Problem 16 - College Algebra

Question

For the following exercises, use any method to solve the system of nonlinear equations. $$-2 x^{2}+y=-5$$ $$6 x-y=9$$

Step-by-Step Solution

Answer

The solutions are \((1, -3)\) and \((2, 3)\).

Step 1: Express One Variable in Terms of the Other

Start with the second equation, where we have: \[6x - y = 9\]We can express \(y\) in terms of \(x\) by rearranging this equation: \[y = 6x - 9\]

Step 2: Substitute into the First Equation

Take the expression we found for \(y\) and substitute it into the first equation:\[-2x^2 + (6x - 9) = -5\]Simplify the expression:\[-2x^2 + 6x - 9 = -5\]

Step 3: Simplify and Arrange the Equation

Move everything to one side to create a standard quadratic equation:\[-2x^2 + 6x - 9 + 5 = 0\]This simplifies to:\[-2x^2 + 6x - 4 = 0\]

Step 4: Solve the Quadratic Equation

We can solve the quadratic equation \[-2x^2 + 6x - 4 = 0\]using the quadratic formula. The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = -2\), \(b = 6\), and \(c = -4\). Insert these values:\[x = \frac{-6 \pm \sqrt{6^2 - 4(-2)(-4)}}{2(-2)}\]This simplifies to:\[x = \frac{-6 \pm \sqrt{36 - 32}}{-4}\]\[x = \frac{-6 \pm \sqrt{4}}{-4}\]\[x = \frac{-6 \pm 2}{-4}\]Solve for \(x\) to get two solutions:\[x = 1, \quad x = 2\]

Step 5: Find Corresponding y-values

Now use the expression for \(y\):\[y = 6x - 9\]For \(x = 1\):\[y = 6(1) - 9 = -3\]For \(x = 2\):\[y = 6(2) - 9 = 3\]

Step 6: Verify the Solutions

Verify these solutions with both original equations. For \((x, y) = (1, -3)\): - Check \(-2x^2 + y = -5\): \(-2(1)^2 + (-3) = -5\) is correct - Check \(6x - y = 9\): \(6(1) - (-3) = 9\) is correctFor \((x, y)= (2, 3)\): - Check \(-2x^2 + y = -5\): \(-2(2)^2 + 3 = -5\) is correct - Check \(6x - y = 9\): \(6(2) - 3 = 9\) is correct.

Key Concepts

Understanding Quadratic Equations

Quadratic equations are equations that involve terms with powers of up to two, typically in the form of:

• \[ ax^2 + bx + c = 0 \]

Here, \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. In a system of nonlinear equations, one or more equations might be quadratic.
These quadratic components can lead to various interesting mathematical behaviors, such as curves called parabolas when graphed on a coordinate plane.
Recognizing the structure of a quadratic equation helps in applying methods to solve them, like factoring, completing the square, or especially useful - the quadratic formula.
Understanding the importance of quadratic equations is key in solving nonlinear systems, as they offer two possible solutions for \(x\).

Using the Substitution Method

The substitution method involves solving one equation for one variable and substituting this into the other equation. In our problem, the system of equations was:

• \[-2x^2 + y = -5\]
• \[6x - y = 9\]

By isolating \(y\) in the second equation, \(y = 6x - 9\), we can substitute \(6x - 9\) for \(y\) in the first equation.
This substitution transforms the first nonlinear equation into a quadratic one in terms of \(x\) alone, making it simpler to solve for \(x\).
It's a handy technique especially in systems where one equation can easily be solved for one of the variables, simplifying the complexity of the system.

Solution Verification

Once potential solutions are found, it's crucial to verify them in the context of the original system of equations.
Verification ensures that the solutions truly satisfy both equations in the system.
In our exercise, after finding

• \((x, y) = (1, -3)\)
• \((x, y) = (2, 3)\)

, each pair should be plugged back into both original equations to verify their accuracy.
For example, substitute \((1, -3)\) into \[-2x^2 + y = -5\]and check the result. Doing so confirms that both solutions are valid, establishing confidence and accuracy in the results.
Verification highlights the importance of checking work, especially with nonlinear equations where mistakes are easy to overlook.

Applying the Quadratic Formula

The quadratic formula is a straightforward method to solve quadratic equations, given by:

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

where \(a\), \(b\), and \(c\) are coefficients from the equation \(ax^2 + bx + c = 0\).
This formula can find solutions for practically any quadratic equation, making it a versatile tool.
In our context, the equation \(-2x^2 + 6x - 4 = 0\) was solved using this formula, resulting in solutions \(x = 1\) and \(x = 2\).
It involves calculating the discriminant \(b^2 - 4ac\), which determines the number and type of solutions. A positive discriminant signals two distinct real roots.
The quadratic formula's reliability in delivering solutions makes it indispensable when working with nonlinear systems that include quadratic elements.