Question

Solve the following equations:
\[2xy-4x+y=17,3yz+y-6z=52,6xz+3z+2x=29\]

Answer 1

Hint: We have the equations, \[2xy-4x+y=17,3yz+y-6z=52\] , and \[6xz+3z+2x=29\] .
Add -2 in LHS and RHS of the equation \[\left( 2xy-4x+y=17 \right)\] and then factorize it. Similarly, add -2 in LHS and RHS of the equation \[\left( 3yz+y-6z=52 \right)\] and then factorize it. Now, add 1 in LHS and RHS of the equation \[\left( 6xz+3z+2x=29 \right)\] and then factorize it. Now, multiply the factors of the equations \[2xy-4x+y=17,3yz+y-6z=52\] , and \[6xz+3z+2x=29\] , and then simplify it. After simplifying, use the factors of the equation \[\left( 2xy-4x+y=17 \right)\] and get the value of z. Similarly, solve it further and get the values of x and y.

Complete step-by-step answer:
According to the question, we have three equations which are,
\[2xy-4x+y=17\] ……………………(1)
\[3yz+y-6z=52\] ………………………(2)
\[6xz+3z+2x=29\] …………………….(3)
Now, simplifying equation (1), we get
\[\begin{align}
  & 2xy-4x+y=17 \\
 & \Rightarrow 2x\left( y-2 \right)+y=17 \\
\end{align}\]
Adding -2 in LHS and RHS of the above equation,
\[\begin{align}
  & \Rightarrow 2x\left( y-2 \right)+y-2=17-2 \\
 & \Rightarrow 2x\left( y-2 \right)+1\left( y-2 \right)=17-2 \\
\end{align}\]
Taking \[\left( y-2 \right)\] as common, we get
\[\Rightarrow 2x\left( y-2 \right)+1\left( y-2 \right)=17-2\]
\[\Rightarrow \left( 2x+1 \right)\left( y-2 \right)=15\] …………………………(4)
Now, simplifying equation (2), we get
\[\begin{align}
  & 3yz+y-6z=52 \\
 & \Rightarrow 3z\left( y-2 \right)+y=52 \\
\end{align}\]
Adding -2 in LHS and RHS of the above equation,
\[\begin{align}
  & \Rightarrow 3z\left( y-2 \right)+y-2=52-2 \\
 & \Rightarrow 3z\left( y-2 \right)+1\left( y-2 \right)=50 \\
\end{align}\]
Taking \[\left( y-2 \right)\] as common, we get
\[\Rightarrow 3z\left( y-2 \right)+1\left( y-2 \right)=50\]
\[\Rightarrow \left( 3z+1 \right)\left( y-2 \right)=50\] ……………………….(5)
Now, simplifying equation (3), we get
\[\begin{align}
  & 6xz+3z+2x=29 \\
 & \Rightarrow 3z\left( 2x+1 \right)+2x=29 \\
\end{align}\]
Adding 1 in LHS and RHS of the above equation,
\[\begin{align}
  & \Rightarrow 3z\left( 2x+1 \right)+2x+1=29+1 \\
 & \Rightarrow 3z\left( 2x+1 \right)+1\left( 2x+1 \right)=30 \\
\end{align}\]
Taking \[\left( 2x+1 \right)\] as common, we get
\[\Rightarrow 3z\left( 2x+1 \right)+1\left( 2x+1 \right)=30\]
\[\Rightarrow \left( 3z+1 \right)\left( 2x+1 \right)=30\] ……………………..(6)
Now, multiplying equation (4), equation (5), and equation (6), we get
\[\begin{align}
  & \Rightarrow \left( 2x+1 \right)\left( y-2 \right)\times \left( 3z+1 \right)\left( y-2 \right)\times \left( 3z+1 \right)\left( 2x+1 \right)=50\times 15\times 30 \\
 & \Rightarrow {{\left( 2x+1 \right)}^{2}}{{\left( y-2 \right)}^{2}}{{\left( 3z+1 \right)}^{2}}=50\times 15\times 30 \\
\end{align}\]
Taking square root in LHS and RHS of the above equation,
\[\begin{align}
  & \Rightarrow {{\left( 2x+1 \right)}^{2}}{{\left( y-2 \right)}^{2}}{{\left( 3z+1 \right)}^{2}}=50\times 15\times 30 \\
 & \Rightarrow \left( 2x+1 \right)\left( y-2 \right)\left( 3z+1 \right)=\sqrt{50\times 15\times 30} \\
