Problem 24 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{array}{c} y-x=1 \\ 4 y^{2}-16 x^{2}=64 \end{array}$$

Step-by-Step Solution

Answer

The given system of equations is: \( \begin{cases} y = x + 1 \\ 4y^2 - 16x^2 = 64 \end{cases} \) After solving the linear equation for y and substituting it into the quadratic equation, we find that the quadratic equation simplifies to: \( -12x^2 + 8x - 60 = 0 \) However, using the quadratic formula to solve for x, we find that the discriminant is negative, which means there are no real solutions for x. Therefore, there are no real solutions for the system of equations.

Step 1: Solve the linear equation for y

We choose to solve the first equation for y: \( y - x = 1 \) Add x to both sides: \( y = x + 1 \)

Step 2: Substitute y into the quadratic equation

Now, we will substitute the expression for y into the second equation: \( 4(x + 1)^2 - 16x^2 = 64 \)

Step 3: Expand and simplify the equation

We will expand the equation and simplify it: \[ 4(x^2 + 2x + 1) - 16x^2 = 64 \] Distribute the multiplication: \( 4x^2 + 8x + 4 - 16x^2 = 64 \) Combine like terms: \( -12x^2 + 8x + 4 = 64 \) Subtract 64 from both sides: \( -12x^2 + 8x - 60 = 0 \)

Step 4: Solve the quadratic equation for x

To solve the quadratic equation, we could either factor, complete the square, or apply the quadratic formula. In this case, we use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a = -12\), \(b = 8\), and \(c = -60\). Plugging in these values, we get \( x = \frac{-8 \pm \sqrt{8^2 - 4(-12)(-60)}}{-24} \) Calculate the discriminant: \( x = \frac{-8 \pm \sqrt{64 - 2880}}{-24} \) \( x = \frac{-8 \pm \sqrt{-2816}}{-24} \) Since the discriminant is negative, we have no real solutions for x.

Step 5: Conclusion

Since there are no real solutions for x, there are no real solutions for the system of equations. The system has no real solutions.

Key Concepts

Linear Equations

Linear equations are algebraic expressions where the highest power of the variable is one. They typically take the form of \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants.
For example, in the given exercise, the equation \( y - x = 1 \) is a simple linear equation.
Solving a linear equation involves isolating one variable, which makes it easier to substitute into another equation in a system. This process simplifies finding solutions for the variable.

Important characteristics of linear equations include:

• Graphically represented by a straight line.
• Can have one or infinitely many solutions, depending on the situation.

In this particular exercise, solving for \( y = x + 1 \) helps integrate it into the quadratic equation, showcasing how linear equations can aid in solving systems of equations.

Quadratic Equations

Quadratic equations are expressions where the highest power of the variable is two. They usually come in the form \( ax^2 + bx + c = 0 \).
In the exercise, after substitution, the equation \( -12x^2 + 8x - 60 = 0 \) is a quadratic equation.
Solving quadratic equations can be more complex than linear equations, and typically involves factoring, completing the square, or using the quadratic formula.

Key points about quadratic equations:

• Graphically represented by a parabola.
• Can have two, one, or no real solutions, depending on the discriminant.

Identifying the type of equation at hand determines which method to use, impacting how we approach finding the solution.

Substitution Method

The substitution method is a technique used to solve systems of equations. It involves solving one equation for one of the variables and then replacing that variable in the other equation. This results in a single equation with one variable.
In this exercise, \( y \) from the linear equation is expressed as \( y = x + 1 \). This expression is substituted into the quadratic equation to simplify the system.
This method is particularly helpful when one of the equations is easier to solve for a single variable.

Benefits of the substitution method include:

• Simplifies complex systems by reducing the number of equations.
• Can reveal whether a system has no solutions, one solution, or infinitely many solutions.

However, care must be taken to simplify correctly and check the solution in both equations.

Discriminant in Quadratic Formula

The discriminant in the quadratic formula is crucial for understanding the nature of the solutions. In the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), the expression under the square root \( b^2 - 4ac \) is the discriminant.
Depending on its value, the discriminant tells us whether the solutions are real or complex:

• If positive, there are two distinct real solutions.
• If zero, there is one real solution (roots are the same).
• If negative, as in this exercise with \( \sqrt{-2816} \), the solutions are not real.

Calculating the discriminant provides valuable information on the expected solutions, simplifying the process of assessing the results for real-world problems.