Problem 2 - College Algebra Essentials

Question

In Exercises \(1-18,\) solve each system by the substitution method. $$ \left\\{\begin{array}{l} x-y=-1 \\ y=x^{2}+1 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x=2, y=1\) or \(x=-1, y=-2\).

Step 1: Rewrite first equation

The first equation can be rewritten as \(x = y + 1\).

Step 2: Substitute into the second equation

By substituting the rewritten first equation into the second equation, we get: \(y = (y+1)^2 + 1 = y^2 + 2y + 1 + 1 = y^2 + 2y + 2\). After rearrangement, this turns into \(y^2 + y - 2 = 0\).

Step 3: Solve for y

Factoring the quadratic, we end up with \((y-1)(y+2) = 0\). Setting each factor equal to zero gives \(y = 1 \; and \; y = -2\).

Step 4: Find corresponding x values

Substitute \(y = 1\) into the first equation gives \(x = 1 + 1 = 2\). While substituting \(y = -2\) into the first equation results in \(x = -2 + 1 = -1\)

Step 5: Write final solution

The solution to the system of equations are the points \((2,1) \; and \; (-1, -2)\)

Key Concepts

Understanding System of Equations

A system of equations consists of two or more equations with the same set of unknowns. The goal in solving a system is to find values for these unknowns that satisfy all the equations simultaneously. In this exercise, we have a system of two equations:

• Linear Equation: \( x - y = -1 \)
• Quadratic Equation: \( y = x^2 + 1 \)

To solve the system, we need one solution for both equations, which means finding a common solution pair \((x, y)\). The substitution method is an effective approach, especially when one equation is easy to rearrange in terms of one variable, which we can then substitute into the other equation.
By using substitution, we reduce the number of variables in the equations, making it easier to solve.

Unpacking the Quadratic Equation

A quadratic equation is a second-degree polynomial equation of the form \( ax^2 + bx + c = 0 \). In this exercise, the substitution led to a quadratic equation: \( y^2 + y - 2 = 0 \), which was originally set up by inserting \( x = y + 1 \) into \( y = x^2 + 1 \). The characteristic U-shape of a quadratic function can result in two solutions, one solution, or no real solutions at all.
Quadratics appear in many contexts and often require techniques such as factoring, completing the square, or using the quadratic formula to solve. Here, we are going to factor the quadratic equation to find the solutions for \( y \).

The Art of Factoring

Factoring is breaking down a complex expression into products of simpler expressions. For quadratics like \( y^2 + y - 2 = 0 \), factoring involves identifying two numbers whose product is equal to the constant term \(-2\) and whose sum is equal to the linear coefficient \(1\). These numbers are \(2\) and \(-1\).
This allows us to express the quadratic equation as \((y - 1)(y + 2) = 0\). Solving these factors gives us the solutions for \( y \). Each factor is set to zero, producing the equations \( y - 1 = 0 \) and \( y + 2 = 0 \) leading to \( y = 1 \) and \( y = -2 \). Factoring provides a straightforward way to find solutions, especially when the quadratic is easily factorable.

Finding the Solution of Equations

The final step in the substitution method involves taking the solutions from the factored expressions \( y = 1 \) and \( y = -2 \), and using them to find the corresponding \( x \) values.

• When \( y = 1 \), substituting into the rearranged linear equation \( x = y + 1 \) gives us \( x = 2 \).
• When \( y = -2 \), substituting into the same produces \( x = -1 \).

Thus, the solution set for the system of equations is \( (2, 1) \) and \( (-1, -2) \). These points represent the coordinates where both the linear and quadratic equations intersect. Therefore, they are the values of \( x \) and \( y \) that satisfy the system of equations. By verifying these solutions satisfy both equations, we ensure the solutions are accurate and complete.