How do you solve #(3x-4)/(x^2-10x+21)-(x-8)/(x^2-18x+77)=(x-5)/(x^2-14x+33)# and check for extraneous solutions?

1 Answer
Dec 24, 2016

Answer:

#x=-1 and 15#..

Explanation:

graph{ y-(3x-4)/((x-7)(x-3))+(x-8)/((x-7)(x-11))+(x-5)/((x-11)(x-3))=0 [-15, 15, -20, 20]}
graph{y-(3x-4)/((x-7)(x-3))+(x-8)/((x-7)(x-11))+(x-5)/((x-11)(x-3))=0 [-50, 50, -1, 1]} Here,

#y = (3x-4)/((x-3)(x-7))-(x-8)/((x-7)(x-11))-(x-5)/((x-11)(x-3))=0#.

This is not a problem, in limits. So, x is none of 3, 7 and 11.

( The graph for the limit problem is also inserted.)

Multiplying by #(x-3)(x-7)(x-11)#,

#(3x-8)(x-11)-(x-8)(x-3)-(x-5)(x-7)=0#.

Upon simplification,

x^2-14x-15=0. Solving.

#x=-1 and 15#.

The two inserted graphs for the same function, on different scales,

are for depicting zeros for y that are solutions for the problem,

besides the indeterminate forms #oo-oo#, at x = 3, 7 and 11.