Answer:
x = 0, x = 7
Step-by-step explanation:
Given equation:
[tex]\implies \dfrac{x+3}{x-3}+\dfrac{x}{x-5}=\dfrac{x+5}{x-5}[/tex]
Multiply by (x - 5):
[tex]\implies \dfrac{(x+3)(x-5)}{x-3}+\dfrac{x(x-5)}{x-5}=\dfrac{(x+5)(x-5)}{x-5}[/tex]
Factor out common term (x - 5):
[tex]\implies \dfrac{(x+3)(x-5)}{x-3}+x=x+5[/tex]
Multiply by (x - 3):
[tex]\implies \dfrac{(x+3)(x-5)(x-3)}{x-3}+x(x-3)=(x+5)(x-3)[/tex]
Factor out common term (x - 3):
[tex]\implies (x+3)(x-5)+x(x-3)=(x+5)(x-3)[/tex]
Expand brackets:
[tex]\implies x^2-5x+3x-15+x^2-3x=x^2-3x+5x-15[/tex]
Combine like terms:
[tex]\implies 2x^2-5x-15=x^2+2x-15[/tex]
Simplify:
[tex]\implies x^2-7x=0[/tex]
Factor and solve:
[tex]\implies x(x-7)=0[/tex]
[tex]\implies x=0,7[/tex]
