Question

Help me please. Finals are next week and i’m stuck.

Answer 1

Answer:

x = 0,  x = 7

Step-by-step explanation:

Given equation:

$\implies \dfrac{x+3}{x-3}+\dfrac{x}{x-5}=\dfrac{x+5}{x-5}$

Multiply by (x - 5):

$\implies \dfrac{(x+3)(x-5)}{x-3}+\dfrac{x(x-5)}{x-5}=\dfrac{(x+5)(x-5)}{x-5}$

Factor out common term (x - 5):

$\implies \dfrac{(x+3)(x-5)}{x-3}+x=x+5$

Multiply by (x - 3):

$\implies \dfrac{(x+3)(x-5)(x-3)}{x-3}+x(x-3)=(x+5)(x-3)$

Factor out common term (x - 3):

$\implies (x+3)(x-5)+x(x-3)=(x+5)(x-3)$

Expand brackets:

$\implies x^2-5x+3x-15+x^2-3x=x^2-3x+5x-15$

Combine like terms:

$\implies 2x^2-5x-15=x^2+2x-15$

Simplify:

$\implies x^2-7x=0$

Factor and solve:

$\implies x(x-7)=0$

$\implies x=0,7$