Problem 14 - Precalculus: Mathematical for Calculus

Question

Use the elimination method to find all solutions of the system of equations. $$\left\\{\begin{aligned}2 x^{2}+4 y &=13 \\\x^{2}-y^{2} &=\frac{7}{2}\end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions to the system are pairs of \((x, y)\) where both equations hold true for given values of x and y.

Step 1: Eliminate decimals by clearing denominators

Multiply the second equation by 2 to eliminate the decimal fraction, converting it into an equation with integer coefficients. The new equation becomes: \\[2(x^2 - y^2) = 2\cdot\frac{7}{2}\]\ which simplifies to: \\[2x^2 - 2y^2 = 7\]

Step 2: Set up modified system of equations

Now rewrite the system of equations with integer coefficients: \\[\begin{aligned}2x^2 + 4y &= 13 \2x^2 - 2y^2 &= 7 \\end{aligned}\]

Step 3: Add the equations to eliminate one variable

Add the two equations together to eliminate the term containing \(x^2\): \\[(2x^2 + 4y) + (2x^2 - 2y^2) = 13 + 7\] which simplifies to: \\[4x^2 + 4y - 2y^2 = 20\]

Step 4: Simplify the resulting equation

Divide the resulting equation by 2 to reduce coefficients: \\[2x^2 + 2y - y^2 = 10\]

Step 5: Rearrange for a quadratic in terms of y

Rearrange the equation to form a quadratic in terms of \(y\): \\[-y^2 + 2y = 10 - 2x^2\] which simplifies to: \\[y^2 - 2y + (2x^2 - 10) = 0\]

Step 6: Solve the quadratic equation for y

Treat \(2x^2 - 10\) as a constant and solve the quadratic equation for \(y\): \\[y = \frac{2 \pm \sqrt{4 + 4(2x^2 - 10)}}{2}\] \Simplify further to obtain expressions for \(y\) in terms of \(x^2\).

Step 7: Substitute back to solve for x and y

Once \(y\) is expressed in terms of \(x\), substitute back into one of the original equations to solve for \(x\).

Step 8: Verify solutions

Verify the solutions by substituting back into both original equations to ensure both equations are satisfied.

Key Concepts

System of Equations

A system of equations consists of multiple equations with the same set of unknowns. Here, we have two equations, each involving variables in different forms. If a solution exists, it would be a set of values for these variables that satisfy all the given equations simultaneously. This is done by finding common values for all the variables involved.For this problem, the system is:

• \(2x^2 + 4y = 13\)
• \(x^2 - y^2 = \frac{7}{2}\)

The elimination method is a common approach to solve systems of equations. It works by adding or subtracting the equations to eliminate one variable, making it easier to solve the remaining expressions. Once one variable is eliminated, it reduces the complexity and allows you to solve for the other variable. In this case, multiplying or rearranging equations to remove terms like decimals simplifies the solution process further.

Quadratic Equation

Quadratic equations are polynomial equations of degree two. They take the general form of \(ax^2 + bx + c = 0\). Solving these equations usually involves finding the roots or values of the variable that make the equation true.In our system, especially at step 5, we transformed one equation into a quadratic form in terms of \(y\):

• \(y^2 - 2y + (2x^2 - 10) = 0\)

This equation is a quadratic in \(y\), and it can be solved using the quadratic formula:

• \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, in terms of our expression, solving for \(y\) involves substituting calculated or simplified values into this formula. This step breaks down the complexity by isolating \(y\) as the subject, allowing you to then solve for potential values for \(x\) in subsequent steps.

Integer Coefficients

Having integer coefficients in equations makes them simpler to handle. In mathematics, integers are whole numbers and when equations have integer coefficients, they are easier to solve as compared to those with fractions or decimals.To achieve this in the given exercise, we multiplied the second equation by 2, eliminating the fraction. The original equation:

• \(x^2 - y^2 = \frac{7}{2}\)

becomes:

• \(2x^2 - 2y^2 = 7\)

This method simplifies the system significantly. It helps in the elimination process, by ensuring all terms are in whole numbers, making manipulations like addition, subtraction, and algebraic operations straightforward and error-free. Working with integer coefficients often leads to easier simplification and clearer results.