Problem 7 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system of nonlinear equations using substitution. $$ \begin{array}{c} y=x-3 \\ x^{2}+y^{2}=9 \end{array} $$

Step-by-Step Solution

Answer

The solutions are (0, -3) and (3, 0).

Step 1: Identify the Equations

We have two equations to solve:1. \( y = x - 3 \)2. \( x^2 + y^2 = 9 \)

Step 2: Substitute 'y' from the First Equation

From the first equation, we have \( y = x - 3 \). Substitute \( y \) in the second equation with \( x - 3 \). This gives us:\( x^2 + (x-3)^2 = 9 \).

Step 3: Expand the Substituted Equation

Now, expand the expression \((x-3)^2\) in the equation from Step 2:\[ x^2 + (x^2 - 6x + 9) = 9 \]

Step 4: Simplify the Equation

Combine like terms:\[ 2x^2 - 6x + 9 = 9 \]

Step 5: Solve the Quadratic Equation

Subtract 9 from both sides to get:\[ 2x^2 - 6x = 0 \]Factor out the common term, \(2x\):\[ 2x(x - 3) = 0 \]This gives two solutions: \( x = 0 \) or \( x = 3 \).

Step 6: Find Corresponding 'y' Values

Substitute \( x = 0 \) and \( x = 3 \) back into \( y = x - 3 \):- For \( x = 0 \), \( y = 0 - 3 = -3 \)- For \( x = 3 \), \( y = 3 - 3 = 0 \).

Step 7: Verify Solutions

Substitute \( (x, y) = (0, -3) \) and \( (x, y) = (3, 0) \) back into the original second equation:- \( 0^2 + (-3)^2 = 9 \) which is true.- \( 3^2 + 0^2 = 9 \) which is true.Both sets of values satisfy both original equations.

Key Concepts

Substitution Method

The substitution method is a straightforward technique for solving systems of equations, whether linear or nonlinear. It involves expressing one variable in terms of another and substituting this expression into another equation. In the given exercise, the substitution method was used to simplify the system of two equations:

• Equation 1: \( y = x - 3 \)
• Equation 2: \( x^2 + y^2 = 9 \)

By substituting the expression for \( y \) from the first equation into the second equation, we create a single equation in terms of one variable. This makes it easier to solve, as we're reducing the complexity from working with two variables to one. This method is particularly useful for equations that appear to be complex at first glance but can be broken down step by step.

Quadratic Equations

Quadratic equations often come into play when solving systems of nonlinear equations. In this problem, after substitution, we arrived at the quadratic equation:\[ 2x^2 - 6x = 0 \]Quadratic equations are typically in the form \( ax^2 + bx + c = 0 \), and the solutions can be found by factoring, completing the square, or using the quadratic formula. For simplicity, when the equation is factorable as in this problem, we can solve by identifying factors that multiply to give zero. Here, we factored the quadratic as:\[ 2x(x - 3) = 0 \]This reveals the solutions \( x = 0 \) and \( x = 3 \). These solutions can then be plugged back into the original equations to find corresponding values of \( y \) and verify them.

Algebraic Expressions

Working with algebraic expressions requires a keen eye for identifying and manipulating components to simplify the problem. In this exercise, we substituted \( y = x - 3 \) to form the algebraic expression \( x^2 + (x-3)^2 \). Properly expanding \((x-3)^2\) involves using the formula \((a-b)^2 = a^2 - 2ab + b^2\), yielding \( x^2 - 6x + 9 \).

• This expansion helps in simplifying the quadratic equation by combining like terms.
• Such manipulations are fundamental techniques in algebra, aiding our understanding of equation structure.

By meticulously handling these expressions, we reduce the problem to simpler parts, making it feasible to solve complex systems of equations.