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
363.3k+ views
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.
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.
Last updated date: 24th Sep 2023
•
Total views: 363.3k
•
Views today: 5.63k
Recently Updated Pages
What do you mean by public facilities

Slogan on Noise Pollution

Paragraph on Friendship

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

What is the Full Form of ILO, UNICEF and UNESCO

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Difference Between Plant Cell and Animal Cell

What is the basic unit of classification class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

One cusec is equal to how many liters class 8 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers
