Question

Solve for the value of $x,y$ and $z$ from the following equations:
$\begin{aligned} & {{y}^{2}}+yz+{{z}^{2}}=49......\left( 1 \right) \\\ & {{z}^{2}}+zx+{{x}^{2}}=19.......\left( 2 \right) \\\ & {{x}^{2}}+xy+{{y}^{2}}=39......\left( 3 \right) \\\ \end{aligned}$

Answer 1

Hint: For solving this question first we will try to get the linear equation in terms of the given three equations then we will solve to find the value of each variable.

Complete step-by-step answer:
Given:
We have three variables and three equations. The equations are written below:
$\begin{aligned} & {{y}^{2}}+yz+{{z}^{2}}=49......\left( 1 \right) \\\ & {{z}^{2}}+zx+{{x}^{2}}=19.......\left( 2 \right) \\\ & {{x}^{2}}+xy+{{y}^{2}}=39......\left( 3 \right) \\\ \end{aligned}$
Now, subtract the equation (2) from the equation (1). Then,

& \left( {{y}^{2}}+yz+{{z}^{2}} \right)-\left( {{z}^{2}}+zx+{{x}^{2}} \right)=49-19 \\\ & \Rightarrow {{y}^{2}}-{{x}^{2}}+yz-zx=30 \\\ & \Rightarrow \left( y-x \right)\left( y+x \right)+z\left( y-x \right)=30 \\\ & \Rightarrow \left( y-x \right)\left( x+y+z \right)=30..................\left( 4 \right) \\\ \end{aligned}$$ Now, subtract the equation (2) from the equation (3). Then, $\begin{aligned} & \left( {{x}^{2}}+xy+{{y}^{2}} \right)-\left( {{z}^{2}}+zx+{{x}^{2}} \right)=39-19 \\\ & \Rightarrow {{y}^{2}}-{{z}^{2}}+xy-zx=20 \\\ & \Rightarrow \left( y-z \right)\left( y+z \right)+x\left( y-z \right)=20 \\\ & \Rightarrow \left( y-z \right)\left( x+y+z \right)=20....................\left( 5 \right) \\\ \end{aligned}$ Now, divide the equation (4) by equation (5). Then, $\dfrac{\left( y-x \right)\left( x+y+z \right)}{\left( y-z \right)\left( x+y+z \right)}=\dfrac{30}{20}$ $\begin{aligned} & \Rightarrow \dfrac{\left( y-x \right)}{\left( y-z \right)}=\dfrac{3}{2} \\\ & \Rightarrow 2y-2x=3y-3z \\\ & \Rightarrow 3z-2x=y \\\ & \Rightarrow y=3z-2x.................\left( 6 \right) \\\ \end{aligned}$ Now, substitute $y=3z-2x$ from equation (6) into equation (1). Then, $\begin{aligned} & {{y}^{2}}+yz+{{z}^{2}}=49 \\\ & \Rightarrow {{\left( 3z-2x \right)}^{2}}+\left( 3z-2x \right)z+{{z}^{2}}=49 \\\ & \Rightarrow 9{{z}^{2}}+4{{x}^{2}}-12zx+3{{z}^{2}}-2zx+{{z}^{2}}=49 \\\ & \Rightarrow 13{{z}^{2}}+4{{x}^{2}}-14zx=49.................\left( 7 \right) \\\ \end{aligned}$ Now, let $z=mx$ . So, substituting $z=mx$ equation (7) and equation (2). Then, $\begin{aligned} & 13{{z}^{2}}+4{{x}^{2}}-14zx=49 \\\ & \Rightarrow 13{{m}^{2}}{{x}^{2}}+4{{x}^{2}}-14m{{x}^{2}}=49 \\\ & \Rightarrow {{x}^{2}}\left( 13{{m}^{2}}+4-14m \right)=49..............\left( 8 \right) \\\ & {{z}^{2}}+zx+{{x}^{2}}=19 \\\ & \Rightarrow {{m}^{2}}{{x}^{2}}+m{{x}^{2}}+{{x}^{2}}=19 \\\ & \Rightarrow {{x}^{2}}\left( {{m}^{2}}+m+1 \right)=19......................