Question

4 Write as a single fraction in its simplest form. 5/x-3+3/x-7+1/2

Answer 1

Answer:

$\dfrac{x^2+6x+-67}{2x^2-20x+42} = \dfrac{x^2+6x+-67}{2(x^2-10x+21)}$

Step-by-step explanation:

$\dfrac{5}{x-3} + \dfrac{3}{x-7} +\dfrac{1}{2}$

In order to perform the sum, the denominator of the fraction must be the same.

So, it will be $2(x-3)(x-7)$

Therefore,

$\dfrac{5\cdot2(x-7)}{2(x-3)(x-7)} + \dfrac{3\cdot2(x-3)}{2(x-3)(x-7)} +\dfrac{(x-3)(x-7)}{2(x-3)(x-7)}$

$\dfrac{10(x-7)}{2(x-3)(x-7)} + \dfrac{6(x-3)}{2(x-3)(x-7)} +\dfrac{(x-3)(x-7)}{2(x-3)(x-7)}$

$\dfrac{10(x-7)+6(x -3)+(x-3)(x-7)}{2(x-3)(x-7)}$

$\dfrac{10x-70+6x -18+x^2-10x+21}{2(x-3)(x-7)}$

$\dfrac{-67+6x+x^2}{2(x-3)(x-7)}$

$2(x-3)(x-7)= 2x^2-20x+42$

So,

$\dfrac{x^2+6x+-67}{2x^2-20x+42} = \dfrac{x^2+6x+-67}{2(x^2-10x+21)}$