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How do you factor #2/3 = 2 - (5x-3)/(x-1)#?

Answer 1

Avery Alexander

#11x - 5= 0# does not factor, but can be solved to give #x = 5/11#

Initially, we must simplify this as it is not in a form that can be recognized for factoring.

#2/3 = 2 - (5x-3)/(x-1)" "color(blue)(xx 3(x-1))#

Take the denominators out:

#2/cancel3color(blue)(xx cancel3(x-1)) = 2color(blue)(xx 3(x-1)) - (5x-3)/cancel((x-1)) color(blue)(xx 3cancel((x-1))#

#2(x-1) = 6(x-1) - 3(5x-3)#

#2x-2 = 6x-6 -15x+9#

#2x-6x+15x = -6+9+2#

This does not factorize and reduces to a straightforward linear equation.

#11x = 5#

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Answer 2

Eva Abramson

To factor the equation 2/3 = 2 - (5x-3)/(x-1), we can start by multiplying both sides of the equation by 3 to eliminate the fraction. This gives us 2 = 6 - (5x-3)/(x-1).

Next, we can multiply both sides of the equation by (x-1) to eliminate the fraction in the denominator. This gives us 2(x-1) = 6(x-1) - (5x-3).

Expanding the equation, we have 2x - 2 = 6x - 6 - 5x + 3.

Combining like terms, we get 2x - 2 = x - 3.

Moving all the terms to one side of the equation, we have 2x - x = -3 + 2.

Simplifying further, we get x = -1.

Therefore, the solution to the equation is x = -1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.