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

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Answer 1 · 2018-02-03T06:18:17

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)#