91Ó°ÊÓ

Linear equations are equations that create straight lines when graphed. They have variables to the first power and look like the equation \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. In the given problem, the second equation \(x - 2y = -2\) is a linear equation. This equation is simple and provides a great starting point for solving the system.
To solve linear equations, we usually solve for one variable and use that expression to find the value of another variable later on. This is why we first chose this equation, making it easiest to solve for \(x\) in terms of \(y\), resulting in \(x = 2y - 2\).
Linear equations form the backbone of many algebraic concepts and are straightforward compared to other equation types.

Quadratic equations involve variables raised to the second power, represented by \(ax^2 + bx + c = 0\). They form parabolic shapes when graphed. In the problem, the first equation \(2x^2 - 3y = 2\) can be rewritten and expressed in terms of \(y\) after substituting \(x\).
In this process, the equation is simplified to a quadratic: \(8y^2 - 11y - 6 = 0\). Quadratics often have up to two real solutions, which require specific methods like the quadratic formula to solve.
Understanding how to transform more complex equations into quadratic forms is a valuable skill in algebra, as it allows you to use structured methods for finding exact solutions.

The substitution method is a powerful technique in solving systems of equations. It involves solving one equation for a variable and substituting this expression into another equation. This helps reduce the complexity of solving multiple equations simultaneously.
In the provided solution, we first isolated \(x\) from the linear equation and used that expression to replace \(x\) in the quadratic equation. This gave us a single equation in terms of \(y\), making it much easier to solve.
Substitution is particularly useful when one equation is simpler and can be easily rearranged, aiding in methods like solving quadratic equations through known formulas.

In algebra, solutions to equations represent values that satisfy the equations simultaneously. When solving systems of equations like the one provided, we often look for real solutions, meaning solutions that are real numbers as opposed to complex numbers.
Real solutions can be determined by checking each potential solution against the original equations to ensure that they hold true. In the provided problem, using the quadratic formula, we calculate potential values for \(y\). Each possible \(y\) value is then substituted back to find corresponding \(x\) values.
Assessing the solutions in this manner confirms whether the pair of values indeed provides real solutions to the system or reveals if no real solution exists.