Problem 19 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{aligned} &y=6 x^{2}-1\\\ &2 x^{2}+5 y=-5 \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the given system of equations is \(x = 0\) and \(y = -1\), which can be written as a coordinate point (0, -1).

Step 1: Identify the equations

We have a system of two equations as follows: 1) \(y = 6x^2 - 1\) 2) \(2x^2 + 5y = -5\) We will use the substitution method to solve the system.

Step 2: Substitute equation 1 into equation 2

Substitute the expression of y from equation 1 into equation 2: \(2x^2 + 5(6x^2 - 1) = -5\)

Step 3: Simplify and solve for x

Simplify the equation: \(2x^2 + 30x^2 - 5 = -5\) Combine like terms: \(32x^2 = 0\) Now, solve for x: \(x^2 = 0 \Rightarrow x = 0\) We find that \(x = 0\)

Step 4: Substitute the value of x into equation 1

Now, substitute the value of x we found in step 3 into equation 1: \(y = 6(0)^2 - 1\) Solve for y: \(y = -1\) So, we find that \(y=-1\)

Step 5: Solution

Thus, the solution to the given system of equations is \(x = 0\) and \(y = -1\), which can be written as a coordinate point (0, -1).

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They are typically in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. In our system of equations:

• The equation \(y = 6x^2 - 1\) is a quadratic equation in terms of \(x\).

Quadratic equations can take multiple forms, but they generally describe a parabola when graphed on a coordinate system. Here, "6x^2" indicates the "a" value of 6 signifies a parabolic shape opening upward. Understanding these basics helps in solving quadratic systems effectively.

Substitution Method

The substitution method provides a strategic approach to solving systems of equations by replacing variables with equivalent expressions. Let's break down the process:

• You start with two equations, and select one to express one variable in terms of the other.
• Next, substitute the expression into the other equation.

In our exercise:

• From the equation \(y = 6x^2 - 1\), we express \(y\) in terms of \(x\).
• Substituting \(y\) in the second equation gives us: \(2x^2 + 5(6x^2 - 1) = -5\).

This step transforms our system into a solvable quadratic equation. Using substitution is especially effective when one equation is already isolated for a variable.

Solving Equations

Solving equations involves finding the values for variables that satisfy all equations simultaneously. Here's how this applies to our system:

• The substituted equation simplifies to \(2x^2 + 30x^2 - 5 = -5\).
• Combine like terms to get \(32x^2 = 0\).

To solve this:

• Divide both sides by 32 resulting in \(x^2 = 0\).
• This leads to \(x = 0\).

To find \(y\), substitute \(x = 0\) back into \(y = 6x^2 - 1\), leading to \(y = -1\). So, the solution is \((0, -1)\), meeting the conditions of both equations.

Coordinate Systems

A coordinate system allows us to visually plot solutions and better understand the relationships between equations. In a 2D coordinate system:

• Points are defined as \((x, y)\) coordinates.
• The graphical representation shows how solutions interact.

For our system:

• The quadratic equation \(y = 6x^2 - 1\) creates a parabola on the coordinate plane.
• The solution \((0, -1)\) is the point where both equations intersect, showing the system's only solution graphically.

Understanding coordinate systems enriches our comprehension of how and where solutions manifest, making it a vital tool in solving and verifying systems of equations.