Answer:
See below.
Step-by-step explanation:
Given equation:
[tex]\dfrac{1}{60}=\dfrac{2x+50}{x(x+50)}[/tex]
Multiply both sides by 60:
[tex]\implies 60 \cdot \dfrac {1}{60}=60 \cdot \dfrac{2x+50}{x(x+50)}[/tex]
[tex]\implies 1=\dfrac{60(2x+50)}{x(x+50)}[/tex]
Multiply both sides by x(x+50):
[tex]\implies 1\cdot x(x+50)=\dfrac{60(2x+50)}{x(x+50)} \cdot x(x+50)[/tex]
[tex]\implies x(x+50)=60(2x+50)[/tex]
Distribute both sides:
[tex]\implies x \cdot x + x \cdot 50=60 \cdot 2x+60 \cdot 50[/tex]
[tex]\implies x^2+50x=120x+3000[/tex]
Subtract 120x from both sides:
[tex]\implies x^2+50x-120x=120x+3000-120x[/tex]
[tex]\implies x^2-70x=3000[/tex]
Subtract 3000 from both sides:
[tex]\implies x^2-70x-3000=3000-3000[/tex]
[tex]\implies x^2-70x-3000=0[/tex]
A rational function is undefined when its denominator is equal to zero.
Therefore, to find the excluded x-values of the domain, set the denominator of the original function to zero and solve for x:
[tex]\implies x(x+50)=0[/tex]
[tex]\implies x=0[/tex]
[tex]\implies x+50=0 \implies x=-50[/tex]
Thus proving that
[tex]\dfrac{1}{60}=\dfrac{2x+50}{x(x+50)}\;\;\; \textsf{is equivalent to}\;\;\;x^2-70x-3000=0[/tex]
for all values of x not equal to zero or -50.