How do you solve $(x-7)/(x-4)+(x+8)/(x-3)=(x+158)/(x^2-7x+12)$ and check for extraneous solutions?

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Answer 1 · 2016-10-11T09:06:32

$x=(7+-sqrt1401)/4$

As $(x-4)(x-3)=x^2-7x+12$ - note that as they are in denominator $x!=3$ and $x!=4$

$(x-7)/(x-4)+(x+8)/(x-3)=(x+158)/(x^2-7x+12)$

$hArr((x-7)(x-3)+(x+8)(x-4))/(x^2-7x+12)=(x+158)/(x^2-7x+12)$

or $((x-7)(x-3)+(x+8)(x-4))=(x+158)$

or $x^2-10x+21+x^2+4x-32=x+158$

or $2x^2-6x-11-x-158=0$

or $2x^2-7x-169=0$

and using quadratic formula $x=(-b+-sqrt(b^2-4ac))/(2a)$ as olution for quadratic equation $ax^2+bx+c=0$

$x=(-(-7)+-sqrt((-7)^2-4*2^*(-169)))/4$

= $(7+-sqrt(49+1352))/4=(7+-sqrt1401)/4$

How do you solve (x-7)/(x-4)+(x+8)/(x-3)=(x+158)/(x^2-7x+12)(x-7)/(x-4)+(x+8)/(x-3)=(x+158)/(x^2-7x+12) and check for extraneous solutions?