Answer:
[tex]\dfrac{x^2+6x+-67}{2x^2-20x+42} = \dfrac{x^2+6x+-67}{2(x^2-10x+21)}[/tex]
Step-by-step explanation:
[tex]\dfrac{5}{x-3} + \dfrac{3}{x-7} +\dfrac{1}{2}[/tex]
In order to perform the sum, the denominator of the fraction must be the same.
So, it will be [tex]2(x-3)(x-7)[/tex]
Therefore,
[tex]\dfrac{5\cdot2(x-7)}{2(x-3)(x-7)} + \dfrac{3\cdot2(x-3)}{2(x-3)(x-7)} +\dfrac{(x-3)(x-7)}{2(x-3)(x-7)}[/tex]
[tex]\dfrac{10(x-7)}{2(x-3)(x-7)} + \dfrac{6(x-3)}{2(x-3)(x-7)} +\dfrac{(x-3)(x-7)}{2(x-3)(x-7)}[/tex]
[tex]\dfrac{10(x-7)+6(x -3)+(x-3)(x-7)}{2(x-3)(x-7)}[/tex]
[tex]\dfrac{10x-70+6x -18+x^2-10x+21}{2(x-3)(x-7)}[/tex]
[tex]\dfrac{-67+6x+x^2}{2(x-3)(x-7)}[/tex]
[tex]2(x-3)(x-7)= 2x^2-20x+42[/tex]
So,
[tex]\dfrac{x^2+6x+-67}{2x^2-20x+42} = \dfrac{x^2+6x+-67}{2(x^2-10x+21)}[/tex]
