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Magazine Chapter 10: Problem 27
Find all solutions of the system of equations. \(\left\\{\begin{aligned} x-y &=4 \\ x y &=12 \end{aligned}\right.\)
Short Answer
Expert verified
Solutions are (6, 2) and (-2, -6).
Step by step solution
01
Understand the System of Equations
We are given a system of equations with two equations. The first equation is \( x - y = 4 \) and the second equation is \( x y = 12 \). Our task is to find all possible values of \( x \) and \( y \) that satisfy both equations simultaneously.
02
Express y in terms of x
From the first equation \( x - y = 4 \), we can express \( y \) in terms of \( x \). By rearranging, we get:\[y = x - 4\]
03
Substitute y in the Second Equation
Substitute \( y = x - 4 \) from Step 2 into the second equation \( x y = 12 \). This gives us:\[x(x - 4) = 12\]Which simplifies to:\[x^2 - 4x = 12\]
04
Rearrange and Simplify the Equation
Rearrange the equation from Step 3 to form a standard quadratic equation:\[x^2 - 4x - 12 = 0\]
05
Solve the Quadratic Equation
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -4 \), and \( c = -12 \):\[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times (-12)}}{2 \times 1}\]Calculate the discriminant:\[= \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2}\]Since \( \sqrt{64} = 8 \), we find:\[x = \frac{4 \pm 8}{2}\]This results in two solutions for \( x \): \( x = 6 \) and \( x = -2 \).
06
Find Corresponding y Values
Using \( y = x - 4 \), find \( y \) for each value of \( x \):- For \( x = 6 \):\[y = 6 - 4 = 2\]- For \( x = -2 \):\[y = -2 - 4 = -6\]
07
Verify Both Solutions
Substitute each pair \((x, y)\) back into the original equations:- For \( (x, y) = (6, 2) \): - First equation: \( 6 - 2 = 4 \), true. - Second equation: \( 6 \times 2 = 12 \), true.- For \( (x, y) = (-2, -6) \): - First equation: \( -2 - (-6) = 4 \), true. - Second equation: \( -2 \times (-6) = 12 \), true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations are vital in solving complex problems involving two variables. A quadratic equation is typically written in the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable we need to find. The equation exhibits a degree of 2 because the highest exponent of \( x \) is 2. The term quadratic comes from "quad," meaning square, as it involves variables squared.
In our exercise, by substituting \( y = x-4 \) into the equation \( xy = 12 \), we derived the quadratic equation \( x^2 - 4x - 12 = 0 \). Solving this equation involved finding values of \( x \) that satisfy the equation, which can sometimes be achieved using factoring, completing the square, or using the quadratic formula. In this problem, we used the quadratic formula, a reliable method especially when other methods are less straightforward.
In our exercise, by substituting \( y = x-4 \) into the equation \( xy = 12 \), we derived the quadratic equation \( x^2 - 4x - 12 = 0 \). Solving this equation involved finding values of \( x \) that satisfy the equation, which can sometimes be achieved using factoring, completing the square, or using the quadratic formula. In this problem, we used the quadratic formula, a reliable method especially when other methods are less straightforward.
- Characteristics of Quadratic Equations: They can have two, one, or sometimes no real solutions depending on the discriminant.
- Applications: Quadratic equations frequently appear in various fields such as physics, engineering, and finance due to their ability to describe parabolic relationships.
Substitution Method
The substitution method is a technique used for solving systems of equations. When you have more than one equation to deal with, like in our system \( x - y = 4 \) and \( xy = 12 \), substitution becomes a handy tool.
This method involves solving one of the equations for one of the variables and then substituting that expression into the other equation. It effectively reduces a system of equations to a single equation with one variable. In our exercise, we took the first equation, \( x - y = 4 \), and expressed \( y \) as \( y = x - 4 \).
This method involves solving one of the equations for one of the variables and then substituting that expression into the other equation. It effectively reduces a system of equations to a single equation with one variable. In our exercise, we took the first equation, \( x - y = 4 \), and expressed \( y \) as \( y = x - 4 \).
- Simplification Process: By substituting \( y = x - 4 \) into the second equation, we simplify the system to a single quadratic equation in terms of \( x \): \( x(x - 4) = 12 \).
- Advantages: This method simplifies the solving process by eliminating one variable early on.
- Limitations: It can get complex if the initial substitution leads to tricky expressions to solve. However, it's straightforward and ideal for systems with simple terms.
Discriminant in Quadratic Equations
The discriminant is a specific value calculated from a quadratic equation that helps determine the number and type of solutions (roots) the equation has. It's expressed in the quadratic formula as \( b^2 - 4ac \). This value is crucial because it reveals much about the roots without having to solve the equation completely.
For our equation \( x^2 - 4x - 12 = 0 \), the discriminant is calculated as follows: \( b = -4 \), \( a = 1 \), and \( c = -12 \). The formula \( b^2 - 4ac \) becomes \( (-4)^2 - 4 \times 1 \times (-12) = 16 + 48 = 64 \).
For our equation \( x^2 - 4x - 12 = 0 \), the discriminant is calculated as follows: \( b = -4 \), \( a = 1 \), and \( c = -12 \). The formula \( b^2 - 4ac \) becomes \( (-4)^2 - 4 \times 1 \times (-12) = 16 + 48 = 64 \).
- Interpretation of the Discriminant:
- If \( b^2 - 4ac > 0 \), there are two distinct real roots.
- If \( b^2 - 4ac = 0 \), there is one real root (a perfect square).
- If \( b^2 - 4ac < 0 \), there are no real roots but two complex roots.
- Conclusion in Our Case: The positive discriminant (64) confirms that our quadratic equation has two distinct real solutions. Thus, it aligns with our final findings of \( x = 6 \) and \( x = -2 \).
