Question

How do you solve \[\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{{{x}^{2}}-3x+2}\]?

Answer 1

Hint: This question is from the topic of algebra and quadratic equations. We will first solve and then factor out the term \[{{x}^{2}}-3x+2\]. After that, we will multiply the term \[{{x}^{2}}-3x+2\] to the both sides of the equation. After solving, we will take all the terms of x to the left side of the equation and all the constants to the right side of the equation. After solving the further equation, we will get the value of x.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to solve the term \[\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{{{x}^{2}}-3x+2}\]. Or, we can say that we have to solve and find the value of x from the equation \[\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{{{x}^{2}}-3x+2}\].
The equation which we have to solve is
\[\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{{{x}^{2}}-3x+2}\]
Let us first find out the factor of the term \[{{x}^{2}}-3x+2\].
The term can also be written as
\[{{x}^{2}}-x-2x+2\]
The above can also be written as
\[x\left( x-1 \right)-2\left( x-1 \right)\]
The above can also be written as
\[\left( x-2 \right)\left( x-1 \right)\]
Now, the equation \[\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{{{x}^{2}}-3x+2}\] can also be written as
\[\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{\left( x-2 \right)\left( x-1 \right)}\]
Now, multiplying \[\left( x-2 \right)\left( x-1 \right)\] (or, we can say \[{{x}^{2}}-3x+2\]) to the both side of equation, we will get
\[\Rightarrow \left( x-2 \right)\left( x-1 \right)\dfrac{3}{\left( x-2 \right)}=\left( x-2 \right)\left( x-1 \right)\dfrac{1}{\left( x-1 \right)}+\left( x-2 \right)\left( x-1 \right)\dfrac{7}{\left( x-2 \right)\left( x-1 \right)}\]
Now, after cancelling out the terms which are same in numerator and denominator, we can write the above equation as
\[\Rightarrow \left( x-1 \right)3=\left( x-2 \right)+7\]
The above equation can also be written as
\[\Rightarrow 3x-3=x-2+7\]
Now, taking all the terms of x to the left side of the equation and the constants to the right side of the equation, we get
\[\Rightarrow 3x-x=3-2+7\]
The above equation can also be written as
\[\Rightarrow 2x=8\]
After dividing 2 to both side of the equation, we get
\[\Rightarrow \dfrac{2x}{2}=\dfrac{8}{2}\]
As we know that if we divide 8 by 2, then we will get the division as 4. So, we can the write the above equation as
\[\Rightarrow x=4\]
Now, we have solved the equation \[\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{{{x}^{2}}-3x+2}\] and found the value of x as 4.

Note: We should have a better knowledge in the topic of algebra and quadratic equations to solve this type of question easily. We should know how to solve the quadratic equation and find the factor of the quadratic equation.
In case, if we want to check if our answer is correct or not. Then, we can cross-check our answer.
So, for cross-checking we will substitute the found value of x (which is 4) in the given equation \[\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{{{x}^{2}}-3x+2}\]
\[\Rightarrow \dfrac{3}{4-2}=\dfrac{1}{4-1}+\dfrac{7}{{{4}^{2}}-3\times 4+2}\]
The above equation can also be written as
\[\Rightarrow \dfrac{3}{2}=\dfrac{1}{3}+\dfrac{7}{16-12+2}\]
\[\Rightarrow \dfrac{3}{2}=\dfrac{1}{3}+\dfrac{7}{6}\]
Now, multiplying 2 to the numerator and denominator in the term \[\dfrac{1}{3}\], we get
\[\Rightarrow \dfrac{3}{2}=\dfrac{2}{6}+\dfrac{7}{6}\]
\[\Rightarrow \dfrac{3}{2}=\dfrac{2+7}{6}\]
\[\Rightarrow \dfrac{3}{2}=\dfrac{9}{6}\]
\[\Rightarrow \dfrac{3}{2}=\dfrac{3\times 3}{2\times 3}\]
The above can also be written as
\[\Rightarrow \dfrac{3}{2}=\dfrac{3}{2}\]
Hence, both sides of the equation are equal.
So, we can say that our answer is correct.