# Find the value of x which satisfies the equation \[\dfrac{5}{{x{\text{ }} - {\text{ }}5}}{\text{ + }}\dfrac{2}{{x{\text{ }} - {\text{ 2}}}}{\text{ = }}\dfrac{3}{{x{\text{ }} - {\text{ }}3}}{\text{ + }}\dfrac{4}{{x{\text{ }} - {\text{ }}4}}\].

A. 0, \[\dfrac{5}{2}\]

B. 1, \[\dfrac{9}{2}\]

C. 2, \[\dfrac{{11}}{2}\]

D. None of these

Answer

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Hint: Let us try to reduce the given equation to a perfect polynomial equation of x. So, we can find the roots of that equation easily.

Complete step-by-step answer:

As we know that the given equation is,

$\Rightarrow$ \[\dfrac{5}{{x{\text{ }} - {\text{ }}5}}{\text{ + }}\dfrac{2}{{x{\text{ }} - {\text{ 2}}}}{\text{ = }}\dfrac{3}{{x{\text{ }} - {\text{ }}3}}{\text{ + }}\dfrac{4}{{x{\text{ }} - {\text{ }}4}}\] (1)

Now, we have to solve equation 1.

First, we had to reduce equation 1 to a perfect polynomial equation.

For that we had to take LCM on LHS and RHS of equation 1 and then cross-multiply both sides of that equation.

So, taking LCM on LHS and RHS of equation 1. We get,

$\Rightarrow$ \[\dfrac{{5\left( {x{\text{ }} - {\text{ 2}}} \right){\text{ }} + {\text{ 2}}\left( {x{\text{ }} - {\text{ 5}}} \right)}}{{\left( {x{\text{ }} - {\text{ 5}}} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)}}{\text{ = }}\dfrac{{3\left( {x{\text{ }} - {\text{ 4}}} \right){\text{ }} + {\text{ }}4\left( {x{\text{ }} - {\text{ 3}}} \right)}}{{\left( {x{\text{ }} - {\text{ 3}}} \right)\left( {x{\text{ }} - {\text{ }}4} \right)}}\] (2)

On solving the numerator of LHS and RHS of equation 2. We get,

$\Rightarrow$ \[\dfrac{{5x{\text{ }} - {\text{ }}10{\text{ }} + {\text{ 2}}x{\text{ }} - {\text{ }}10}}{{\left( {x{\text{ }} - {\text{ 5}}} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)}}{\text{ = }}\dfrac{{3x{\text{ }} - {\text{ }}12{\text{ }} + {\text{ }}4x{\text{ }} - {\text{ }}12}}{{\left( {x{\text{ }} - {\text{ 3}}} \right)\left( {x{\text{ }} - {\text{ }}4} \right)}}\]

$\Rightarrow$ \[\dfrac{{\left( {7x{\text{ }} - {\text{ }}20} \right)}}{{\left( {x{\text{ }} - {\text{ 5}}} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)}}{\text{ = }}\dfrac{{\left( {7x{\text{ }} - {\text{ }}24} \right)}}{{\left( {x{\text{ }} - {\text{ 3}}} \right)\left( {x{\text{ }} - {\text{ }}4} \right)}}\] (3)

Now cross-multiplying LHS and RHS of equation 3. To get the required polynomial equation of x.

$\Rightarrow$ \[\left( {7x{\text{ }} - {\text{ }}20} \right)\left( {x{\text{ }} - {\text{ 3}}} \right)\left( {x{\text{ }} - {\text{ }}4} \right){\text{ = }}\left( {7x{\text{ }} - {\text{ }}24} \right)\left( {x{\text{ }} - {\text{ 5}}} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)\] (4)

Now we had to solve equation 3.

So, solving equation 3 step by step.

Multiplying \[\left( {7x{\text{ }} - {\text{ }}20} \right)\] and \[\left( {x{\text{ }} - {\text{ 3}}} \right)\] in LHS of equation 3.

