How do you solve # 7/(u - 3) - 42/(u^2 - 9) =1# and find any extraneous solutions?

Question

Ethan Adkins · Accepted Answer

Roots are #color(green)(u = 3, 4)# #7 / (u-3) - 42 / (u^2-9) = 1# Taking LCM fo LHS as #(u^2 -9)#, #(7 (u+3) - 42) = 1 * (u^2 - 9)# #7u + 21 - 42 = u^2 - 9# #u^2 - 7u -21 + 42 -9 = 0# #u^2 - 7u + 12 = 0# #u^2 - 3u - 4u + 12 = 0# #u(u-3) -4 (u-3) = 0# #(u-3) * (u-4) = 0# Roots are #color(green)(u = 3, 4)#

Nathan Axel · Answer

undefined To solve the equation 7/(u - 3) - 42/(u^2 - 9) = 1 and find any extraneous solutions, follow these steps:

1. Start by finding a common denominator for the fractions on the left side of the equation. The common denominator is (u - 3)(u + 3), as it includes both denominators.

2. Rewrite the equation with the common denominator: (7(u + 3))/[(u - 3)(u + 3)] - (42)/(u^2 - 9) = 1.

3. Simplify the equation by multiplying both sides by the common denominator to eliminate the fractions: 7(u + 3) - 42 = (u - 3)(u + 3).

4. Expand the right side of the equation: 7u + 21 - 42 = u^2 - 9.

5. Combine like terms: 7u - 21 = u^2 - 9.

6. Rearrange the equation to bring all terms to one side: u^2 - 7u + 12 = 0.

7. Factor the quadratic equation: (u - 3)(u - 4) = 0.

8. Set each factor equal to zero and solve for u: u - 3 = 0 or u - 4 = 0.

9. Solve for u: u = 3 or u = 4.

10. Check for extraneous solutions by substituting the values of u back into the original equation. In this case, both solutions, u = 3 and u = 4, are valid and not extraneous.

Therefore, the solutions to the equation 7/(u - 3) - 42/(u^2 - 9) = 1 are u = 3 and u = 4, with no extraneous solutions.