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Question: How do you solve \(\dfrac{{x - 2}}{{x + 1}} = \dfrac{{x + 1}}{{x - 2}}\) ?...

How do you solve x2x+1=x+1x2\dfrac{{x - 2}}{{x + 1}} = \dfrac{{x + 1}}{{x - 2}} ?

Explanation

Solution

To solve the equation, first we need to simplify the equation using cross multiplication, where the denominator of Left Hand Side is multiplied with the numerator of Right Hand Side and the denominator of Right Hand Side is multiplied with the numerator of Left Hand Side.

Complete step-by-step answer:
To simplify the equation lets us multiply the denominator on the left-hand side with the numerator on the right-hand side and the same process with the denominator on the right-hand side and the numerator on the left-hand side.
From x2x+1=x+1x2\dfrac{{x - 2}}{{x + 1}} = \dfrac{{x + 1}}{{x - 2}} , we get
\Rightarrow (x2)×(x2)=(x+1)×(x+1)(x - 2) \times (x - 2) = (x + 1) \times (x + 1)
To simplify the equation, let’s open the brackets by the Distributive law.
\Rightarrow x×(x2)2×(x2)=x×(x+1)+1×(x+1)x \times (x - 2) - 2 \times (x - 2) = x \times (x + 1) + 1 \times (x + 1)
Further simplifying the equation and applying the Distributive Law.
\Rightarrow x×xx×22×x+2×2=x×x+x×1+1×x+1×1x \times x - x \times 2 - 2 \times x + 2 \times 2 = x \times x + x \times 1 + 1 \times x + 1 \times 1
x22x2x+4=x2+x+x+1\Rightarrow {x^2} - 2x - 2x + 4 = {x^2} + x + x + 1
Now, let’s shift all the terms on the right-hand side to the left-hand side
x22x2x+4x2xx1=0\Rightarrow {x^2} - 2x - 2x + 4 - {x^2} - x - x - 1 = 0
Rearranging the terms
x2x22x2xxx+41=0\Rightarrow {x^2} - {x^2} - 2x - 2x - x - x + 4 - 1 = 0
To simplify the term 2  x2\;x ,
\Rightarrow 2x=x+x2x = x + x
Therefore, simplifying the equation
x2x2(x+x)(x+x)xx+41=0\Rightarrow {x^2} - {x^2} - (x + x) - (x + x) - x - x + 4 - 1 = 0
x2x2xxxxxx+41=0\Rightarrow {x^2} - {x^2} - x - x - x - x - x - x + 4 - 1 = 0
Now, separating the terms with the same power,
(x2x2)(xxxxxx)+(41)=0\Rightarrow ({x^2} - {x^2})( - x - x - x - x - x - x) + (4 - 1) = 0
06x+3=0\Rightarrow 0 - 6x + 3 = 0
Now, shifting the term 6  x6\;x from R.H.S. to L.H.S.
3=6x\Rightarrow 3 = 6x
6x=3\Rightarrow 6x = 3
Now, making xx as the subject of the equation
x=36\Rightarrow x = \dfrac{3}{6}
Converting the numerator and denominator in their prime factors
x=3×13×2\Rightarrow x = \dfrac{{3 \times 1}}{{3 \times 2}}
Hence, the final answer by omitting the common factor from numerator and denominator can be shown
as

x=12 \Rightarrow x = \dfrac{1}{2}

Additional information: If the equations multiplying are the same (like in this question) then you can directly use the formula as shown (a+b)×(a+b)=(a+b)2=a2+2ab+b2(a + b) \times (a + b) = {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}

Note:
Remember the fact, that the terms can only be added or subtracted if they are of the same power or degree, for eg., the term x2{x^2} can only be added or subtracted from the term with the same power i.e. x2{x^2} . You cannot add or subtract a term like 2  x2\;x into the term like x2{x^2}.