2/(2x-1)+3/(x-3)=(x+1)/(x-3)+x/(2x-1)

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вопрос

To solve this equation, first find a common denominator for all fractions:

2/(2x-1) + 3/(x-3) = (x+1)/(x-3) + x/(2x-1)

Common denominator = (2x-1)(x-3)

Rewrite each fraction with the common denominator:

(2(x-3))/[(2x-1)(x-3)] + (3(2x-1))/[(2x-1)(x-3)] = [(x+1)(2x-1)]/[(2x-1)(x-3)] + (x(x-3))/[(2x-1)(x-3)]

Now simplify the fractions:

[2x-6]/[(2x-1)(x-3)] + [6x-3]/[(2x-1)(x-3)] = [2x^2-x+2x-1]/[(2x-1)(x-3)]

Combine like terms:

[2x-6 + 6x-3]/[(2x-1)(x-3)] = [2x^2+x-1]/[(2x-1)(x-3)]

Now combine the fractions on the left side:

[8x-9]/[(2x-1)(x-3)] = [2x^2+x-1]/[(2x-1)(x-3)]

Now, cross multiply to simplify and solve for x:

(8x-9)(2x-1)(x-3) = (2x^2+x-1)(2x-1)(x-3)

After multiplying out, you will have a quadratic equation that can be solved for x.