Problem 2 - Algebra A Combined Function

Question

Solve each system of equations by the substitution method. $$ \left\\{\begin{array}{l} x+y=20 \\ x=3 y \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 15 \) and \( y = 5 \).

Step 1: Express One Variable in Terms of the Other

In this system of equations, the second equation is already solved for \( x \). This means we can use the equation \( x = 3y \) directly for substitution in the first equation.

Step 2: Substitute the Expression in the Other Equation

Take the expression for \( x \) from the second equation, \( x = 3y \), and substitute it into the first equation: \( x + y = 20 \).Replace \( x \) with \( 3y \), resulting in:\[ 3y + y = 20 \]

Step 3: Solve the Equation for y

We now have a single equation in terms of \( y \):\[ 3y + y = 20 \]Combine like terms:\[ 4y = 20 \]Divide both sides by 4 to solve for \( y \):\[ y = 5 \]

Step 4: Substitute the Value of y Back into the Expression for x

Now that we have \( y = 5 \), substitute it back into the expression for \( x \) (from Step 1):\[ x = 3y = 3(5) \]Calculate the value:\[ x = 15 \]

Step 5: Verify the Solution

Verify the solution by plugging \( x = 15 \) and \( y = 5 \) back into the original equations:1. For \( x + y = 20 \): \[ 15 + 5 = 20 \] which is true.2. For \( x = 3y \): \[ 15 = 3(5) \] which also holds true.Both equations are satisfied, so \( x = 15 \) and \( y = 5 \) is the correct solution.

Key Concepts

System of Equations

A system of equations is a set of equations with multiple variables. The solution to the system is the set of values for the variables that satisfy all the equations simultaneously. In our example, we have two equations that form a system:

• First equation: \( x + y = 20 \)
• Second equation: \( x = 3y \)

These equations are related to each other through the variables \( x \) and \( y \). To solve the system means to find values of \( x \) and \( y \) that make both equations true at the same time.

Systems of equations can be solved in various ways, like graphing, substitution, or elimination methods. Each method has its own advantages and is useful in different scenarios.

Solving Equations

Solving equations is about finding the unknown values that make the equations true. With the system of equations example \( \{ x + y = 20, x = 3y \} \), we employed the substitution method to find these unknowns.

The substitution method involves replacing one variable with an expression from another equation. This simplifies solving as it reduces the number of variables. Here's how it works:

• Identify an equation where a variable is already expressed in terms of the other variable. In our case, \( x = 3y \).
• Substitute this expression into the other equation. Here, we replaced \( x \) in \( x + y = 20 \) with \( 3y \).
• This results in a simpler equation: \( 3y + y = 20 \).
• Simplify and solve the resulting equation for the variable \( y \).

Using substitution helps to untangle complex relationships between variables, making it easier to find the solutions.

Algebraic Expressions

Algebraic expressions consist of variables, numbers, and operations. In our system of equations, the expressions used are:

• \( x + y = 20 \) is an equation where the expression sums two variables.
• \( x = 3y \) expresses \( x \) as a multiple of \( y \).
• \( 3y + y = 20 \), an intermediary step, combines the coefficients of \( y \).

Understanding how to manipulate these expressions is key in solving systems of equations. You rearrange terms, combine like ones, and perform arithmetic operations to isolate variables.

In our solution, acknowledging that \( x = 3y \) simplified the process because it allowed one expression (\( 3y + y = 20 \)) to replace a more complex system momentarily. Once one variable was found, substituting back to find the other completed the solution. This manipulation of expressions to isolate and solve for a variable is the crux of algebraic problem-solving.