Problem 2 - Algebra 2

Question

Find the exact solution(s) of each system of equations. $$ \begin{array}{l}{y-x=1} \\ {x^{2}+y^{2}=25}\end{array} $$

Step-by-Step Solution

Answer

The solutions are \((3, 4)\) and \((-4, -3)\).

Step 1: Isolate one variable

From the first equation, \( y - x = 1 \), solve for \( y \). Add \( x \) to both sides to get \( y = x + 1 \).

Step 2: Substitute isolated variable

Use the expression for \( y \) from Step 1 and substitute it into the second equation: \( x^2 + (x + 1)^2 = 25 \).

Step 3: Expand and simplify the quadratic equation

Expand the substituted equation:\[ x^2 + (x^2 + 2x + 1) = 25 \]Combine like terms:\[ 2x^2 + 2x + 1 = 25 \]Subtract 25 from both sides to set the equation to zero:\[ 2x^2 + 2x - 24 = 0 \].

Step 4: Simplify the coefficients

Divide every term by 2 to simplify the equation:\[ x^2 + x - 12 = 0 \].

Step 5: Solve quadratic equation

Factor the quadratic equation:\[ (x - 3)(x + 4) = 0 \]This gives the solutions:\( x = 3 \) or \( x = -4 \).

Step 6: Find corresponding y-values

Substitute the \( x \) values back into \( y = x + 1 \):- If \( x = 3 \), then \( y = 3 + 1 = 4 \).- If \( x = -4 \), then \( y = -4 + 1 = -3 \).

Step 7: Verify solutions

Check each pair \((x, y)\) in the original equations:- For \((x, y) = (3, 4)\): 1. \( y - x = 4 - 3 = 1 \) 2. \( x^2 + y^2 = 3^2 + 4^2 = 9 + 16 = 25 \) - For \((x, y) = (-4, -3)\): 1. \( y - x = -3 - (-4) = 1 \) 2. \( x^2 + y^2 = (-4)^2 + (-3)^2 = 16 + 9 = 25 \)Both pairs satisfy the original equations.

Key Concepts

Solving Quadratic Equations

Solving quadratic equations is an essential skill in mathematics, especially when dealing with complex systems of equations. A quadratic equation is any equation that can be rearranged in the standard form:

• \[ ax^2 + bx + c = 0 \]

where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). To find the solutions, we can use several methods - the quadratic formula, factoring, or by completing the square.
The quadratic formula is a universal method that can always be used to find the solutions. It is given by:

• \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]

This calculates the roots of any quadratic equation, where the expression inside the square root (\( b^2 - 4ac \)) is known as the discriminant. The value of the discriminant tells us about the nature of the roots:

• If positive, there are two distinct solutions.
• If zero, there is exactly one solution (a repeated root).
• If negative, there are no real solutions (complex roots).

Recognizing these properties is crucial for understanding the nature of quadratic equations in different contexts.

Substitution Method

The substitution method is a powerful technique for solving systems of equations, specifically when one of the equations is easier to manipulate than the other. Typically, this method involves two main steps:

• **Step 1**: Solve one of the equations for one variable in terms of the other variable.
• **Step 2**: Substitute this expression into the other equation to solve for the remaining variable.

For example, consider the system given:

• \(y - x = 1\)
• \(x^2 + y^2 = 25\)

By solving the first equation for \( y \), we get \( y = x + 1 \). We then substitute \( y = x + 1 \) into the second equation to simplify and solve for \( x \). This substitution reduces the system to a single quadratic equation, making it easier to solve.
The substitution method can be used not only in linear systems but also with more complex equations, like those involving quadratics, allowing for greater flexibility in solving various types of mathematical problems.

Factoring Quadratics

Factoring quadratics is a straightforward and efficient method for solving some quadratic equations. It involves writing the quadratic expression as a product of two linear factors, which can then be used to find the solutions by setting each factor to zero. For instance, consider the quadratic equation:

• \[ x^2 + x - 12 = 0 \]

The factoring process searches for two numbers that multiply to the constant term, \(-12\), and add up to the coefficient of the linear term, \(1\). Here, these numbers are \(3\) and \(-4\). Therefore, we can factor the quadratic as:

• \[(x - 3)(x + 4) = 0\]

Setting each factor to zero gives:

• \[ x - 3 = 0 \Rightarrow x = 3 \]
• \[ x + 4 = 0 \Rightarrow x = -4 \]

Thus, the solutions to the quadratic equation \( x^2 + x - 12 = 0 \) are \( x = 3 \) and \( x = -4 \). Factoring quadratics is a handy tool when both factorable and recognizable patterns are present, making it efficient to solve without the need for the quadratic formula.