How do you solve the equation (2x+3)/(x+2)=(x-1)/(x-2)?

3 Answers

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Before solving the equation, we'll impose the following constraint: the roots of the equation must not be the roots of denominators of the fractions. Therefore, the values of x are not allowed to be -2 or 2.

With respect to this constraint, we'll solve the equation by cross multiplying.

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

We'll remove the brackets:

2x^2 - 4x + 3x - 6 = x^2 - x + 2x - 2

We'll shift all terms to the left:

x^2 - x - x - 6 + 2 = 0

x^2 - 2x - 4 = 0

We'll apply quadratic formula:

x1 = [2+sqrt(4+16)]/2

x1 = (2+2sqrt5)/2

x1 = 1 + sqrt 5

x2 = 1 - sqrt 5

Since the values of x does not cancel the denominators of the given fractions, we'll accept them as solutions: x1 = 1 + sqrt 5 and x2 = 1 - sqrt 5.

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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To solve the equation `(2x+3)/(x+2)=(x-1)/(x-2)` first eliminate the terms from the denominator. This gives two quadratic terms on either side of the equation. The resulting quadratic equation is is then solved for x.


-> `(2x+3)(x-2) = (x-1)(x+2)`

-> `2x^2-4x+3x-6=x^2-x+2x-2`

-> `x^2-2x-4=0`

Now use the quadratic formula for the roots of the equation:`(-b+-sqrt(b^2-4ac))/(2a)`

The roots are `(2+-sqrt(4+16))/2`

= `(2+-sqrt 20)/2`

= `(2+-2sqrt 5)/2`

= `1+sqrt5` and `1-sqrt 5`

The solutions of the given equation are `1+sqrt5` and `1-sqrt 5`

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nerdrafi | Student, Grade 9 | (Level 2) eNoter

Posted on

First of all , warm greetings to you

What we need to do here is'Cross Multiply' . What I mean by cross multiply is that we multiply the Numerator with the next Denomenator . For instance : 2x/2x=3x/3x then 2x will be multiplied with the next denomenator i.e 3x so eventually it is 2x(3x) = 6x^2

So lets do this applying the rules of cross multiply

Our Numerator : (1)12x+3 and(2) x-1 Denominator : (1)1x+2 and (2)x-2


so lets multiply with the 1st Numerator with the second Denominator and the 2nd Numerator with the 1st Denominator (JUST AS CROSS-CROSS) :-

Step 1 : 2x+3(x-2) =x+2(x-1)

Step 2 (Break Them) : 2x(x-2)+3(x-2) =x(x-1)+2(x-1)

Step 3 (Multiply 'em) : 2x^2-4x+3x-6  =x^2-x+2x-2x

Step 4 (Cancel out)   :2x^2-x-6          =x^2+x-2

Step 5 ( Bring 'em Together): 2x^2-x^2-x-x-6+2

Step 6 ( Do the Calculation) : x^2-2x-4


Hope This Helps


Reply Me back Dude !!!