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a_sh-v [17]
3 years ago
11

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

Mathematics
1 answer:
Brrunno [24]3 years ago
7 0

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

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