How do you solve #(x+2)/(x^2-6x-27)+(x-17)/(x^2-2x-63)=(x+1)/(x^2+10x+21)# and check for extraneous solutions?

Question

3853 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-10T16:46:11

Multiply everything by the least command denominator to remove the fractions The solve for x

#( x^2 -6x -27)# can be factored into ( x -9) x ( x +3)

#( x^2 -2x -62)# can be factored into ( x -9) x ( x+7)

#( x^2 +10x +21) # can be factored into ( x + Mu7) ( x +3)

The least common denominator for all three is
( x-9) x ( x+7) x (x +3)

multiply all the terms by the least common denominator

# (x-9)(x+7)(x+3) xx {(x+2)/(x^2 -6x - 27) + (x-17)/(x^2-2x -63)=(x +1)/(x^2) +10x +21}# This gives

(x + 7) ( x+2) + (x+3)(x -7) = (x + 1) Multiplying this out gives

# x^2 +9x + 14 + x^2 -4x -7 = x+1# Adding the common terms gives

# 2x^2 + 5x + 7 = x + 1 # subtract 1 x from each side gives

# 2x^2 + 5x - 1x + 7 = x - x + 1 # resulting in

# 2x^2 + 4x +7 = 1 # subtract one from each side gives

# 2x^2 + 4x +7 -1 = 1 -1 # resulting in

# 2x^2 + 4x +6 = 0 #

Use the quadratic equation to solve for the roots