Problem 22 - Precalculus: Functions and Graphs

Question

Use the method of substitution to solve the system. $$\left\\{\begin{aligned} x &=y^{2}-4 y+5 \\ x-y &=1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \((x, y) = (5, 4)\) and \((x, y) = (2, 1)\).

Step 1: Express First Equation

Start by observing the first equation: \[ x = y^2 - 4y + 5 \]This equation expresses \(x\) in terms of \(y\). We will use this expression to substitute \(x\) in the second equation.

Step 2: Substitute in the Second Equation

Since \(x = y^2 - 4y + 5\), substitute this expression for \(x\) in the second equation:\[ x - y = 1 \]which becomes:\[ (y^2 - 4y + 5) - y = 1 \]Simplify to solve for \(y\).

Step 3: Simplify and Solve for y

Simplify the equation:\[ y^2 - 4y + 5 - y = 1 \]Combine like terms:\[ y^2 - 5y + 5 = 1 \]Subtract 1 from both sides:\[ y^2 - 5y + 4 = 0 \]Solve the quadratic equation.

Step 4: Solve Quadratic Equation

The quadratic equation \( y^2 - 5y + 4 = 0 \) can be solved by factoring:\[ (y - 4)(y - 1) = 0 \]Thus, \( y = 4 \) or \( y = 1 \).

Step 5: Find Corresponding x Values

Use the values of \(y\) found in Step 4 to find corresponding \(x\) values using the expression \(x = y^2 - 4y + 5\):1. For \(y = 4\), \(x = 4^2 - 4(4) + 5 = 16 - 16 + 5 = 5\).2. For \(y = 1\), \(x = 1^2 - 4(1) + 5 = 1 - 4 + 5 = 2\).

Step 6: Verify Solutions

Verify each \((x, y)\) pair in both equations:1. For \((x, y) = (5, 4)\): - First equation: \(5 = 4^2 - 4(4) + 5\) is true. - Second equation: \(5 - 4 = 1\) is true.2. For \((x, y) = (2, 1)\): - First equation: \(2 = 1^2 - 4(1) + 5\) is true. - Second equation: \(2 - 1 = 1\) is true.Both pairs satisfy the system.

Key Concepts

System of Equations

When we talk about a system of equations, we are looking at a set of equations with multiple variables that are all interrelated. In this exercise, we have two equations involving two variables, \(x\) and \(y\): The first equation is \(x = y^2 - 4y + 5\) and the second equation is \(x - y = 1\). The goal is to find values of \(x\) and \(y\) that satisfy both equations simultaneously.

Systems of equations can be solved using various methods, such as graphing, substitution, or elimination. Here, the method of substitution is used. The idea is to solve one of the equations for one variable and use that expression to replace the corresponding variable in the other equation. This simplifies the process as it reduces the number of variables in one of the equations, making it easier to solve.

Quadratic Equation

Quadratic equations are polynomial equations of degree two. They generally have the form \(ax^2 + bx + c = 0\). In our solution, we encounter a quadratic equation after the substitution step: \(y^2 - 5y + 4 = 0\).

Quadratic equations can have up to two solutions, which represent the points where the parabola (the graph of the equation) crosses the \(x\)-axis. Solving this type of equation can be done using different methods, such as factoring, using the quadratic formula, or completing the square. In this scenario, factoring was chosen as the simplest and most straightforward method.

Factoring

Factoring involves writing the quadratic expression as a product of its binomials. To solve \(y^2 - 5y + 4 = 0\) through factoring, we express it as \((y - 4)(y - 1) = 0\).

This factorization is then set to zero, because for the entire expression to equal zero, at least one of the factors must be zero. Hence, we solve each factor set to zero:

• \(y - 4 = 0\), results in \(y = 4\)
• \(y - 1 = 0\), results in \(y = 1\)

These are the solutions for \(y\). Factoring allows us to break down complex expressions into simpler parts, making it easier to identify the solutions of the equation.

Verification of Solutions

Verification is a crucial step in any calculation as it ensures that the solutions calculated actually satisfy the original system of equations. It involves substituting the solutions back into the original equations to check for validity.

For this problem, two solutions were found, \(y = 4\) and \(y = 1\).

First, for \((x, y) = (5, 4)\):

• The first equation becomes \(5 = 4^2 - 4(4) + 5\) which simplifies correctly.
• The second equation shows \(5 - 4 = 1\), confirming our solutions are correct.

Then, for \((x, y) = (2, 1)\):

• The first equation becomes \(2 = 1^2 - 4(1) + 5\), which holds true.
• The second equation is \(2 - 1 = 1\), which is also satisfied.

By verifying both solution pairs, we ensure the integrity and accuracy of the answer, confirming that both pairs indeed solve the system.