Problem 41 - Intermediate Algebra

Question

Give a general description of how to use the substitution method to solve a system of two linear equations in two variables.

Step-by-Step Solution

Answer

Use substitution by solving one equation for a variable, then substitute into the other equation and solve.

Step 1: Identify the System of Equations

Start with the system of two linear equations in two variables. For example, let's consider the equations:\[ \begin{align*} y & = 3x + 5 \ \ 2x + y & = 11 \end{align*} \]

Step 2: Express One Variable in Terms of the Other

Choose one of the equations and solve for one variable in terms of the other. In our example, the first equation is already solved for \( y \), so we have \( y = 3x + 5 \).

Step 3: Substitute into the Other Equation

Replace the expression for \( y \) found in Step 2 into the other equation. Substituting \( y = 3x + 5 \) into the second equation gives:\[ 2x + (3x + 5) = 11 \]

Step 4: Solve the Resulting Equation

Now, solve the resulting equation for the remaining variable. Simplifying the equation from Step 3, we have:\[ 2x + 3x + 5 = 11 \]Combine like terms:\[ 5x + 5 = 11 \]Subtract 5 from both sides:\[ 5x = 6 \]Divide by 5:\[ x = \frac{6}{5} \]

Step 5: Substitute Back to Find Other Variable

Use the value found in Step 4 and substitute back into the equation used in Step 2 to find the other variable. Substituting \( x = \frac{6}{5} \) into \( y = 3x + 5 \):\[ y = 3\left( \frac{6}{5} \right) + 5 \]Calculate:\[ y = \frac{18}{5} + 5 \]Convert 5 to \( \frac{25}{5} \):\[ y = \frac{18}{5} + \frac{25}{5} = \frac{43}{5} \]

Step 6: Check the Solution

Verify the solution by substituting \( x = \frac{6}{5} \) and \( y = \frac{43}{5} \) back into the original equations to ensure they satisfy both equations. For the second equation:\[ 2\left( \frac{6}{5} \right) + \frac{43}{5} = 11 \]Calculate both sides:\[ \frac{12}{5} + \frac{43}{5} = \frac{55}{5} = 11 \]Both sides are equal, confirming the solution is correct.

Key Concepts

Understanding Linear Equations

Linear equations are equations where each term is either a constant or the product of a constant with a single variable. These equations form straight lines when graphed on a coordinate plane.

In their simplest form, linear equations in two variables can be written as: \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are variables.

This type of equation describes a line where each solution is a point on the line.

• If you change the values of \( x \) and \( y \), the equation remains balanced.
• Every linear equation can have one solution, no solution, or infinitely many solutions.
• Understanding linear equations helps in graphing them and visualizing solutions.

The goal when working with linear equations, especially in a system, is to determine the values of \( x \) and \( y \) that make the equation true.

Solving a System of Equations

A system of equations consists of two or more equations with the same set of variables. Solving the system means finding the values of these variables that satisfy all the equations simultaneously.

There are several methods to solve systems of equations, such as substitution, elimination, and graphing. Each method has its advantages, and choosing the right method depends on the specific equations you have.

The substitution method involves expressing one variable in terms of another and then replacing it in the other equation. This method is often straightforward when one of the equations is already solved for one variable.

• Start by looking at the system to determine which method seems easiest based on the given equations.
• Always double-check each solution by substituting it back into the original equations to verify its correctness.

When you solve a system using any method, it's like solving two puzzles, each connected to a point where the lines intersect in a graph.

Algebraic Solution via Substitution Method

The substitution method is a popular algebraic technique used for solving systems of equations. It involves a series of strategic steps that make finding an intersecting point more accessible.

Begin by solving one of the equations for one variable. This expression will then be substituted into the other equation. Here are some clear steps:

• First, express one variable from one equation in terms of the other variable. If one equation is already simplified, that's a great start.
• Next, substitute this expression into the other equation. Doing so allows us to eliminate one variable, making it possible to solve for the other.
• Once you have the value of one variable, substitute it back into the expression found in the first step to find the second variable.

These steps are systematic and ensure that each equation is used to its full potential in finding the intersection point. This point represents the solution to the system of equations, shown algebraically rather than graphically.