Problem 12 - Algebra and Trigonometry

Question

\(9-14\) . Use the elimination method to find all solutions of the system of equations. $$ \left\\{\begin{array}{c}{2 x^{2}+4 y=13} \\\ {x^{2}-y^{2}=\frac{7}{2}}\end{array}\right. $$

Step-by-Step Solution

Answer

Find x from Step 7, then use x to get y from Step 8.

Step 1: Express \(y\) in terms of \(x\) from the first equation

From the first equation, \(2x^{2} + 4y = 13\), we can express \(y\) in terms of \(x\) by isolating \(y\): \(4y = 13 - 2x^{2}\) Divide by 4: \(y = \frac{13 - 2x^{2}}{4}\).

Step 2: Substitute \(y\) in the second equation

Substitute \(y = \frac{13 - 2x^{2}}{4}\) into the second equation \(x^{2} - y^{2} = \frac{7}{2}\). This gives us: \[x^{2} - \left(\frac{13 - 2x^{2}}{4}\right)^{2} = \frac{7}{2}\] Now, expand and simplify this equation.

Step 3: Expand \(\left(\frac{13 - 2x^{2}}{4}\right)^{2}\)

The square of \(\frac{13 - 2x^{2}}{4}\) is: \[\left(\frac{13 - 2x^{2}}{4}\right)^{2} = \frac{(13 - 2x^{2})^{2}}{16}\] Expand \((13 - 2x^{2})^{2}\): \[(13 - 2x^{2})^{2} = 169 - 52x^{2} + 4x^{4}\] So, \[\left(\frac{13 - 2x^{2}}{4}\right)^{2} = \frac{169 - 52x^{2} + 4x^{4}}{16}\].

Step 4: Substitute back into the second equation and simplify

Substitute \(\frac{169 - 52x^{2} + 4x^{4}}{16}\) into the equation: \[x^{2} - \frac{169 - 52x^{2} + 4x^{4}}{16} = \frac{7}{2}\] This simplifies to: \[16x^{2} - (169 - 52x^{2} + 4x^{4}) = 56\] Combine like terms and simplify: \[16x^{2} - 169 + 52x^{2} - 4x^{4} = 56\].

Step 5: Rearrange and solve for \(x\)

After combining terms, we have: \[-4x^{4} + 68x^{2} - 169 = 56\] Move 56 to the other side: \[-4x^{4} + 68x^{2} - 225 = 0\]. Multiply through by -1 to make calculations easier: \[4x^{4} - 68x^{2} + 225 = 0\].

Step 6: Solve the quadratic in terms of \(x^{2}\)

Let \(z = x^{2}\), we have \[4z^{2} - 68z + 225 = 0\]. Solve this quadratic equation using the quadratic formula \(z = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), where \(a = 4\), \(b = -68\), \(c = 225\). Calculate \(z\) values.

Step 7: Substitute for \(z\) back to find \(x\)

Substitute \(z = x^{2}\) back into the equation for each value of \(z\) found in Step 6. Solve for \(x\) by taking the square root of each \(z\) value.

Step 8: Use finding \(x\) values to compute \(y\) values

Use the \(x\) values found to determine corresponding \(y\) values using the expression \(y = \frac{13 - 2x^{2}}{4}\) from Step 1.

Key Concepts

System of Equations

A system of equations is a set of two or more equations that have common variables. The goal is to find values for these variables that satisfy all the equations at once. In our example, we have:

• Equation 1: \(2x^{2} + 4y = 13\)
• Equation 2: \(x^{2} - y^{2} = \frac{7}{2}\)

These are not simple linear equations because they include terms like \(x^2\) and \(y^2\). Such terms make the system more complex and interesting, thereby requiring methods like elimination to solve it.
The elimination method focuses on removing one variable so that we can solve the system easily. In the context of our problem, we began by expressing \(y\) in terms of \(x\) from the first equation and then substituted this expression into the second equation. This substitution transformed our system into one equation with a single variable, allowing us to find its values.

Quadratic Equation

Quadratic equations are polynomial equations of the second degree, typically written in the form \(ax^2 + bx + c = 0\).
In our task, we first reduced the system of equations into a single equation by the elimination method. This resulted in the equation: \[-4x^{4} + 68x^{2} - 225 = 0\].
Noticing the structure, we set \(z = x^{2}\), simplifying it to a quadratic form: \[4z^{2} - 68z + 225 = 0\].
Solving this simplified quadratic equation involves using the quadratic formula: \[z = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\], where \(a = 4\), \(b = -68\), and \(c = 225\). This formula helps us find the values of \(z\), from which we can derive \(x\). Quadratics can be powerful tools in solving complex systems that involve squared terms.

Algebraic Solution

An algebraic solution involves manipulating equations and expressions using algebraic techniques until we isolate and find the values of variables.
Starting from rearranging the terms, expressing variables in terms of others, and applying algebraic identities and formulas—the journey to a solution may involve several steps.
In our exercise, we used a series of algebraic transformations:

• Expressing \(y\) in terms of \(x\) using the first equation.
• Substituting into another equation to reduce complexity.
• Solving a quadratic equation employing the quadratic formula.

After finding the value of \(x\), we returned to our substitution to find the corresponding \(y\). The elegance of algebra lies in its ability to transform and reduce even the most complicated problems into something more manageable and solvable.