Use the substitution method to find all solutions of the system of equations. \[ \begin{array}{r}x^{2}+y^{2}=49 \\ y=2 x\end{array} \] smaller \( x \)-value \( \quad \begin{array}{r}(x, y)=(\square)\end{array} \) \( \begin{array}{ll}\text { larger } x \text {-value } & (x, y)=(\square)\end{array} \)

To solve the system of equations using the substitution method, we start with the equations: 1. \( x^2 + y^2 = 49 \) 2. \( y = 2x \) We can substitute \( y \) from the second equation into the first equation: \( x^2 + (2x)^2 = 49 \) This simplifies to: \( x^2 + 4x^2 = 49 \) \( 5x^2 = 49 \) Dividing both sides by 5 gives: \( x^2 = \frac{49}{5} \) Taking the square root of both sides, we find: \( x = \pm \sqrt{\frac{49}{5}} \) \( x = \pm \frac{7}{\sqrt{5}} = \pm \frac{7\sqrt{5}}{5} \) Next, we substitute these values back to find \( y \): For \( x = \frac{7\sqrt{5}}{5} \): \( y = 2x = 2 \cdot \frac{7\sqrt{5}}{5} = \frac{14\sqrt{5}}{5} \) For \( x = -\frac{7\sqrt{5}}{5} \): \( y = 2x = 2 \cdot -\frac{7\sqrt{5}}{5} = -\frac{14\sqrt{5}}{5} \) Thus, the solutions are: Smaller \( x \)-value: \( (x, y) = \left(-\frac{7\sqrt{5}}{5}, -\frac{14\sqrt{5}}{5}\right) \) Larger \( x \)-value: \( (x, y) = \left(\frac{7\sqrt{5}}{5}, \frac{14\sqrt{5}}{5}\right) \)