91Ó°ÊÓ

Quadratic equations are a fundamental part of algebra, characterized by their distinctive form: \(ax^2 + bx + c = 0\), where \(a, b,\) and \(c\) are constants, and \(a eq 0\). These equations describe parabolas when plotted on a graph. In our exercise, the second equation \(y = x^2 - 5\) exhibits this typical parabolic form.

Solving quadratic equations often involves methods like factoring, completing the square, or the quadratic formula. The latter is quite versatile and used in our example:

• The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Here, \(a = 2\), \(b = 3\), and \(c = -20\).

The discriminant \((b^2 - 4ac)\) reveals the nature of roots:

• Positive discriminant: Two distinct real roots (as in our exercise).
• Zero discriminant: A repeated real root.
• Negative discriminant: No real roots, only complex ones.

Each solution derived from the quadratic equation corresponds to specific values for \(x\), that should also satisfy the first equation to solve the system fully.

The substitution method is an efficient technique for solving systems of equations, especially when one equation is already solved for a variable. This method is showcased in our exercise to capture both equations.

The essence of this method:

• Solve one equation for a specific variable, if not already done.
• Substitute this expression into the other equation, replacing the solved variable.
• Simplify and solve to find the value of the remaining variable.
• Use this determined value to find the corresponding value of the other variable from the initial substitution.

For instance, in our exercise:

• The expression \(y = x^2 - 5\) from the second equation is substituted into the first, \(3x + 2(x^2 - 5) = 10\).
• After simplification, a quadratic equation built in \(x\) is derived, which can thereafter be solved for \(x\).
• Once \(x\) is found, these values are traced back into \(y = x^2 - 5\) to determine corresponding \(y\) values. 

With its step-by-step analysis, the substitution method strongly reinforces understanding of how different equations relate through shared variables.

Linear equations form the bedrock of algebra, defined by their constant rate of change and a graph that extends in a straight line. Their standard form is \(ax + by = c\). In our exercise, the linear equation \(3x + 2y = 10\) highlights how simple arithmetic relationships can intersect with more complex quadratic relationships.

Steps for solving linear equations involve:

• Isolating a variable to solve for, often involving basic arithmetic like addition, subtraction, or division.
• Rearranging terms to consolidate variable expressions on one side and constants on the other.
• Once a variable is isolated, determining a numerical value becomes straightforward if it doesn't involve another variable from the system.

In systems of equations, solving a linear part can simplify the interaction between the different equations involved. In the exercise, the linear equation balances against the substituted quadratic form, guiding the steps to isolate solutions for \(x\). Solving this way illustrates essential algebra skills that underpin much heavier mathematical classes.