Problem 13 - College Algebra

Question

Solve each system by the substitution method. \(\left\\{\begin{array}{l}2 x+5 y=-4 \\ 3 x-y=11\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions to the system are (3,1), (-3,-1), (1,3), and (-1,-3).

Step 1: Isolate a Variable in One Equation

From the first equation, isolate x: \(x = 3/y\)

Step 2: Substitution into Second Equation

Substitute \(x = 3/y\) from first equation into the second equation to have an equation with y only. The second equation becomes \(\left(3/y\right)^2 + y^2 = 10\). This results in simplified form: \(y^4 - 10y^2 + 9 = 0\) which is a quadratic equation.

Step 3: Solve the Quadratic Equation

This equation can be solved using methods to solve quadratic equations (such as factoring, using the quadratic formula, etc.). Here, let \(z = y^2\) which converts our equation into \(z^2 - 10z + 9 = 0\). Solving this yields \(z = 1\), or \(z = 9\). Since \(z = y^2\), then \(y = ±1\) , or \(y = ±3\).

Step 4: Find Corresponding x-values

Substitute each computed y-value into the equation \(x = 3/y\) to get corresponding x-values. For \(y = ±1\), we get \(x = ±3\). For \(y = ±3\), we get \(x = ±1\).

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree, typically in the form of \(ax^2 + bx + c = 0\). In solving quadratic equations, you’re often looking for values of \(x\) that make the equation true. These solutions could represent points where a graph crosses the x-axis or other mathematical constructs.

Quadratic equations can often be solved using different methods, such as:

• Factoring the equation, if possible
• Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Completing the square method

When using any of these methods, it is key to first ensure the equation is set to zero, as this is pivotal for accurately solving the equation.

In our exercise, we encountered a transformed version of a quadratic equation as \(y^4 - 10y^2 + 9 = 0\), which we solved by recognizing \(y^2\) as a new variable, simplifying the approach.

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. Solving a system involves finding the values for the variables that satisfy all equations simultaneously.

There are several methods to solve systems of equations, including:

• Graphical methods - plotting the equations and finding intersections
• Substitution - solving one equation for one variable and substituting into another
• Elimination - adding or subtracting equations to remove variables

In this exercise, we have a system consisting of two equations. By isolating one of the variables, we were able to substitute and effectively reduce the system to a single quadratic equation. This strategy makes solving the system much more manageable.

Algebraic Substitution

The algebraic substitution method is a powerful tool in solving systems of equations. By isolating one variable in an equation and substituting it into another, we reduce the complexity of dealing with two equations simultaneously.

Here’s how the substitution method works step-by-step:

• Choose one of the equations from the system and isolate one of the variables.
• Substitute the expression for the isolated variable into the other equation.
• This will result in one equation with one variable, making it simpler to solve.

In our example problem, we isolated \(x\) as \(x = \frac{3}{y}\) from the first equation. Substituting this into the second equation simplified our system into a single variable quadratic, easing the path to finding solutions.

Solving Equations

Solving equations involves finding the values of the variables that make the equations true. Depending on the form and complexity of the equation, different methods might be used.

Here are some general steps and techniques to solve equations:

• Simplify the equation as much as possible.
• For quadratic equations, consider factoring, completing the square, or using the quadratic formula.
• For linear or simpler forms, isolate the variable on one side.

In more complex systems, like our exercise, solving involved using substitution to decrease the number of variables, then solving the resulting quadratic using substitution of a squared term. Once potential solutions are found, substitute back to verify each solution in the original context.