How do you solve #\frac { 1} { y + 3} = \frac { 7} { y - 3} - \frac { 2} { y ^ { 2} - 9}#?

Question

Ethan Adkins · Accepted Answer

#y=-11/3# #1/(y+3)=7/(y-3)-2/(y^2-9)# #1/(y+3)=7/(y-3)-2/((y-3)(y+3))# #(y-3)/((y-3)(y+3))=(7(y+3))/((y-3)(y+3))-2/((y-3)(y+3))# #y-3=7y+21-2# #6y=-3-21+2# #6y=-22# #y=-11/3#

Eva Abramson · Answer

undefined To solve the equation $\frac{1}{y+3} = \frac{7}{y-3} - \frac{2}{y^2-9}$, we can follow these steps:

1. Start by finding a common denominator for the fractions on the right side of the equation. The common denominator in this case is $(y-3)(y+3)$.

2. Multiply both sides of the equation by the common denominator to eliminate the fractions. This gives us:
$(y-3)(y+3) \cdot \frac{1}{y+3} = (y-3)(y+3) \cdot \left(\frac{7}{y-3} - \frac{2}{y^2-9}\right)$.

3. Simplify both sides of the equation. On the left side, the $(y+3)$ terms cancel out, leaving us with just 1. On the right side, distribute the $(y-3)(y+3)$ to each term within the parentheses.

4. After simplifying, we are left with a quadratic equation. Rearrange the terms to bring all terms to one side of the equation, resulting in $y^2 - 7y - 6 = 0$.

5. Factor the quadratic equation or use the quadratic formula to find the values of $y$ that satisfy the equation.

6. Solve for $y$ by setting each factor equal to zero: $y-8=0$ or $y+1=0$.

7. The solutions to the equation are $y=8$ and $y=-1$.