Solve the following equations:
\[\begin{align}
  & 2xy-4x+y=17 \\
 & 3yz+y-6z=62 \\
 & 6xz+3z+2x=29 \\
\end{align}\]
\[\begin{align}
  & A.\left( 2,5,\dfrac{3}{7} \right). \\
 & B.\left( 1,7,\dfrac{-11}{3} \right) \\
 & C.\left( \dfrac{1}{2},3,-2 \right) \\
 & D.\text{None of these} \\
\end{align}\]

Hint: Here, we are given three different equations. Hence, we can apply the substitution method here. We will find the value of y in terms of x and z from the first equation. Then, we will find the value of z in terms of x and y from the third equation. Then, we will put both values in the second equation to obtain an equation in terms of x only. Solving, we will get the value of x which can be put in previously obtained equations to find the value of y and z.

Complete step by step answer:
We are given three equations which are:
\[\begin{align}
  & 2xy-4x+y=17 \\
 & 3yz+y-6z=62 \\
 & 6xz+3z+2x=29 \\
\end{align}\]
Let us take first equation which is \[2xy-4x+y=17\]
Taking y on one side and rest terms on other side to find value of y in terms of x and z, we get:
\[\begin{align}
  & y\left( 2x+1 \right)=17+4x \\
 & \Rightarrow y=\dfrac{17+4x}{2x+1}\cdots \cdots \cdots \cdots \left( i \right) \\
\end{align}\]
Let us take third equation which is \[6xz+3z+2x=29\]
Taking z on one side and rest of the terms on other side to obtain value of z in terms of x and y, we get:
\[\begin{align}
  & z\left( 6x+3 \right)=29-2x \\
 & \Rightarrow z=\dfrac{29-2x}{3\left( 2x+1 \right)}\cdots \cdots \cdots \cdots \left( ii \right) \\
\end{align}\]
Now, let us take second equation which is \[3yz+y-6z=62\]
Substituting value of y and z from (i) and (ii) we get:
\[3\left( \dfrac{17+4x}{2x+1} \right)\left( \dfrac{29-2x}{3\left( 2x+1 \right)} \right)+\dfrac{17+4x}{2x+1}-6\left( \dfrac{29-2x}{3\left( 2x+1 \right)} \right)=62\]
Simplifying the above equation by taking LCM we get:
\[\dfrac{493-34x+116-8{{x}^{2}}+\left( 2x+1 \right)\left[ \left( 17+4x \right)-2\left( 29-2x \right) \right]}{{{\left( 2x+1 \right)}^{2}}}=62\]
Further simplifying the equation by taking ${{\left( 2x+1 \right)}^{2}}$ on right hand side we get:
\[\begin{align}
  & 493+82x-8{{x}^{2}}+\left( 2x+1 \right)\left( 8x-41 \right)=62{{\left( 2x+1 \right)}^{2}} \\
 & \Rightarrow 493+82x-8{{x}^{2}}+16{{x}^{2}}-82x+8x-41=248{{x}^{2}}+62+248x \\
 & \Rightarrow 240{{x}^{2}}+240x-480=0 \\
\end{align}\]
Dividing both sides by 240 we get:
\[{{x}^{2}}+x-2=0\]
As we can see, we have obtained a quadratic equation in x. Hence, we will find two values of x using splitting the middle term method. In splitting the middle term, as we can see 2, -1 are two numbers whose sum is 1 and product is -2 therefore, x can be written as 2x-x.
\[{{x}^{2}}+2x-x-2=0\]
Taking x common from first two terms and -1 common from last two terms, we get:
\[\begin{align}
  & x\left( x+2 \right)-1\left( x+2 \right)=0 \\
 & \Rightarrow \left( x-1 \right)\left( x+2 \right)=0 \\
\end{align}\]
Since, two factor are equal to zero therefore, either $\left( x-1 \right)=0\Rightarrow \left( x+2 \right)=0$
\[\begin{align}
  & x-1=0\Rightarrow x+2=0 \\
 & x=1\Rightarrow x=-2 \\
\end{align}\]
We have obtained two values of x, which are 1 and -2. Similarly, we will obtain two values of y and z as well.
Let us substitute both values of x one by one in (i) to obtain two values of y which are:
\[\begin{align}
  & y=\dfrac{17+4\left( 1 \right)}{2\left( 1 \right)+1}\text{ and }y=\dfrac{17+4\left( -2 \right)}{2\left( -2 \right)+1} \\
 & \Rightarrow y=\dfrac{21}{3}\text{ and }y=\dfrac{9}{-3} \\
 & \Rightarrow y=7\text{ and }y=-3 \\
\end{align}\]
Now let us substitute both values of x one by one in (ii) to obtain two values of z which are:
\[\begin{align}
  & z=\dfrac{29-2\left( 1 \right)}{3\left( 2\left( 1 \right)+1 \right)}\text{ and }z=\dfrac{29-2\left( -2 \right)}{3\left( 2\left( -2 \right)+1 \right)} \\
 & \Rightarrow z=\dfrac{27}{9}\text{ and }z=\dfrac{33}{-9} \\
 & \Rightarrow z=3\text{ and }z=\dfrac{11}{-3} \\
\end{align}\]
As we can see, we have obtained two pair of solutions which can satisfy given equation that are $\left( 1,7,3 \right)\text{ and }\left( -2,-3,\dfrac{-11}{3} \right)$
As there is no option which satisfies all three equations, hence D. None of these is the correct option.

Note:
 Students should take care while substituting values of y and z in other equations as the calculation part is a little tricky and difficult. We can solve these equations using the elimination method also.