 & \Rightarrow \left( 2x+1 \right)\left( y-2 \right)\left( 3z+1 \right)=\sqrt{50\times 15\times 15\times 2} \\
 & \Rightarrow \left( 2x+1 \right)\left( y-2 \right)\left( 3z+1 \right)=\sqrt{{{\left( 10 \right)}^{2}}{{\left( 15 \right)}^{2}}} \\
\end{align}\]
\[\Rightarrow \left( 2x+1 \right)\left( y-2 \right)\left( 3z+1 \right)=\pm 150\] ………………………(7)
From equation (4), we have \[\left( 2x+1 \right)\left( y-2 \right)=15\] .
Now, from equation (4) and equation (7), we have
\[\begin{align}
  & \left( 2x+1 \right)\left( y-2 \right)\left( 3z+1 \right)=\pm 150 \\
 & \Rightarrow 15\left( 3z+1 \right)=\pm 150 \\
 & \Rightarrow \left( 3z+1 \right)=\dfrac{\pm 150}{15} \\
 & \Rightarrow \left( 3z+1 \right)=\pm 10 \\
\end{align}\]
So, \[\left( 3z+1 \right)=10\] and \[\left( 3z+1 \right)=-10\] .
Taking \[\left( 3z+1 \right)=10\] and solving it we get,
\[\begin{align}
  & \left( 3z+1 \right)=10 \\
 & \Rightarrow 3z=10-1 \\
 & \Rightarrow 3z=9 \\
\end{align}\]
\[\Rightarrow z=3\] …………….(8)
Taking \[\left( 3z+1 \right)=-10\] and solving it we get,
\[\begin{align}
  & \left( 3z+1 \right)=-10 \\
 & \Rightarrow 3z=-10-1 \\
 & \Rightarrow 3z=-11 \\
\end{align}\]
\[\Rightarrow z=\dfrac{-11}{3}\] ………………..(9)
Now, from equation (5) and equation (7), we have
\[\begin{align}
  & \left( 2x+1 \right)\left( y-2 \right)\left( 3z+1 \right)=\pm 150 \\
 & \Rightarrow \left( 2x+1 \right)50=\pm 150 \\
 & \Rightarrow \left( 2x+1 \right)=\dfrac{\pm 150}{50} \\
 & \Rightarrow \left( 2x+1 \right)=\pm 3 \\
\end{align}\]
So, \[\left( 2x+1 \right)=3\] and \[\left( 2x+1 \right)=-3\] .
Taking \[\left( 2x+1 \right)=3\] and solving it we get,
\[\begin{align}
  & \left( 2x+1 \right)=3 \\
 & \Rightarrow 2x=3-1 \\
 & \Rightarrow 2x=2 \\
\end{align}\]
\[\Rightarrow x=1\] …………….(10)
Taking \[\left( 2x+1 \right)=-3\] and solving it we get,
\[\begin{align}
  & \left( 2x+1 \right)=-3 \\
 & \Rightarrow 2x=-3-1 \\
 & \Rightarrow 2x=-4 \\
\end{align}\]
\[\Rightarrow x=-2\] ………………..(11)
Now, from equation (6) and equation (7), we have
\[\begin{align}
  & \left( 2x+1 \right)\left( y-2 \right)\left( 3z+1 \right)=\pm 150 \\
 & \Rightarrow \left( y-2 \right)30=\pm 150 \\
 & \Rightarrow \left( y-2 \right)=\dfrac{\pm 150}{30} \\
 & \Rightarrow \left( y-2 \right)=\pm 5 \\
\end{align}\]
So, \[\left( y-2 \right)=5\] and \[\left( y-2 \right)=-5\] .
Taking \[\left( y-2 \right)=5\] and solving it we get,
\[\begin{align}
  & \left( y-2 \right)=5 \\
 & \Rightarrow y-2=5 \\
 & \Rightarrow y=5+2 \\
\end{align}\]
\[\Rightarrow y=7\] …………….(12)
Taking \[\left( y-2 \right)=-5\] and solving it we get,
\[\begin{align}
  & \left( y-2 \right)=5 \\
 & \Rightarrow y-2=-5-1 \\
 & \Rightarrow y=-5+2 \\
\end{align}\]
\[\Rightarrow y=-3\] ………………..(13)
Hence, the solution sets are \[x=1,y=7,z=3\text{ or }x=-2,y=-3,z=\dfrac{-11}{3}\] .

Note: In this question, one might think to get the value of y from equation (2) and put it in equation (3). Now, we have two equations in the variables x and y and two equations. One can think to get the value of x from one equation and substitute the value of x in other equations.