Question

How do you solve \[\dfrac{2}{3x}+\dfrac{2}{3}=\dfrac{8}{x+6}\]?

Answer 1

Hint: To solve the given equation we will need given properties, first is the addition of fractions. Sum of fractions \[\dfrac{a}{b}\] and \[\dfrac{c}{d}\] is evaluated as \[\Rightarrow \dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}\]. The expression \[(a+b)(c+d)\] can be written in expanded form as \[ac+ad+bc+bd\]. We should also remember that for a quadratic equation \[a{{x}^{2}}+bx+c=0\], here \[a,b,c\in \] Real numbers. Using the formula method, the roots of the equation are \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step by step answer:
The given equation is \[\dfrac{2}{3x}+\dfrac{2}{3}=\dfrac{8}{x+6}\], we have to find the value of x which satisfies this equality.
\[\Rightarrow \dfrac{2}{3x}+\dfrac{2}{3}=\dfrac{8}{x+6}\]
Using the addition of fractions in LHS of the above equation, we get
\[\Rightarrow \dfrac{2\times 3+2\times 3x}{3x\times 3}=\dfrac{8}{x+6}\]
\[\Rightarrow \dfrac{6+6x}{9x}=\dfrac{8}{x+6}\]
Multiplying \[9x(x+6)\] to both sides of the equation, we get,
\[\Rightarrow \left( \dfrac{6+6x}{9x} \right)\times 9x(x+6)=\left( \dfrac{8}{x+6} \right)\times 9x(x+6)\]
\[\Rightarrow (6+6x)(x+6)=8\times 9x\]
We know that the expression \[(a+b)(c+d)\] can be written in expanded form \[ac+ad+bc+bd\]. Using this method, the LHS of the above equation can be written as follows
\[\Rightarrow (6+6x)(x+6)=8\times 9x\]
\[\begin{align}
  & \Rightarrow 6x+6\times 6+6{{x}^{2}}+6x\times 6=72x \\
 & \Rightarrow 6{{x}^{2}}+42x+36=72x \\
\end{align}\]
\[\Rightarrow 6{{x}^{2}}+42x+36=72x\]
Subtracting \[72x\] from both sides of the above equation, we get
\[\begin{align}
  & \Rightarrow 6{{x}^{2}}+42x+36-72x=72x-72x \\
 & \Rightarrow 6{{x}^{2}}-30x+36=0 \\
\end{align}\]
Multiplying \[\dfrac{1}{6}\] to both sides of the equation, we get
\[\begin{align}
  & \Rightarrow \left( 6{{x}^{2}}-30x+36 \right)\dfrac{1}{6}=0\times \dfrac{1}{6} \\
 & \Rightarrow \left( 6{{x}^{2}} \right)\dfrac{1}{6}-\left( 30x \right)\dfrac{1}{6}+\left( 36 \right)\dfrac{1}{6}=0 \\
 & \Rightarrow {{x}^{2}}-5x+6=0 \\
\end{align}\]
Comparing with the general equation of quadratic \[a{{x}^{2}}+bx+c=0\], here \[a=1,b=-5\And c=6\]
So, using the formula method the roots of the equation are \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] , we substitute the value of a, b, and c here. We get
\[\Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 1\times 6}}{2\times 1}\]
\[\begin{align}
  & \Rightarrow x=\dfrac{5\pm \sqrt{25-24}}{2} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{1}}{2} \\
 & \Rightarrow x=\dfrac{5\pm 1}{2} \\
\end{align}\]
\[\Rightarrow x=\dfrac{5+1}{2}\] or \[x=\dfrac{5-1}{2}\]
\[\Rightarrow x=3\] or \[x=2\]

The solution of the given equation is \[x=3\] or \[x=2\].

Note: We can check whether our answer is correct or not by substituting the values of \[x\]in the given equation.
Substitute \[x=3\] in the given equation, LHS =\[\dfrac{2}{3\times 3}+\dfrac{2}{3}=\dfrac{2}{9}+\dfrac{2}{3}=\dfrac{8}{9}\] and RHS = \[\dfrac{8}{3+6}=\dfrac{8}{9}\]. \[\therefore LHS=RHS\], hence it is the solution of the equation.
Substitute \[x=2\] in the given equation, LHS = \[\dfrac{2}{3\times 2}+\dfrac{2}{3}=\dfrac{2}{6}+\dfrac{2}{3}=1\] and RHS = \[\dfrac{8}{2+6}=1\]. \[\therefore LHS=RHS\], hence it is the solution of the equation.