Question

Solve the system using any method. Explain your choice of method. \(y=x^2-1\) \(-y=2 x^2+1\)

Short answer

The solutions are \(x = \sqrt{2/3}, y = -1/3\) and \(x = -\sqrt{2/3}, y = -1/3\)

Step-by-step solution

Combine the Equations

Since the equations are both solved for y, but opposite in sign, we can combine the equations using addition which will remove the need for substitution and simplify the equation. It would result in the new equation: \(0 = x^2 -1 - (-2x^2 -1) \) = \(0 = 3x^2 -2\).

Solve Combined Equation

Our task now is to solve this quadratic equation. To do this, first, rearrange the equation by adding 2 to both sides which gives us \(3x^2 = 2\). Then, divide both sides by 3 to isolate \(x^2\), resulting in \(x^2 = 2/3\). Finally, square root both sides to obtain \(x = \sqrt{2/3}\) and \(x = - \sqrt{2/3}\), as there are two possible square roots of any given number.

Find y Values

Substitute the found x values into one of the original equations to find the corresponding y values. For \(x = \sqrt{2/3}\), it is \(y = (\sqrt{2/3})^2 - 1 = -1/3\). For \(x = -\sqrt{2/3}\), it is \(y = (-\sqrt{2/3})^2 - 1 = -1/3\). Therefore, the solutions are \(x = \sqrt{2/3}, y = -1/3\) and \(x = -\sqrt{2/3}, y = -1/3\).

Key concepts

Understanding Quadratic Equations

Quadratic equations are fundamental in algebra and are characterized by their standard form, \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These equations represent parabolas when graphed, with the solutions (or roots) of these equations corresponding to the x-intercepts of the parabola.

Quadratic equations may have:

• Two real and distinct solutions
• One real, repeated solution
• No real solutions (complex roots)

To solve the quadratic equation found in our exercise, \( 3x^2 - 2 = 0 \), we first rearrange it to find \( x^2 = \frac{2}{3} \). The next step is to take the square root of both sides, which yields two potential values for \( x \): \( \pm \sqrt{\frac{2}{3}} \).

Whenever solving quadratic equations, always remember these can yield two potential solutions due to their parabolic nature.

Exploring the Substitution Method

The substitution method is a popular technique to solve systems of equations, particularly when one equation is already solved for one variable. In our case, substitution is not directly used, but it’s important to grasp this method for similar problems.

Here's how it generally works:

• Solve one of the equations for one variable.
• Substitute this expression into the other equation.
• Solve for the remaining variable.
• Back-solve for the first variable.

This method can be particularly helpful when an equation is already expressed in terms of one variable, like \( y = x^2 - 1 \) from our exercise. The flexibility of this method makes it suitable for both linear and nonlinear systems of equations.

In scenarios where equations do not align well for elimination, substitution becomes a key technique, helping you plug one expression into the other to reduce the number of equations and variables involved.

Utilizing the Addition Method

The addition method, occasionally referred to as the elimination method, involves adding or subtracting equations in a system to eliminate one of the variables. This method becomes powerful when equations can be combined so that one or more variables cancel out.

In our exercise, notice how combining the two equations: \( y = x^2 - 1 \) and \( -y = 2x^2 + 1 \), using addition simplifies to \( 0 = 3x^2 - 2 \).

This elimination of \( y \) was possible because the equations were conveniently arranged, reducing the system down to a single quadratic equation. This approach aids in quickly reducing complex systems to simpler forms that are easier to solve.

Keep in mind, the addition method hinges on:

• Aligning the equations for elimination.
• Combining them to remove one variable.
• Solving the resulting simple equation.

Its efficiency shines in cases where substitution would otherwise prove complex or cumbersome.