And multiplying \[\left( {7x{\text{ }} - {\text{ }}24} \right)\] and \[\left( {x{\text{ }} - {\text{ 5}}} \right)\] in RHS of equation 3. We get,

$\Rightarrow$ \[\left( {7{x^2}{\text{ }} - {\text{ }}21x{\text{ }} - {\text{ }}20x{\text{ }} + {\text{ }}60} \right)\left( {x{\text{ }} - {\text{ }}4} \right) = \left( {7{x^2}{\text{ }} - {\text{ }}35x{\text{ }} - {\text{ }}24x{\text{ }} + {\text{ }}120} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)\]

$\Rightarrow$ \[\left( {7{x^2}{\text{ }} - {\text{ 4}}1x{\text{ }} + {\text{ }}60} \right)\left( {x{\text{ }} - {\text{ }}4} \right) = \left( {7{x^2}{\text{ }} - {\text{ }}59x{\text{ }} + {\text{ }}120} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)\]

Now solving the LHS and RHS of the above equation. We get,

$\Rightarrow$ \[\left( {7{x^3}{\text{ }} - {\text{ 4}}1{x^2}{\text{ }} + {\text{ }}60x} \right){\text{ }} - {\text{ }}\left( {28{x^2}{\text{ }} - {\text{ 164}}x{\text{ + 24}}0} \right){\text{ }} = {\text{ }}\left( {7{x^3}{\text{ }} - {\text{ }}59{x^2}{\text{ }} + {\text{ }}120x} \right){\text{ }} - {\text{ }}\left( {14{x^2}{\text{ }} - {\text{ 118}}x{\text{ + }}240} \right)\]

On solving the above equation. We get,

$\Rightarrow$ \[7{x^3}{\text{ }} - {\text{ 69}}{x^2}{\text{ + 224}}x{\text{ }} - {\text{ 24}}0{\text{ }} = {\text{ }}7{x^3}{\text{ }} - {\text{ 73}}{x^2}{\text{ }} + {\text{ }}238x{\text{ }} - {\text{ }}240\] (5)

Now subtracting RHS from the LHS of equation 5. We get,

$\Rightarrow$ \[4{x^2}{\text{ }} - {\text{ }}14x{\text{ }} = {\text{ }}0\]

Now we have to solve the above quadratic equation of x. To get the required value of x.

Taking x common from LHS of the above equation. We get,

$\Rightarrow$ x(4x – 14) = 0

So, from the above equation we get x = 0, \[\dfrac{{14}}{4}\].

Hence, the correct value of x that satisfies the given equation will be x = 0, \[\dfrac{7}{2}\].

So, the correct option will be D.

Note: Whenever we come up with this type of problem where we are given with an equation in which variable x is in the denominator of the equation. Then first we take the LCM of LHS and RHS of the given equation. And then we cross-multiply both sides of the equation. After that we try to manipulate the equation. So, that we can get a perfect polynomial equation. After that it will be easy to find the solution of a quadratic or cubic.

Complete step-by-step answer:

As we know that the given equation is,

$\Rightarrow$ \[\dfrac{5}{{x{\text{ }} - {\text{ }}5}}{\text{ + }}\dfrac{2}{{x{\text{ }} - {\text{ 2}}}}{\text{ = }}\dfrac{3}{{x{\text{ }} - {\text{ }}3}}{\text{ + }}\dfrac{4}{{x{\text{ }} - {\text{ }}4}}\] (1)

Now, we have to solve equation 1.

First, we had to reduce equation 1 to a perfect polynomial equation.

For that we had to take LCM on LHS and RHS of equation 1 and then cross-multiply both sides of that equation.

So, taking LCM on LHS and RHS of equation 1. We get,

$\Rightarrow$ \[\dfrac{{5\left( {x{\text{ }} - {\text{ 2}}} \right){\text{ }} + {\text{ 2}}\left( {x{\text{ }} - {\text{ 5}}} \right)}}{{\left( {x{\text{ }} - {\text{ 5}}} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)}}{\text{ = }}\dfrac{{3\left( {x{\text{ }} - {\text{ 4}}} \right){\text{ }} + {\text{ }}4\left( {x{\text{ }} - {\text{ 3}}} \right)}}{{\left( {x{\text{ }} - {\text{ 3}}} \right)\left( {x{\text{ }} - {\text{ }}4} \right)}}\] (2)

On solving the numerator of LHS and RHS of equation 2. We get,

$\Rightarrow$ \[\dfrac{{5x{\text{ }} - {\text{ }}10{\text{ }} + {\text{ 2}}x{\text{ }} - {\text{ }}10}}{{\left( {x{\text{ }} - {\text{ 5}}} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)}}{\text{ = }}\dfrac{{3x{\text{ }} - {\text{ }}12{\text{ }} + {\text{ }}4x{\text{ }} - {\text{ }}12}}{{\left( {x{\text{ }} - {\text{ 3}}} \right)\left( {x{\text{ }} - {\text{ }}4} \right)}}\]

