Problem 2 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{aligned} y &=x^{2}+1 \\ x+y &=3 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system has solutions \((x, y) = (-2, 5)\) and \((x, y) = (1, 2)\).

Step 1: Substitute y in the second equation

We are given the first equation as \( y = x^2 +1 \). We will substitute \( y \) in the second equation \( x + y = 3 \) with \( x^2 + 1 \). Therefore, the new equation becomes \( x + (x^2 + 1) = 3 \).

Step 2: Simplify the equation

Simplify the substituted equation \( x + x^2 + 1 = 3 \). First, combine like terms and rearrange, which results in \( x^2 + x + 1 = 3 \) becoming \( x^2 + x - 2 = 0 \) by moving all terms to one side.

Step 3: Solve the quadratic equation

We now solve the equation \( x^2 + x - 2 = 0 \) by factoring or using the quadratic formula. Factoring, we find \( (x + 2)(x - 1) = 0 \), giving the solutions \( x = -2 \) and \( x = 1 \).

Step 4: Find corresponding values of y

For each value of \( x \), substitute back into \( y = x^2 + 1 \). When \( x = -2 \), \( y = (-2)^2 + 1 = 5 \). When \( x = 1 \), \( y = (1)^2 + 1 = 2 \).

Step 5: Solution of the system

Therefore, we have two solutions for the system of equations: \( (x, y) = (-2, 5) \) and \( (x, y) = (1, 2) \).

Key Concepts

System of Equations

A system of equations involves two or more equations that share the same set of variables. In this exercise, we are working with a system consisting of one linear equation and one quadratic equation. The solution to the system is the point or points where the equations intersect.

To solve a system of equations, we often use substitution or elimination. Substitution is particularly useful when one equation is easily solved for one variable, as in our given example.

• First, one equation is solved for a single variable. In this case, we use the equation for \( y \), which is already expressed as \( y = x^2 + 1 \).
• Next, this expression is substituted into the other equation, replacing the variable.
• This creates a single equation with one variable, simplifying the process to find solutions.

This method lets us break down complex systems into more manageable single-variable equations.

Quadratic Equation

A quadratic equation is any equation that can be rearranged into the form \( ax^2 + bx + c = 0 \). In this context, the equation we're solving is quadratic because of the \( x^2 \) term. Br>
Quadratic equations can have two solutions because the equation represents a parabola, which can intersect the x-axis at two, one, or no points. In our example, simplifying the system through substitution gives us the quadratic equation \( x^2 + x - 2 = 0 \).

• Quadratics can be solved by several methods, such as factoring, completing the square, or using the quadratic formula.
• Factorization is often the easiest method if the equation can be expressed as a product of binomials.

Being able to solve quadratic equations is essential since it appears in numerous mathematical and practical applications.

Factoring

Factoring involves breaking an expression down into simpler substances, or factors, that when multiplied together give the original expression. In our problem, we have the quadratic \( x^2 + x - 2 = 0 \). Factoring transforms this equation into a product: \((x + 2)(x - 1) = 0\).

The solutions for the equation are the values of \( x \) that make each factor zero.

• For \(x + 2 = 0\), the solution is \(x = -2\).
• For \(x - 1 = 0\), the solution is \(x = 1\).

Understanding how to factor is helpful as it provides an efficient way to solve quadratic equations and helps in understanding the underlying structure of mathematical expressions.

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating these symbols. In these exercises, we use algebraic skills to manipulate and solve for variables within equations.

In our exercise, algebra comes into play in several ways:

• Simplifying expressions by combining like terms.
• Rearranging equations to form a standard quadratic equation format.
• Substituting expressions to find solutions.

Algebraic manipulation is crucial in solving not just systems of equations, but a diverse range of mathematical problems, from simple arithmetic to complex calculus contexts.