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

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giorgiana1976's profile pic

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+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

p.s : ( THE NUMBER IN BRACKTET REPRESTENTS THAT THEY ARE IN THE SAME FRACTION)

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 !!!

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