Problem 23 - Algebra and Trigonometry

Question

Find all solutions of the system of equations. \(\left\\{\begin{array}{l}{y+x^{2}=4 x} \\ {y+4 x=16}\end{array}\right.\)

Step-by-Step Solution

Answer

The solution is \((x, y) = (4, 0)\).

Step 1: Rewrite the System

First, we need to express the system of equations in a simpler form for substitution. The second equation is already straightforward: \( y + 4x = 16 \). For the first equation, we can write it as \( y = 4x - x^2 \).

Step 2: Substitute y in the Second Equation

Substitute the expression for \( y \) from the first equation \( y = 4x - x^2 \) into the second equation: \( (4x - x^2) + 4x = 16 \).

Step 3: Simplify and Solve for x

Combine like terms in the equation: \( -x^2 + 8x = 16 \). Rearrange it to form a quadratic equation \( x^2 - 8x + 16 = 0 \).

Step 4: Solve the Quadratic Equation

Solve the quadratic equation \( x^2 - 8x + 16 = 0 \) by using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -8 \), and \( c = 16 \).Compute: \[x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 1 \times 16}}{2 \times 1} \]Calculate the discriminant: \[64 - 64 = 0\]Since the discriminant is zero, there is one unique solution for \( x \): \[x = \frac{8}{2} = 4\].

Step 5: Find Corresponding y Values

Substitute the found value of \( x = 4 \) back into the equation \( y = 4x - x^2 \): \( y = 4(4) - 4^2 = 16 - 16 = 0 \). Thus, the pair \((x, y) = (4, 0)\) satisfies both equations.

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree. The general form is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). The solutions to a quadratic equation are also known as the roots of the equation.
Quadratic equations can represent various phenomena, such as the trajectory of a projectile. Understanding how to solve them is important in algebra and calculus. There are multiple methods to solve them, including factoring, completing the square, and using the quadratic formula.
If you use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), you need to calculate the discriminant, \( b^2 - 4ac \), first. This provides information on the nature and number of solutions:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution.
• If it is negative, there are no real solutions (but two complex solutions).

Substitution Method

The substitution method is a fundamental technique in solving a system of equations. This method involves solving one of the equations for one of the variables and then substituting that expression into the other equation.
In the first step, isolate one variable in one of the equations. For example, if you have \( y = 4x - x^2 \), you've isolated \( y \). Then substitute this expression for \( y \) into the other equation to eliminate \( y \). This turns the system of equations into a single equation, which is usually easier to solve.
Advantages of the substitution method include:

• It provides a straightforward way to solve linear and non-linear systems of equations.
• It helps to reduce errors by working consistently on one variable at a time.

However, it can be complex if the expressions involved are complicated, as the substitution may lead to a more complex equation than initially expected.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions and equations to solve for unknowns. This is a vital skill in mathematics, especially when dealing with multiple equations like in systems of equations.
Consider the step where we had to rearrange \( -x^2 + 8x = 16 \) to the standard form of a quadratic equation, \( x^2 - 8x + 16 = 0 \). This manipulation involved moving terms around to set the equations equal to zero, which is necessary for using the quadratic formula.
Key tasks in algebraic manipulation include:

• Combining like terms.
• Using the distributive property to expand or factor expressions.
• Adding, subtracting, multiplying, or dividing both sides by the same number without changing the equality.

Ultimately, mastering these manipulations allows you to streamline complex algebraic problems and make them manageable.