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6/(x^2-3x)=12/x+1/(x-3) with full expanation from the internet

Sagot :

semsee45

Answer:

[tex]x=\dfrac{42}{13}[/tex]

Step-by-step explanation:

Given equation:

[tex]\dfrac{6}{(x^2-3x)}=\dfrac{12}{x}+\dfrac{1}{(x-3)}[/tex]

Factor the denominator on the left side:

[tex]\implies \dfrac{6}{x(x-3)}=\dfrac{12}{x}+\dfrac{1}{(x-3)}[/tex]

Combine the fractions on the right side by multiplying by the LCM:

[tex]\begin{aligned}\implies \dfrac{6}{x(x-3)} & =\dfrac{12}{x} \cdot \dfrac{(x-3)}{(x-3)}+\dfrac{1}{(x-3)} \cdot \dfrac{x}{x}\\\\& =\dfrac{12(x-3)}{x(x-3)}+\dfrac{x}{x(x-3)}\\\\& =\dfrac{12(x-3)+x}{x(x-3)} \end{aligned}[/tex]

As the denominators are now the same on both sides:

[tex]\begin{aligned}\implies \dfrac{6}{x(x-3)} & =\dfrac{12(x-3)+x}{x(x-3)}\\\\ \implies 6 & = 12(x-3)+x\end{aligned}[/tex]

Simplify and solve for x:

[tex]\begin{aligned}\implies 6 & = 12(x-3)+x\\\\6 & =12x-36+x\\\\6 & =13x-36\\\\13x & =42\\\\\implies x & =\dfrac{42}{13}\end{aligned}[/tex]

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