Solve for n (4n+8)/(n^2+n-72)+8/(n^2+n-72)=1/(n+9)

Problem

(4*n+8)/(n^2+n−72)+8/(n^2+n−72)=1/(n+9)

Solution

1. Combine the fractions on the left side since they share a common denominator.

(4*n+8+8)/(n^2+n−72)=1/(n+9)

(4*n+16)/(n^2+n−72)=1/(n+9)

2. Factor the quadratic expression in the denominator of the left side.

n^2+n−72=(n+9)*(n−8)

(4*n+16)/((n+9)*(n−8))=1/(n+9)

3. Identify the restricted values for n to avoid division by zero.

n≠−9

n≠8

4. Multiply both sides of the equation by the least common denominator (n+9)*(n−8) to clear the fractions.

4*n+16=1*(n−8)

5. Solve the resulting linear equation for n

4*n+16=n−8

3*n+16=−8

3*n=−24

n=−8

6. Verify that the solution n=−8 is not one of the restricted values. Since −8≠−9 and −8≠8 the solution is valid.

Final Answer

(4*n+8)/(n^2+n−72)+8/(n^2+n−72)=1/(n+9)⇒n=−8

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