Problem 6 - Precalculus: Mathematical for Calculus

Question

Use the substitution method to find all solutions of the system of equations. $$\left\\{\begin{aligned}x^{2}+y &=9 \\\x-y+3 &=0\end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \((x, y) = (-3, 0)\) and \((x, y) = (2, 5)\).

Step 1: Express 'x' in terms of 'y'

Starting with the second equation: \(x - y + 3 = 0\), we solve for \(x\). Add \(y\) to both sides: \(x = y - 3\).

Step 2: Substitute 'x' in the first equation

Substitute \(x = y - 3\) into the first equation: \((y - 3)^2 + y = 9\).

Step 3: Expand and simplify the equation

Expand \((y - 3)^2\):\(y^2 - 6y + 9 + y = 9\).Combine like terms:\(y^2 - 5y + 9 = 9\).

Step 4: Solve the quadratic equation

Subtract 9 from both sides:\(y^2 - 5y = 0\).Factor the equation:\(y(y - 5) = 0\).Set each factor to zero:\(y = 0\) or \(y = 5\).

Step 5: Solve for 'x' using found values of 'y'

For \(y = 0\), substitute back into \(x = y - 3\):\(x = 0 - 3 = -3\).For \(y = 5\), substitute back into \(x = y - 3\):\(x = 5 - 3 = 2\).

Step 6: Write the solutions

The solutions to the system of equations are \((x, y) = (-3, 0)\) and \((x, y) = (2, 5)\).

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree 2. It typically takes the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) represent constants, and \(a eq 0\). One essential feature of quadratic equations is their parabola-shaped graph, which can open upwards or downwards depending on the sign of \(a\).

The solutions to a quadratic equation are known as its roots, which can be found using several methods:

• Factoring
• Completing the square
• Quadratic formula

Each method has its particular centricity and may be chosen based on the equation's complexity. In our specific problem, after substitution, we ended up with the quadratic equation \( y^2 - 5y = 0 \). Notice that one of the simplest ways to solve this is by factoring, which is a straightforward method when the equation can be easily returned to the form \( y(y - 5) = 0 \). These roots correspond to the possible values of \(y\) that satisfy the equation.

Substitution Method

The substitution method is a technique used to solve systems of equations. It is especially powerful for linear systems but can also help solve nonlinear systems. To use this method, you express one variable in terms of another, then substitute this expression into another equation.

In our exercise, we have the system:

• \( x^2 + y = 9 \)
• \( x - y + 3 = 0 \)

We started with the second equation \( x - y + 3 = 0 \) and expressed \( x \) in terms of \( y \) by rearranging it to \( x = y - 3 \). This expression was then substituted into the first equation to replace \( x \) and thus reduce the two-variable equation into a single equation with one variable, \( y \). This simplification allows for the problem to become a quadratic equation, which can then be solved straightforwardly.

Solving Equations

Solving equations is a fundamental skill in algebra that involves finding all the values of variables that satisfy the equation. The given system of equations can be thought of as a combination of two constraints that need to be simultaneously satisfied.

The process begins with simplifying the equations whenever possible, as shown in the steps we followed. Substitute to reduce complexity, and then solve for one variable first, using methods like factoring, after the equation is simplified to a form that allows those techniques.

For our system, after using the substitution method, we arrived at a quadratic equation \( y(y - 5) = 0 \) where each factor was set to zero individually to give \( y = 0 \) and \( y = 5 \). Solving for \( y\) first within our quadratic factors was efficient. Subsequently, we used these \( y \)-values to determine their respective \( x \)-values from the previously derived linear expression \( x = y - 3 \). Consequently, we arrived at the complete solutions: \( (x, y) = (-3, 0) \) and \( (x, y) = (2, 5) \). Each step is about steadily working through the problem, maintaining the balance required in each equation.