Question

How do you solve \[\dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=-\dfrac{20}{{{x}^{2}}-4x}\]?

Answer 1

Hint: To solve \[\dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=-\dfrac{20}{{{x}^{2}}-4x}\], we need to first consider the LHS. Take the LCM for the LHS and simplify the numerator and denominator with the help of distributive property, that is \[a\left( b+c \right)=ab+ac\]. Then, equate the simplified LHS with the RHS and multiply the whole equation by \[{{x}^{2}}-4x\] and cancel the common terms from the numerator and denominator. We get a quadratic equation with variable x.

Complete step by step solution:
According to the question, we are asked to solve \[\dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=-\dfrac{20}{{{x}^{2}}-4x}\].
We have been given the equation is \[\dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=-\dfrac{20}{{{x}^{2}}-4x}\]. --------(1)
First, we have to consider the LHS.
Take the LCM of both the terms in the LHS. We get
\[\Rightarrow \dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=\dfrac{\left( 2x+1 \right)\left( x-4 \right)-x\left( x+1 \right)}{x\left( x-4 \right)}\]
Let us use the distributive property \[a\left( b+c \right)=ab+ac\] in the denominator.
We get
\[\dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=\dfrac{\left( 2x+1 \right)\left( x-4 \right)-x\left( x+1 \right)}{{{x}^{2}}-4x}\]
Now, we need to simplify the numerator using the distributive property.
\[\Rightarrow \dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=\dfrac{2x\times x-4\times 2x+x-4-{{x}^{2}}-x}{{{x}^{2}}-4x}\]
On further simplifications, we get
\[\dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=\dfrac{2{{x}^{2}}-8x+x-4-{{x}^{2}}-x}{{{x}^{2}}-4x}\]
Let us group all the \[{{x}^{2}}\] terms, x terms and the constants.
\[\Rightarrow \dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=\dfrac{{{x}^{2}}\left( 2-1 \right)+x\left( -8+1-1 \right)-4}{{{x}^{2}}-4x}\]
On simplification, we get
\[\dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=\dfrac{{{x}^{2}}-8x-4}{{{x}^{2}}-4x}\]
Now, let us equate the simplified LHS with the RHS. We get
\[\dfrac{{{x}^{2}}-8x-4}{{{x}^{2}}-4x}=-\dfrac{20}{{{x}^{2}}-4x}\]
Let us now divide the whole equation by \[{{x}^{2}}-4x\].
\[\dfrac{{{x}^{2}}-8x-4}{{{x}^{2}}-4x}\times \left( {{x}^{2}}-4x \right)=-\dfrac{20}{{{x}^{2}}-4x}\times \left( {{x}^{2}}-4x \right)\]
We find that \[{{x}^{2}}-4x\] are common in both the numerator and denominator of the LHS and RHS. On cancelling \[{{x}^{2}}-4x\], we get
\[{{x}^{2}}-8x-4=-20\]
Now, let us add 20 on both the sides of the equation.
\[\Rightarrow {{x}^{2}}-8x-4+20=-20+20\]
We know that terms with same magnitude and opposite signs cancel out. On cancelling 20, we get
\[\Rightarrow {{x}^{2}}-8x-4+20=0\]
On further simplifications, we get
\[\Rightarrow {{x}^{2}}-8x+16=0\]
Now, we have got a quadratic equation. To find the value of x, we can use the quadratic formula.
For a quadratic equation \[a{{x}^{2}}+bx+c=0\], the values of x are
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Here, we find that a=1, b=-8 and c=16.
On substituting the values in the quadratic formula, we get
\[x=\dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\times 1\times 16}}{2\times 1}\]
On further simplifications, we get
\[x=\dfrac{8\pm \sqrt{{{\left( -8 \right)}^{2}}-4\times 1\times 16}}{2}\]
\[\Rightarrow x=\dfrac{8\pm \sqrt{{{\left( -8 \right)}^{2}}-4\times 16}}{2}\]
We know that \[4\times 16=64\] and \[{{\left( -8 \right)}^{2}}=64\].
Therefore, we get
\[x=\dfrac{8\pm \sqrt{64-64}}{2}\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 64, we get
\[x=\dfrac{8\pm \sqrt{0}}{2}\].
But we know that \[\sqrt{0}=0\].
\[\Rightarrow x=\dfrac{8}{2}\]
We can write x as \[x=\dfrac{4\times 2}{2}\].
On cancelling 2 from the numerator and denominator, we get
x=4

Hence, the value of x in the equation \[\dfrac{2x+1}{x}-\dfrac{x+1}{x-4}=-\dfrac{20}{{{x}^{2}}-4x}\] is 4.

Note: We should be very careful with taking LCM and cancelling the common terms. Avoid calculation mistakes based on sign conventions. It is advisable to first simplify the given function and then find the value of x.