91Ó°ÊÓ

A quadratic equation is an equation of the form \[\begin{equation} ax^2 + bx + c = 0 \end{equation}\] where ‘a’, ‘b’, and ‘c’ are constants and 'a' is not equal to zero. To solve such equations, you may use the quadratic formula: \[\begin{equation} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{equation}\] This formula helps find the roots (solutions) of the equation. In the exercise, the equation to solve was \[\begin{equation} 5y^2 - 11y + 6 = 0 \end{equation}\] Plugging 'a', 'b', and 'c' into the formula, we get \[\begin{equation} y = \frac{11 \pm \sqrt{121 - 120}}{10} \end{equation}\]. This results in two possible values: 1.2 and 0.6. Quadratic equations can have 0, 1, or 2 real roots, and understanding how to apply the quadratic formula is crucial for solving these kinds of problems.

The substitution method is a straightforward technique for solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation. In our exercise, after forming the linear equation, we solve for x in terms of y: \[\begin{equation} x = 5 - 3y \end{equation}\]. This expression is then substituted back into one of the original equations to find the value of y. This method works well when one of the equations is already solved for one variable or can be easily rearranged. It simplifies the system, making it easier to solve.

Expanding equations involves removing parentheses in mathematical expressions. This helps simplify and combine like terms. In the given exercise, both equations were expanded: Expand \[\begin{equation} (x-1)^{2} + (y+1)^{2} = 5 \end{equation}\] to \[\begin{equation} x^{2} - 2x + 1 + y^{2} + 2y + 1 = 5 \end{equation}\], and \[\begin{equation} x^{2} + (y+4)^{2} = 29 \end{equation}\] to \[\begin{equation} x^{2} + y^{2} + 8y + 16 = 29 \end{equation}\]. This is an essential skill in algebra, as it transforms complex expressions into forms that are easier to manipulate.

A linear equation is an equation of the first degree, meaning it has no exponents greater than 1. In our exercise, we derived the linear equation \[\begin{equation} x + 3y = 5 \end{equation}\]. Linear equations can be graphed as straight lines and usually take the form \[\begin{equation} ax + by = c \end{equation}\]. Solving linear equations often involves simple rearrangement and algebraic manipulation. They form the foundation of more complex systems of equations in algebra.

A system of nonlinear equations includes at least one equation that is not linear. In our exercise, both equations involved terms squared, making them nonlinear: \[\begin{equation} (x-1)^{2}+(y+1)^{2}=5 \end{equation}\] and \[\begin{equation} x^{2}+(y+4)^{2}=29 \end{equation}\]. Solving such a system requires methods that handle the complexity of quadratic and other nonlinear terms. Techniques like substitution, elimination, or using matrices can be helpful. Nonlinear systems can have multiple or no solutions, and understanding how to approach them is essential for higher-level algebra and calculus.