Question

solve by substitution. $xy = 5 \\ {5x}^{2} + {y}^{2}$

Answer 1

we have the system of equations

$\begin{gathered} xy=5 \\ y=\frac{5}{x}\text{ ----> equation 1} \\ 5x^2+y^2=30\text{ ----> equation 2} \\ \end{gathered}$

Solve by substitution

substitute equation 1 in equation 2

$5x^2+(\frac{5}{x})^2=30$

solve for x

$\begin{gathered} 5x^2+\frac{25}{x^2}=30 \\ \frac{5x^4+25}{x^2}=30 \\ \\ 5x^4+25=30x^2 \\ 5x^4-30x^2+25=0 \\ \end{gathered}$

Let

change of variable

$\begin{gathered} u^2=x^4 \\ u=x^2 \end{gathered}$

substitute

$\begin{gathered} 5x^4-30x^2+25=0 \\ 5u^2-30u+25=0 \end{gathered}$

Solve the quadratic equation

using the formula

a=5

b=-30

c=25

substitute

$u=\frac{-(-30)\pm\sqrt{-30^2-4(5)(25)}}{2(5)}$$u=\frac{30\pm20}{10}$

The values of u are

u=5 and u=1

Now solve for x

Remember that

$\begin{gathered} u=x^{2} \\ For\text{ u=1} \\ x^2=1 \\ x_1=1 \\ x_2=-1 \\ For\text{ u=5} \\ x^2=5 \\ x_3=\sqrt{5} \\ x_4=-\sqrt{5} \end{gathered}$

Find out the values of y

For each value of x find out the value of y

Remember that

$\begin{equation*} y=\frac{5}{x}\text{ } \end{equation*}$

For x_1

$\begin{gathered} x_1=1 \\ y_1=\frac{5}{1}=5 \end{gathered}$

The first solution is the point (1,5)

For x_2

$\begin{gathered} x_2=-1 \\ y_2=\frac{5}{-1}\text{ =-5} \end{gathered}$

The second solution is the point (-1,-5)

For x_3

$\begin{gathered} x_3=\sqrt{5} \\ y_3=\frac{5}{\sqrt{5}}\text{ } \\ \\ y_3=\sqrt{5} \\ The\text{ third solution is the point \lparen}\sqrt{5},\sqrt{5}) \end{gathered}$

For x_4

$\begin{gathered} x_4=-\sqrt{5} \\ y_4=\frac{5}{-\sqrt{5}}\text{ } \\ \\ y_4=-\sqrt{5} \\ The\text{ fourth solution is the point }(-\sqrt{5},-\sqrt{5}) \end{gathered}$