factor equal to 0 and solve.
Set the first factor equal to 0.
y+1=0
Subtract 1 from both sides of the equation.
y=-1
y=-1
Set the next factor equal to 0 and solve.
Set the next factor equal to 0.
y+3=0
Subtract 3 from both sides of the equation.
y=-3
y=-3
The final solution is all the values that make (y+1)(y+3)=0 true.
y=-1,-3
y=-1,-3
Exclude the solutions that do not make y+3y+3=1y+2 true.
y=-1
Solve for y (y+3)/(y+3)=1/(y+2)

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