$\Rightarrow$ \[\dfrac{{\left( {7x{\text{ }} - {\text{ }}20} \right)}}{{\left( {x{\text{ }} - {\text{ 5}}} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)}}{\text{ = }}\dfrac{{\left( {7x{\text{ }} - {\text{ }}24} \right)}}{{\left( {x{\text{ }} - {\text{ 3}}} \right)\left( {x{\text{ }} - {\text{ }}4} \right)}}\] (3)

Now cross-multiplying LHS and RHS of equation 3. To get the required polynomial equation of x.

$\Rightarrow$ \[\left( {7x{\text{ }} - {\text{ }}20} \right)\left( {x{\text{ }} - {\text{ 3}}} \right)\left( {x{\text{ }} - {\text{ }}4} \right){\text{ = }}\left( {7x{\text{ }} - {\text{ }}24} \right)\left( {x{\text{ }} - {\text{ 5}}} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)\] (4)

Now we had to solve equation 3.

So, solving equation 3 step by step.

Multiplying \[\left( {7x{\text{ }} - {\text{ }}20} \right)\] and \[\left( {x{\text{ }} - {\text{ 3}}} \right)\] in LHS of equation 3.

And multiplying \[\left( {7x{\text{ }} - {\text{ }}24} \right)\] and \[\left( {x{\text{ }} - {\text{ 5}}} \right)\] in RHS of equation 3. We get,

$\Rightarrow$ \[\left( {7{x^2}{\text{ }} - {\text{ }}21x{\text{ }} - {\text{ }}20x{\text{ }} + {\text{ }}60} \right)\left( {x{\text{ }} - {\text{ }}4} \right) = \left( {7{x^2}{\text{ }} - {\text{ }}35x{\text{ }} - {\text{ }}24x{\text{ }} + {\text{ }}120} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)\]

$\Rightarrow$ \[\left( {7{x^2}{\text{ }} - {\text{ 4}}1x{\text{ }} + {\text{ }}60} \right)\left( {x{\text{ }} - {\text{ }}4} \right) = \left( {7{x^2}{\text{ }} - {\text{ }}59x{\text{ }} + {\text{ }}120} \right)\left( {x{\text{ }} - {\text{ 2}}} \right)\]

Now solving the LHS and RHS of the above equation. We get,

$\Rightarrow$ \[\left( {7{x^3}{\text{ }} - {\text{ 4}}1{x^2}{\text{ }} + {\text{ }}60x} \right){\text{ }} - {\text{ }}\left( {28{x^2}{\text{ }} - {\text{ 164}}x{\text{ + 24}}0} \right){\text{ }} = {\text{ }}\left( {7{x^3}{\text{ }} - {\text{ }}59{x^2}{\text{ }} + {\text{ }}120x} \right){\text{ }} - {\text{ }}\left( {14{x^2}{\text{ }} - {\text{ 118}}x{\text{ + }}240} \right)\]

On solving the above equation. We get,

$\Rightarrow$ \[7{x^3}{\text{ }} - {\text{ 69}}{x^2}{\text{ + 224}}x{\text{ }} - {\text{ 24}}0{\text{ }} = {\text{ }}7{x^3}{\text{ }} - {\text{ 73}}{x^2}{\text{ }} + {\text{ }}238x{\text{ }} - {\text{ }}240\] (5)

Now subtracting RHS from the LHS of equation 5. We get,

$\Rightarrow$ \[4{x^2}{\text{ }} - {\text{ }}14x{\text{ }} = {\text{ }}0\]

Now we have to solve the above quadratic equation of x. To get the required value of x.

Taking x common from LHS of the above equation. We get,

$\Rightarrow$ x(4x – 14) = 0

So, from the above equation we get x = 0, \[\dfrac{{14}}{4}\].

Hence, the correct value of x that satisfies the given equation will be x = 0, \[\dfrac{7}{2}\].

So, the correct option will be D.

Note: Whenever we come up with this type of problem where we are given with an equation in which variable x is in the denominator of the equation. Then first we take the LCM of LHS and RHS of the given equation. And then we cross-multiply both sides of the equation. After that we try to manipulate the equation. So, that we can get a perfect polynomial equation. After that it will be easy to find the solution of a quadratic or cubic.

Last updated date: 03rd Oct 2023

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