Problem 21 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{array}{l} y=\frac{4}{x+2} \\ y=x+5 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \( (x, y) = (-1, 4) \) and \( (x, y) = (-6, -1) \).

Step 1: Set Equations Equal

Since both equations are equal to \( y \), we can set the right-hand sides of the equations equal to each other. This gives us: \[ \frac{4}{x+2} = x+5 \]

Step 2: Eliminate the Fraction

To eliminate the fraction, multiply both sides by \( x+2 \): \[ 4 = (x+2)(x+5) \]

Step 3: Expand the Right Side

Expand the expression on the right side of the equation: \[ \begin{align*} 4 &= x^2 + 5x + 2x + 10 \ 4 &= x^2 + 7x + 10 \end{align*} \]

Step 4: Rearrange into Quadratic Equation

Bring all terms to one side to form a standard quadratic equation: \[ x^2 + 7x + 10 - 4 = 0 \] which simplifies to \[ x^2 + 7x + 6 = 0 \]

Step 5: Factor the Quadratic Equation

Factor the quadratic equation: \[ (x + 1)(x + 6) = 0 \]

Step 6: Solve for x

Set each factor equal to zero and solve for \( x \): \[ x + 1 = 0 \text{ or } x + 6 = 0 \] implies \[ x = -1 \text{ or } x = -6 \]

Step 7: Solve for y Using the Original Equations

Substitute the values of \( x \) back into the equation \( y = x + 5 \) to find \( y \). For \( x = -1 \): \[ y = -1 + 5 = 4 \] For \( x = -6 \): \[ y = -6 + 5 = -1 \]

Step 8: Verify Solutions

Substitute both sets of \( (x, y) \) values into \( y = \frac{4}{x+2} \) to verify they satisfy both equations: - For \( (x, y) = (-1, 4) \): \[ y = \frac{4}{-1 + 2} = 4 \] (True) - For \( (x, y) = (-6, -1) \): \[ y = \frac{4}{-6 + 2} = -1 \] (True) Both solutions check out.

Key Concepts

substitution method

The substitution method is a handy technique for solving systems of equations. It involves replacing one variable with an expression from another equation. This method is especially useful when we have one of the equations solved for a single variable.
In the exercise provided, we have two equations both set equal to \( y \):

• \( y = \frac{4}{x+2} \)
• \( y = x + 5 \)

We start by setting the right-hand sides of these equations equal, since they both equal \( y \). This gives us an equation with just one variable \( x \):

• \( \frac{4}{x+2} = x+5 \)

The goal is to find \( x \) and then substitute back to find \( y \). By following this approach, the substitution variable simplifies the problem. It turns a system of equations into a single equation to solve. After finding \( x \), we substitute it back into any of the original equations to find \( y \), systematically approaching the solution.

quadratic equations

Quadratic equations are mathematical expressions in the form \( ax^2 + bx + c = 0 \). In solving real-world problems, they frequently appear when we factor polynomials or solve systems of equations.
In the problem provided, after eliminating the fraction, we arrived at the quadratic equation:

• \( x^2 + 7x + 6 = 0 \)

Solving quadratic equations often begins by checking if they can be factored. If they can, it's the most straightforward solution besides using the quadratic formula. In this exercise, the quadratic was factorable:

• \( (x + 1)(x + 6) = 0 \)

Each factor set to zero gives possible solutions for \( x \). Quadratic equations frequently give two solutions. It's essential to check both against the original problem to ensure they make sense in context.

verification of solutions

Once potential solutions are found, verification ensures the answers meet all original conditions. It's a critical step in solving equations, as it confirms our solutions' validity, especially in systems of equations.
After finding \( x = -1 \) and \( x = -6 \), we substituted these back into \( y = x + 5 \) to get corresponding \( y \) values. We found:

• \((x, y) = (-1, 4) \)
• \((x, y) = (-6, -1) \)

Next, we put these pairs into the other original equation \( y = \frac{4}{x+2} \). Both values satisfied this equation, confirming they are solutions. This step ensures our algebraic manipulations didn't introduce errors and that both parts of the system of equations are satisfied.

algebraic manipulation

Algebraic manipulation is the active process of rearranging and simplifying equations. It includes operations like addition, subtraction, multiplication, division, and factoring. This skill is crucial in solving equations, especially when dealing with fractions and polynomial expansions.
In our step-by-step solution, we manipulated equations to eliminate fractions by multiplying through by a common denominator \( (x + 2) \). This cleared up the original fraction in \( \frac{4}{x+2} \):

• \( 4 = (x+2)(x+5) \)

Next, we expanded and rearranged the equation to form the standard quadratic equation. Manipulations like these streamline solving processes and highlight algebra's power. Writing the equation as \( x^2 + 7x + 6 = 0 \) made it easy to factor and find solutions.Mastering algebraic manipulation allows problem solvers to handle complex equations by systematically breaking them into simpler parts, ensuring clarity and understanding throughout the process.