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Solve the following equations:
$\begin{align}
  & 5{{y}^{2}}-7{{x}^{2}}=17 \\
 & 5xy-6{{x}^{2}}=6 \\
\end{align}$
This question has multiple correct options
(a) $x=\pm 2;y=\pm 3$
(b) $x=\pm 3;y=\pm 4$
(c) $x=\pm 1;y=\pm 2$
(d) $x=\pm 4;y=\pm 1$

Answer Verified Verified
Hint: We are given two equations from which we have to find the value of $x,y$. From the given two equations, eliminate the term ${{x}^{2}}$. Then, solve the obtained equation with $5xy-6{{x}^{2}}=6$ to get the value of $\dfrac{y}{x}$. Use this value of $\dfrac{y}{x}$ with the two given equations to get the answer.

We are given two equations having variables $x$ and $y$ and we have to find the value of these variables by solving the two given equations. Let us denote the two given equations as,
$\begin{align}
  & 5{{y}^{2}}-7{{x}^{2}}=17...........\left( 1 \right) \\
 & 5xy-6{{x}^{2}}=6..............\left( 2 \right) \\
\end{align}$
First, we will eliminate ${{x}^{2}}$ from these two equations. For that, we will multiply the equation $\left( 1 \right)$ with $6$ and multiply the equation $\left( 2 \right)$ with $7$. Then we will subtract these two equations.
$\begin{align}
  & 6\left( 5{{y}^{2}}-7{{x}^{2}} \right)-7\left( 5xy-6{{x}^{2}} \right)=17\times 6-6\times 7 \\
 & \Rightarrow 30{{y}^{2}}-42{{x}^{2}}-35xy+42{{x}^{2}}=102-42 \\
 & \Rightarrow 30{{y}^{2}}-35xy=60 \\
 & \Rightarrow 5\left( 6{{y}^{2}}-7xy \right)=5\left( 12 \right) \\
 & \Rightarrow 6{{y}^{2}}-7xy=12..............\left( 3 \right) \\
\end{align}$
Dividing equation $\left( 3 \right)$ and equation $\left( 2 \right)$, we get,
 $\begin{align}
  & \dfrac{6{{y}^{2}}-7xy}{5xy-6{{x}^{2}}}=\dfrac{12}{6} \\
 & \Rightarrow \dfrac{y\left( 6y-7x \right)}{x\left( 5y-6x \right)}=2 \\
\end{align}$
Dividing numerator and denominator with $x^2$ in the left side of the above equation, we get,
$\begin{align}
  & \dfrac{\dfrac{y}{x}\left( \dfrac{6y-7x}{x} \right)}{\left( \dfrac{5y-6x}{x} \right)}=2 \\
 & \Rightarrow \dfrac{\dfrac{y}{x}\left( 6\dfrac{y}{x}-7 \right)}{\left( 5\dfrac{y}{x}-6 \right)}=2 \\
\end{align}$
Let us substitute $\dfrac{y}{x}=t$ in the above equation.
$\begin{align}
  & \Rightarrow \dfrac{t\left( 6t-7 \right)}{\left( 5t-6 \right)}=2 \\
 & \Rightarrow 6{{t}^{2}}-7t=2\left( 5t-6 \right) \\
 & \Rightarrow 6{{t}^{2}}-7t=10t-12 \\
 & \Rightarrow 6{{t}^{2}}-17t+12=0 \\
\end{align}$
To solve this quadratic equation, we will use a quadratic formula. Let us assume a quadratic equation $a{{x}^{2}}+bx+c=0$. From the quadratic formula, the roots of this equation are given by,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Using this quadratic formula in the equation $6{{t}^{2}}-17t+12=0$, we get,
$\begin{align}
  & t=\dfrac{-\left( -17 \right)\pm \sqrt{{{\left( -17 \right)}^{2}}-4\left( 6 \right)\left( 12 \right)}}{2\left( 6 \right)} \\
 & \Rightarrow t=\dfrac{17\pm \sqrt{289-288}}{12} \\
 & \Rightarrow t=\dfrac{17\pm \sqrt{1}}{12} \\
 & t=\dfrac{17\pm 1}{12} \\
 & \Rightarrow t=\dfrac{18}{12},t=\dfrac{16}{12} \\
 & \Rightarrow t=\dfrac{3}{2},t=\dfrac{4}{3} \\
\end{align}$
Resubstituting $t=\dfrac{y}{x}$, we get,
$\dfrac{y}{x}=\dfrac{3}{2},\dfrac{y}{x}=\dfrac{4}{3}$
$\Rightarrow y=\dfrac{3x}{2},y=\dfrac{4x}{3}.............\left( 4 \right)$
Substituting both the values of $y$ from equation $\left( 4 \right)$ in equation $\left( 2 \right)$, we get,
$\begin{align}
  & 5x\left( \dfrac{3x}{2} \right)-6{{x}^{2}}=6,5x\left( \dfrac{4x}{3} \right)-6{{x}^{2}}=6 \\
 & \Rightarrow \left( \dfrac{15{{x}^{2}}}{2} \right)-6{{x}^{2}}=6,\left( \dfrac{20{{x}^{2}}}{3} \right)-6{{x}^{2}}=6 \\
 & \Rightarrow \dfrac{15{{x}^{2}}-12{{x}^{2}}}{2}=6,\dfrac{20{{x}^{2}}-18{{x}^{2}}}{3}=6 \\
 & \Rightarrow 3{{x}^{2}}=12,2{{x}^{2}}=18 \\
 & \Rightarrow {{x}^{2}}=4,{{x}^{2}}=9 \\
 & \Rightarrow x=\pm 2,x=\pm 3 \\
\end{align}$
Substituting $x=\pm 2$ in $y=\dfrac{3x}{2}$ and $x=\pm 3$ in $y=\dfrac{4x}{3}$, we get,
$y=\pm 3,y=\pm 4$
So, the two possible answers are $x=\pm 2;y=\pm 3$ and $x=\pm 3;y=\pm 4$.
Hence, the answer is option (a) and option (b).

Note: There is an alternative approach to solve this problem very quickly. We can try to substitute each option in the equation given in the question. Whichever option satisfies the equation, that option will be considered as an answer.
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