- Isolate \(s\) in the linear equation to get \(s = \frac{t+5}{3}\).
- Substitute into the quadratic equation to form \(19t^2 + 10t - 29 = 0\).
- Apply the quadratic formula to solve for \(t\).
- The values are \(t = \boxed{-\frac{29}{19}}\) and \(t = \boxed{1}\).
Explanation
- Expressing s in terms of t and substitution
We are given the system of equations:
1) \(s^2 + 2t^2 = 6\)
2) \(3s - t = 5\)
To solve for \(t\), we first express \(s\) in terms of \(t\) using the linear equation (2):
\(3s = t + 5 \implies s = \frac{t+5}{3}\).
We then substitute this expression for \(s\) into the quadratic equation (1):
\(\left(\frac{t+5}{3}\right)^2 + 2t^2 = 6\).
- Forming a quadratic equation in t
Expanding the squared term gives:
\(\frac{t^2 + 10t + 25}{9} + 2t^2 = 6\).
To eliminate the fraction, multiply the entire equation by 9:
\(t^2 + 10t + 25 + 18t^2 = 54\).
Combining like terms results in:
\(19t^2 + 10t + 25 - 54 = 0\),
\(19t^2 + 10t - 29 = 0\).
- Solving the quadratic equation for t
We use the quadratic formula \(t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\) with \(A = 19\), \(B = 10\), and \(C = -29\).
First, calculate the discriminant:
\(B^2 - 4AC = 10^2 - 4(19)(-29) = 100 + 2204 = 2304\).
The square root of the discriminant is \(\sqrt{2304} = 48\).
The solutions for \(t\) are:
\(t = \frac{-10 \pm 48}{38}\).
Calculating the two roots:
\(t_1 = \frac{-10 - 48}{38} = \frac{-58}{38} = -\frac{29}{19}\),
\(t_2 = \frac{-10 + 48}{38} = \frac{38}{38} = 1\).
- Conclusion
The two values for \(t\) are \(-\frac{29}{19}\) and \(1\). Ordering them from smallest to largest, we have \(t = -\frac{29}{19}\) and \(t = 1\). The system is thus solved. The final values are \(t = \boxed{-\frac{29}{19}}\) and \(t = \boxed{1}\).
Examples
This system of equations can be used to model the intersection of a circle (or ellipse) and a line in geometry. For example, if an object follows a path defined by the ellipse \(s^2+2t^2=6\) and a line of sensors detects its position along the path \(3s-t=5\), finding the values of \(t\) helps determine the specific moments or locations when the object crosses the sensor beam. Solving systems of non-linear equations is a fundamental skill in fields like engineering and physics for locating objects or predicting motion.