Problem 81 - College Algebra with Modeling and Visualization

Question

Solve each system. $$ \begin{aligned} &x^{2}=2 y\\\ &x^{2}=y+1 \end{aligned} $$

Step-by-Step Solution

Answer

The solutions are \((\sqrt{2}, 1)\) and \((-\sqrt{2}, 1)\).

Step 1: Express Both Equations in Terms of y

The system of equations given is \( x^2 = 2y \) and \( x^2 = y + 1 \). Since both equations equal \( x^2 \), we can set them equal to each other. Thus, we have: \[ 2y = y + 1 \] This gives us a simpler equation to solve for \( y \).

Step 2: Solve for y

From the equation \( 2y = y + 1 \), subtract \( y \) from both sides to isolate \( y \):\[ 2y - y = y + 1 - y \]\[ y = 1 \]Now, we have found that \( y = 1 \).

Step 3: Substitute y into Original Equation to Solve for x

Substitute \( y = 1 \) back into one of the original equations, such as \( x^2 = 2y \):\[ x^2 = 2(1) \]\[ x^2 = 2 \]Take the square root of both sides to solve for \( x \):\[ x = \pm \sqrt{2} \]

Step 4: Write the Solution as Ordered Pairs

The solutions for \( x \) and \( y \) are two cases since \( x = \pm \sqrt{2} \). Thus, the system of equations has the solutions as ordered pairs: \[ (\sqrt{2}, 1) \] and \[ (-\sqrt{2}, 1) \].

Key Concepts

Substitution Method

The substitution method is a technique for solving systems of equations where one equation is solved for one variable and then this expression is substituted into the other equation. In the case presented, both equations are equivalent to eliminating the need to express one variable explicitly. Instead, we directly equate the equations to simplify our system.

To use the substitution method effectively, follow these steps:

• Identify the equations you want to work with (in this problem: \( x^2 = 2y \) and \( x^2 = y + 1 \)).
• Set the equations equal to each other since both contain \( x^2 \) (i.e., \( 2y = y + 1 \)).
• Solve the resulting simpler equation to find the value of one variable (here, \( y = 1 \)).

This method helps streamline the process of solving systems by reducing it to simpler algebraic operations. It's particularly useful for systems where expression for substitution can easily be manipulated.

Solving Quadratic Equations

Quadratic equations form an essential part of solving systems that involve second-degree variables. In our solution, finding \( x \) involves solving the quadratic equation \( x^2 = 2 \) once \( y \) is determined.

Here's how you approach solving a simple quadratic equation like \( x^2 = 2 \):

• Isolate the quadratic term if it's not already isolated (here, it already is as \( x^2 \)).
• Take the square root of both sides to solve for the variable (\( x = \pm \sqrt{2} \)).

Remember, quadratic equations can have two solutions (as in this case: \( x = \sqrt{2} \) and \( x = -\sqrt{2} \)) since both the positive and negative roots satisfy \( x^2 = 2 \). Quadratic equations often yield two solutions, accounting for scenarios in both directions of the quadratic curve.

Ordered Pairs Solution

An ordered pair solution is a way of expressing the precise values of variables that satisfy a system of equations. Once individual variables have been solved, results are typically presented as ordered pairs, following the format \((x, y)\).

For our equation system:

• After determining the solutions for \( x \) and \( y \) (\( y = 1 \), \( x = \pm \sqrt{2} \)), we combine them into pairs.
• The pairs \((\sqrt{2}, 1)\) and \((-\sqrt{2}, 1)\) encompass the solutions for the system, reflecting the relationship between both variables that achieves equality in both original equations.

Ordered pairs give clarity to solutions, particularly in graphs, where they represent points on a coordinate plane. This ensures any graphical interpretation aligns with the numeric solutions, presenting a comprehensive picture of the solution set for the system.