\left( 9 \right) \\\ \end{aligned}$ Now, divide the equation (8) by equation (9) and then solve for the value of $m$ . Then, $\begin{aligned} & \dfrac{{{x}^{2}}\left( 13{{m}^{2}}+4-14m \right)}{{{x}^{2}}\left( {{m}^{2}}+m+1 \right)}=\dfrac{49}{19} \\\ & \Rightarrow 247{{m}^{2}}+76-266m=49{{m}^{2}}+49m+49 \\\ & \Rightarrow 198{{m}^{2}}-315m+27=0 \\\ & \Rightarrow 22{{m}^{2}}-35m+3=0 \\\ & \Rightarrow 22{{m}^{2}}-33m-2m+3=0 \\\ & \Rightarrow 11m\left( 2m-3 \right)-\left( 2m-3 \right)=0 \\\ & \Rightarrow \left( 11m-1 \right)\left( 2m-3 \right)=0 \\\ & \Rightarrow m=\dfrac{1}{11},\dfrac{3}{2} \\\ \end{aligned}$ Now, we got two values of $m$ . So, we can substitute these values in equation (8) and equation (9) to get the value of $x$ . Now, substituting $m=\dfrac{3}{2}$ in equation (8) and equation (9). Then, $\begin{aligned} & {{x}^{2}}\left( 13{{m}^{2}}+4-14m \right)=49 \\\ & \Rightarrow {{x}^{2}}\left( 13\times \dfrac{9}{4}+4-14\times \dfrac{3}{2} \right)=49 \\\ & \Rightarrow {{x}^{2}}\times \dfrac{49}{4}=49 \\\ & \Rightarrow {{x}^{2}}=4 \\\ & \Rightarrow x=\pm 2 \\\ & {{x}^{2}}\left( {{m}^{2}}+m+1 \right)=19 \\\ & \Rightarrow {{x}^{2}}\left( \dfrac{9}{4}+\dfrac{3}{2}+1 \right)=19 \\\ & \Rightarrow {{x}^{2}}\times \dfrac{19}{4}=19 \\\ & \Rightarrow {{x}^{2}}=4 \\\ & \Rightarrow x=\pm 2 \\\ \end{aligned}$ Now, as for $m=\dfrac{3}{2}$ the value of $x=\pm 2$ . So, $z=mx=\pm 3$ . Now, when we put $x=\pm 2$ and $z=\pm 3$ in equation (6) to find the value of $y$ . Then, $\begin{aligned} & y=3z-2x \\\ & \Rightarrow y=9-4 \\\ & \Rightarrow y=5 \\\ & y=3z-2x \\\ & \Rightarrow y=-9+4 \\\ & \Rightarrow y=-5 \\\ \end{aligned}$ Thus, $x=\pm 2,z=\pm 3$ and $y=\pm 5$ will be the solution of the given set of equations. Now, substituting $m=\dfrac{1}{11}$ in equation (8) and equation (9). Then, $$\begin{aligned} & {{x}^{2}}\left( 13{{m}^{2}}+4-14m \right)=49 \\\ & \Rightarrow {{x}^{2}}\left( \dfrac{13}{121}+4-\dfrac{14}{11} \right)=49 \\\ & \Rightarrow {{x}^{2}}\times \dfrac{343}{121}=49 \\\ & \Rightarrow {{x}^{2}}=\dfrac{121}{7} \\\ & \Rightarrow x=\pm \dfrac{11}{\sqrt{7}} \\\ \end{aligned}$$ $$\begin{aligned} & {{x}^{2}}\left( {{m}^{2}}+m+1 \right)=19 \\\ & \Rightarrow {{x}^{2}}\left( \dfrac{1}{121}+\dfrac{1}{11}+1 \right)=19 \\\ & \Rightarrow {{x}^{2}}\times \dfrac{133}{121}=19 \\\ & \Rightarrow {{x}^{2}}=\dfrac{121}{7} \\\ & \Rightarrow x=\pm \dfrac{11}{\sqrt{7}} \\\ \end{aligned}$$ Now, as for $m=\dfrac{1}{11}$ the value of $x=\pm \dfrac{11}{\sqrt{7}}$ . So, $z=mx=\pm \dfrac{1}{\sqrt{7}}$ . Now, we put $x=\pm \dfrac{11}{\sqrt{7}}$ and $z=\pm \dfrac{1}{\sqrt{7}}$ in equation (6) to find the value of $y$ . Then, $\begin{aligned} & y=3z-2x \\\ & \Rightarrow y=\dfrac{3}{\sqrt{7}}-\dfrac{22}{\sqrt{7}} \\\ & \Rightarrow y=-\dfrac{19}{\sqrt{7}} \\\ & y=3z-2x \\\ & \Rightarrow y=-\dfrac{3}{\sqrt{7}}+\dfrac{22}{\sqrt{7}} \\\ & \Rightarrow y=\dfrac{19}{\sqrt{7}} \\\ \end{aligned}$ Thus, $x=\pm \dfrac{11}{\sqrt{7}},z=\pm \dfrac{1}{\sqrt{7}}$ and $y=\mp \dfrac{19}{\sqrt{7}}$ will be the solutions of the given set of equations. For the given set of equations we will have the following solutions: $\begin{aligned} & x=2,z=3,y=5 \\\ & x=-2,z=-3,y=-5 \\\ & x=\dfrac{11}{\sqrt{7}},z=\dfrac{1}{\sqrt{7}},y=-\dfrac{19}{\sqrt{7}} \\\ & x=-\dfrac{11}{\sqrt{7}},z=-\dfrac{1}{\sqrt{7}},y=\dfrac{19}{\sqrt{7}} \\\ \end{aligned}$ Note: Here, the student must proceed stepwise while solving the equations and must do the correct substitutions to get the correct answer easily. Moreover, avoid making calculation mistakes while solving the question.