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Mathematics 14 Online
OpenStudy (anonymous):

Solve: \[\frac{2x}{x+1} = \frac{2x-1}{x}\]

Parth (parthkohli):

Cross - multiply.

OpenStudy (freckles):

2x(x)=(2x-1)(x+1)

OpenStudy (freckles):

and x cannot be -1 or 0 :)

OpenStudy (anonymous):

uhmm okay...

Parth (parthkohli):

\(\Large \color{purple}{\rightarrow 2x(x) = (2x - 1)(x + 1) }\)

Parth (parthkohli):

Do you know igba, why we cross multiply?

OpenStudy (anonymous):

so \[2x62 = 2x^2 + x - 1\]

OpenStudy (anonymous):

that's \(2x^2\)

OpenStudy (anonymous):

that right?

OpenStudy (diyadiya):

yes

Parth (parthkohli):

\(\Large \color{purple}{\rightarrow 2x^2 = 2x(x + 1) - 1(x + 1) }\) \(\Large \color{purple}{\rightarrow 2x^2 = 2x^2 + 2x - x - 1 }\) \(\Large \color{purple}{\rightarrow 2x^2 = 2x^2 + x - 1 }\)

OpenStudy (anonymous):

yay so what's next @Diyadiya

Parth (parthkohli):

\(\Large \color{purple}{\rightarrow 0 = x - 1 }\) \(\Large \color{purple}{\rightarrow x = 1 }\)

Parth (parthkohli):

Note how I subtracted \(2x^2\) from both sides.

OpenStudy (diyadiya):

got it ?

OpenStudy (anonymous):

i think so..ill try more problems thanks :)

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