Chapter 11.6, Problem 31AYU

To determine

The solution of the system $\{\begin{array}{l}9x^2-8xy+4y^2=70\\ 3x+2y=10\end{array}$

Answer to Problem 31AYU

Solution:

The solutions are $\left(3,1/2\right)$ and $\left(1/3,9/2\right)$.

Explanation of Solution

Given information:

The system is $\{\begin{array}{l}9x^2-8xy+4y^2=70\\ 3x+2y=10\end{array}$.

Explanation:

Here, the system contains two variable $x$ and $y$.

The system is $\{\begin{array}{l}9x^2-8xy+4y^2=70\left(1\right)\\ 3x+2y=10\left(2\right)\end{array}$.

From the equation (1),

$9x^2-8xy+4y^2=70$

$9x^2+12xy+4y^2-20xy=70$

${\left(3x+2y\right)}^2-20xy=70$

Now from the equation (2),

$3x+2y=10$

$\Rightarrow {\left(3x+2y\right)}^2=100$

Substitute this value in the equation (1)

$100-20xy=70$

$-20xy=70-100$

$-20xy=-30$

$xy=3/2$

$\Rightarrow x=3/2y$

Substitute this value in the equation (2),

$3\left(3/2y\right)+2y=10$

$\Rightarrow \frac{9}{2y}+2y=10$

$\Rightarrow 9+4y^2=20y$

$\Rightarrow 4y^2-20y+9=0$

The solution of the equation $a{x}^2+bx+c=0$ is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a=4,b=-20,c=9$.

$\Rightarrow y=\frac{-\left(-20\right)\pm \sqrt{{\left(-20\right)}^2-4\times 4\times 9}}{2\times 4}$

$\Rightarrow y=\frac{20\pm \sqrt{400-144}}{8}$

$\Rightarrow y=\frac{20\pm \sqrt{256}}{8}$

$\Rightarrow y=\frac{20\pm 16}{8}$

$\Rightarrow y=\frac{36}{8}$ or $y=\frac{4}{8}$

$\Rightarrow y=\frac{9}{2}$ or $y=\frac{1}{2}$.

Now, for $y=1/2$,

$x=\frac{3}{2\left(1/2\right)}=3$

Now, for $y=9/2$,

$x=\frac{3}{2\left(9/2\right)}=\frac{1}{\left(3/1\right)}=\frac{1}{3}$.

Thus, the solutions are $\left(3,1/2\right)$ and $\left(1/3,9/2\right)$.