Question Find all solutions, including complex, for the nonlinear system: $7x^2 - 9y^2 = 42$ , $21x^2 + 5y^2 = 126$ .

Studdy Solution

STEP 1

Assumptions
1. We have a system of two nonlinear equations with two variables, $x$ and $y$ .
2. The system includes real and nonreal complex solutions.
3. The equations are: \begin{align} 7x^2 - 9y^2 &= 42, \\ 21x^2 + 5y^2 &= 126. \end{align}

STEP 2

First, we will express each equation in terms of $x^2$ and $y^2$ to facilitate elimination.
For the first equation: $x^2 = \frac{42 + 9y^2}{7}$
For the second equation: $x^2 = \frac{126 - 5y^2}{21}$

STEP 3

Now, set the two expressions for $x^2$ equal to each other to eliminate $x$ and solve for $y^2$ .
$\frac{42 + 9y^2}{7} = \frac{126 - 5y^2}{21}$

STEP 4

Multiply both sides by the least common multiple of the denominators, which is $21$ , to clear the fractions.
$21 \cdot \frac{42 + 9y^2}{7} = 21 \cdot \frac{126 - 5y^2}{21}$

STEP 5

Simplify both sides of the equation.
$3(42 + 9y^2) = 126 - 5y^2$

STEP 6

Expand the left side of the equation.
$126 + 27y^2 = 126 - 5y^2$

STEP 7

Move all terms involving $y^2$ to one side of the equation and constant terms to the other side.
$27y^2 + 5y^2 = 126 - 126$

STEP 8

Combine like terms.
$32y^2 = 0$

STEP 9

Solve for $y^2$ .
$y^2 = \frac{0}{32}$

STEP 10

Simplify the expression for $y^2$ .
$y^2 = 0$

STEP 11

Since $y^2 = 0$ , we can find the value of $y$ .
$y = \pm\sqrt{0}$

STEP 12

Calculate the value of $y$ .
$y = 0$

STEP 13

Now that we have the value of $y$ , we can substitute it back into one of the original equations to solve for $x^2$ . We will use the first equation.
$7x^2 - 9(0)^2 = 42$

STEP 14

Simplify the equation.
$7x^2 = 42$

STEP 15

Solve for $x^2$ .
$x^2 = \frac{42}{7}$

STEP 16

Calculate the value of $x^2$ .
$x^2 = 6$

STEP 17

Find the value of $x$ .
$x = \pm\sqrt{6}$

STEP 18

Calculate the values of $x$ .
$x = \sqrt{6}, \quad x = -\sqrt{6}$

STEP 19

Combine the values of $x$ and $y$ to give the solutions to the system of equations.
The solutions are: $(x, y) = (\sqrt{6}, 0), \quad (x, y) = (-\sqrt{6}, 0)$

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