Problem 23 - College Algebra Essentials

Question

In Exercises \(19-28,\) solve each system by the addition method. $$ \left\\{\begin{array}{l} 3 x^{2}+4 y^{2}-16=0 \\ 2 x^{2}-3 y^{2}-5=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions to the system of equations are \(x = ±\sqrt{58/13}\) and \(y = ±\sqrt{50/52}\).

Step 1: Multiply the Equations

First, we aim to eliminate one variable by properly multiplying the equations. Here, if we multiply the first equation by 3 and the second by 2, the terms containing \(y^2\) will get cancelled when added. So, the multiplied equations are: \[9x^2 + 12y^2 - 48 = 0\] and \[4x^2 - 6y^2 - 10 = 0\]

Step 2: Add the multiplied equations

Adding these two equations yields: \[13x^2 - 58 = 0\].

Step 3: Solve for x

Rearrange the equation and solve for \(x\): \[x^2 = 58 / 13\] which gives \(x = ±\sqrt{58/13}\), so \(x = ±\sqrt{58/13}\).

Step 4: Substitute x in either equation

Substituting \(x\) in the first equation gives: \[3(58/13) + 4y^2 - 16 = 0\]. Rearranging and simplifying for \(y\), we get: \[4y^2 = 16 - 174/13\], which further simplifies to: \[y^2 = 50/52\], so \(y = ±\sqrt{50/52}\). Hence, \(y = ±\sqrt{50/52}\).

Step 5: Present the solutions

So the solutions to the system are \(x = ±\sqrt{58/13}\), \(y = ±\sqrt{50/52}\).

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations with the same set of variables. The goal is to find values for these variables that satisfy all equations in the system simultaneously. In the given exercise, we have two equations involving quadratic terms.

• The first equation is: \(3x^2 + 4y^2 - 16 = 0\)
• The second equation is: \(2x^2 - 3y^2 - 5 = 0\)

Solving a system of equations can reveal points of intersection between the curves represented by these equations.
Depending on the equations, solutions might consist of ordered pairs \((x, y)\) that satisfy all equations.

Solving Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving them involves finding the values of \(x\) that make the equation true. In many systems of equations, one or more of the equations are quadratic, requiring specific solution methods.
For the exercise given, after using the addition method to eliminate \(y^2\), we have a quadratic equation:
\[13x^2 - 58 = 0\]To solve for \(x\), rearrange it as \[x^2 = \frac{58}{13}\], and take the square root to find \(x = \pm\sqrt{\frac{58}{13}}\). Often, quadratic solutions include positive and negative roots.

Variables in Algebra

Variables are symbols or letters used to represent numbers or values in equations. They are the unknowns that we aim to solve. In algebra, variables are crucial because they allow us to write general rules and relationships.
In the exercise, the variables \(x\) and \(y\) are used:

• \(x\) and \(y\) represent unknown values we need to determine.
• Quadratic terms, such as \(x^2\) and \(y^2\), indicate the equations are quadratic in nature.

By finding specific values of \(x\) and \(y\), we solve the equations given.

Substitution Method

The substitution method involves solving one of the equations for one variable and substituting this expression into the other equation. This process helps in reducing the system of equations to a single equation in one variable.
In our exercise, once we had \(x = \pm\sqrt{\frac{58}{13}}\), we substituted \(x\) back into one of the original equations. This allowed us to solve for the variable \(y\):

• Substitute the value of \(x\) into the first equation to find \(y\).
• This substitution transformed our problem into solving a simpler equation \(4y^2 = 16 - \frac{174}{13}\).

This method simplifies the process of finding solutions to variables.