Problem 41 - College Algebra Essentials

Question

In Exercises \(29-42,\) solve each system by the method of your choice.$$ \left\\{\begin{array}{l} x^{2}+y^{2}+3 y=22 \\ 2 x+y=-1 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions to the system of equations are \((x, y) = (-2, 3)\) and \((x, y)= (3, -7)\)

Step 1 - Solve the linear equation for one variable

The linear equation in this case is \(2x + y = -1\). Let's solve this for \(x\), to do this, subtract \(y\) from both sides of the equation, yielding \(x = (-1 - y) / 2\).

Step 2 - Substitute \(x\) into the non-linear equation

Substitute \(x = (-1 - y) / 2 \) into the first equation \(x^{2}+y^{2}+3 y=22\). This gives us \(((-1 - y) / 2)^{2} + y^{2} + 3y = 22\).

Step 3 - Simplify and solve the resulting equation

Simplify the equation to \(0.25y^{2} - y + 1 + y^{2} + 3y - 22 = 0\). This simplifies further to \(1.25y^{2} +2y - 21 = 0\). Solving this quadratic equation we get \(y = 3\) and \(y = -7\).

Step 4 - Solve for \(x\)

Substitute \(y = 3\) into the equation \(x = (-1 - y) / 2\) to get \(x = -2\), then substitute \(y = -7\) into the same equation, to get \(x = 3\)

Key Concepts

Substitution Method

The substitution method is a standard technique used to solve systems of equations, which consists of two or more equations involving the same set of variables. The principle behind this method is quite straightforward: solve one of the equations for one variable in terms of the others, and then substitute this expression into the other equations. This reduces the number of equations, as well as the number of variables, simplifying the system.

When using substitution, it's usually best to solve the simplest equation for the easiest variable first. In the given example, the linear equation (which typically is the simpler one) \(2x + y = -1\) is solved for \(x\) to find \(x = (-1 - y) / 2\). This expression for \(x\) is then substituted into the nonlinear equation. It is key to perform substitutions precisely to prevent errors in subsequent steps. Let's look at a few points to remember when using this method:

System of Linear and Non-linear Equations

A system of equations may consist of linear equations, non-linear equations, or a combination of both. Linear equations form straight lines when graphed and their highest power of variables is always one. Non-linear equations can form curves, circles, ellipses, and other shapes on the graph, and they include powers higher than one, or variables multiplied together.

In the given exercise, we see a system composed of a linear equation, \(2x + y = -1\), and a non-linear equation, \(x^{2} + y^{2} + 3y = 22\), which represents a circle. Solving such a mixed system often requires using the linear equation to express one variable in terms of the other and then substituting this into the non-linear equation. The challenges include simplifying the substituted equation and solving for the remaining variable, usually resulting in a quadratic equation that needs to be solved.

Solving Quadratic Equations

After substitution and simplification, the problem often boils down to solving a quadratic equation. Quadratic equations are of the form \(ax^{2} + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and the highest power of the variable is two. There are various methods to solve these, including factoring, using the quadratic formula, completing the square, or graphing.

In our example, the resulting quadratic equation \(1.25y^{2} + 2y - 21 = 0\) can be solved using any of these methods. The solutions to this equation, \(y = 3\) and \(y = -7\), provide the values of \(y\) needed to solve for \(x\) in the next step of the substitution method. It is essential to ensure that all possible solutions for the equation are found since omitting a solution can lead to incomplete answers to the system of equations.