- Isolate \(x\) from the linear equation: \(x = \frac{3y+12}{2}\).
- Substitute into the quadratic equation to get \(9y^2 + 80y + 124 = 0\).
- Apply the quadratic formula to solve for \(y\).
- Final values: \(\boxed{y = -2, y = -\frac{62}{9}}\).
Explanation
- Problem Analysis
We are given the system of equations:
1) \(x^2 + 2y = 5\)
2) \(2x - 3y = 12\)
Our objective is to solve this system for \(y\). We have a quadratic equation and a linear equation, which means we can express \(x\) in terms of \(y\) using the second equation and then substitute it into the first.
- Substitution and Expansion
From the second equation, \(2x - 3y = 12\), we can isolate \(x\):
\(2x = 3y + 12\)
Dividing by 2 gives:
\(x = \frac{3y + 12}{2}\)
Now, we substitute this expression for \(x\) into the first equation:
\(\left(\frac{3y + 12}{2}\right)^2 + 2y = 5\)
Expanding the squared term:
\(\frac{(3y + 12)^2}{4} + 2y = 5\)
\(\frac{9y^2 + 72y + 144}{4} + 2y = 5\)
- Forming the Quadratic Equation
To eliminate the fraction, multiply every term by 4:
\(9y^2 + 72y + 144 + 8y = 20\)
Combining like terms:
\(9y^2 + 80y + 144 = 20\)
Subtract 20 from both sides to form a standard quadratic equation:
\(9y^2 + 80y + 124 = 0\)
- Solving the Quadratic Equation
Now we solve the quadratic equation \(9y^2 + 80y + 124 = 0\) for \(y\) using the quadratic formula:
\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Here, \(a = 9\), \(b = 80\), and \(c = 124\).
Substituting the values:
\(y = \frac{-80 \pm \sqrt{80^2 - 4(9)(124)}}{2(9)}\)
\(y = \frac{-80 \pm \sqrt{6400 - 4464}}{18}\)
\(y = \frac{-80 \pm \sqrt{1936}}{18}\)
Calculating the square root, we get \(\sqrt{1936} = 44\).
Thus, \(y = \frac{-80 \pm 44}{18}\).
We have two solutions:
1) \(y_1 = \frac{-80 + 44}{18} = \frac{-36}{18} = -2\)
2) \(y_2 = \frac{-80 - 44}{18} = \frac{-124}{18} = -\frac{62}{9}\)
The values of \(y\) are \(-2\) and \(-\frac{62}{9}\).
Examples
Systems of equations like these are fundamental in engineering and physics, such as finding the intersection points of a parabolic trajectory and a linear path. For example, if you are modeling the height of a projectile \(h(t) = -t^2 + 5\) and a target sensor moving at a linear rate \(s(t) = \frac{2}{3}t - 4\), solving their system helps determine the exact moments their positions align. Knowing how to solve for these variables is crucial for precise control and design in robotics, aviation, and civil engineering.