Problem 17 - College Algebra

Question

Solve each system by the substitution method. $$\begin{aligned} &x+y=1\\\ &(x-1)^{2}+(y+2)^{2}=10 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions to the system of equations are (0,1) and (4, -3).

Step 1: Isolate a Variable in the Linear Equation

From equation \(x + y = 1\), express \(x\) in terms of \(y\). This gives us \(x = 1 - y\).

Step 2: Substitution into Non-linear Equation

Substitute \(x = 1 - y\) into the second equation: \((1 - y - 1)^2 + (y + 2)^2 = 10\). This simplifies to \((- y)^{2} + (y + 2)^{2} = 10\). Further simplification results in \(y^{2} + y^{2} + 4y + 4 = 10\), which can be reduced to \(2y^2 + 4y - 6 = 0\).

Step 3: Solve Reduced Equation

The reduced equation can then be solved. Divide through by 2, simplifying to \(y^2 + 2y - 3 = 0\). Factor to get \((y - 1)(y + 3) = 0\). From here, we solve for \(y\) and find \(y = 1\) and \(y = -3\).

Step 4: Find Corresponding 'x' Values

Substitute the values \(y = 1\) and \(y = -3\) into \(x = 1 - y\). When \(y = 1\), \(x = 1 - 1 = 0\). When \(y = -3\), \(x = 1 - (-3) = 4\).

Key Concepts

System of Equations

A system of equations is a set of two or more equations that you deal with at the same time. The goal is to find a solution that satisfies all equations in the system simultaneously. In simpler terms, it's about finding the values for the variables that make all the equations true. These systems can consist of:

• Linear equations: Equations that graph as straight lines.
• Non-linear equations: Equations involving curves, circles, or more complex shapes.

In our exercise, we have one linear equation and one non-linear equation, which makes it a mixed-type system. Solving such systems helps us understand how different mathematical relationships interact with each other in various applications.

Linear Equation

A linear equation is any equation that can be written in the form of \[ ax + by = c \]where,

• \(a\) and \(b\) are constants, showing the slope and intercept of the line.
• \(x\) and \(y\) are variables.
• \(c\) is a constant indicating where the line intersects the y-axis.

In our problem, we have the linear equation \(x + y = 1\). This is a straight line equation. It means if you add the values of \(x\) and \(y\), you will always get 1. Knowing how to manipulate linear equations is crucial when solving systems, as they often form the backbone for setting up simpler expressions for substitution.

Non-linear Equation

Non-linear equations, unlike linear ones, do not graph as straight lines. They can be circles, parabolas, ellipses, or any other shape. These equations often include variables raised to powers greater than one or two or use multiplication or division among variables.
In our exercise, the non-linear equation is \[(x-1)^{2} + (y+2)^{2} = 10\]This represents a circle centered at \((1, -2)\) with a radius \(\sqrt{10}\).
To find the solution, you have to understand how this curve interacts with the straight line \(x + y = 1\). The solutions to the system are actually the points where these two intersect. Since they represent different kinds of equations, you often need methods like substitution to tackle them efficiently.

Problem-solving Steps

Solving systems of equations, especially those that involve a mix of linear and non-linear equations, often requires a structured approach. Here's a breakdown of the process as used in the original exercise:

• Isolate a variable: Start by isolating one variable in the simplest equation, usually the linear one. In our example, from \(x + y = 1\), we express \(x = 1 - y\). This sets up a clear path for substitution.
• Substitute: Replace the isolated variable in the other equation, typically the more complex or non-linear one. This substitution helps convert the system into a single-variable equation.
• Simplify and solve: Work through algebraic simplification to solve the resulting equation. Often, this involves factoring or using the quadratic formula for quadratic equations, as we did by simplifying to \(y^2 + 2y - 3 = 0\).
• Find corresponding values: Once you have the solution for one variable, plug it back into the equation used for substitution to find the value of the other variable.

These steps are sequential and logical, making sure you break down complex problems into manageable parts. Practice with these steps can help solidify your equation-solving skills.