Problem 85 - Algebra 1

Question

Use substitution to solve the system. $$\begin{aligned}&x+4 y=300\\\&x-2 y=0\end{aligned}$$

Step-by-Step Solution

Answer

\(y = 150\), \(x = 300\)

Step 1: Substitution

Substitute \(x\) in the first equation with the right-hand side of the second equation. Thus the new first equation will be: \(-2y + 4y = 300\)

Step 2: Simplify and isolate variable \(y\)

Simplify the equation to find the value of \(y\). \(-2y + 4y\) simplifies to \(2y\), thus \(2y = 300\). Lastly, solve for \(y\) by dividing both sides of the equation by 2. So, \(y = 150\).

Step 3: Find \(x\)

Substitute \(y = 150\) in the second equation to find the value for \(x\). Therefore, \(x = 2y = 2*150 = 300\)

Key Concepts

System of equations

A system of equations is a collection of two or more equations with the same set of unknowns. In our exercise, we are dealing with the system: \(\begin{aligned}&x+4y=300\&x-2y=0\end{aligned}\).
The goal is to find the values of the variables that satisfy all equations simultaneously. When solving a system of equations, there are several methods available, such as substitution, elimination, and graphical methods. Substitution is particularly useful when one equation can be easily solved for one variable, which can be substituted into the other equations.

• The primary aim is to reduce the equations down to find one variable at a time.
• Once one variable is known, it can be plugged back into one of the original equations to find the other variables.

Finding solutions to systems of equations is foundational because it allows us to model and solve real-world problems involving multiple conditions or constraints. The solutions identify points of intersection when viewed graphically, indicating where equations meet.

Solving equations

Solving equations generally involves finding values for the unknown variables that make the equation true. In the context of our problem, we first substituted into one of the equations to make the solving process straightforward. The procedure starts with manipulating one of the equations to express one variable in terms of the other. Here, we expressed \(x\) in terms of \(y\) using the second equation to get \(x = 2y\). This allowed the substitution into the first equation, transforming it into a single equation with one variable: \(2y = 300\).
To solve the resulting equation:

• Simplify the terms to consolidate like terms or remove any equal terms.
• Isolate the variable by performing operations such as addition, subtraction, multiplication, or division.

Once resolved, this gives one of the necessary variable values, allowing us to use it to find the others as needed, creating a complete solution to the system.

Algebraic manipulation

Algebraic manipulation refers to the process of rearranging and simplifying equations to make them easier to solve. This often involves using basic algebraic operations such as addition, subtraction, multiplication, and division. In the given exercise, the key algebraic manipulations were fundamental to solve the system of equations smoothly.Starting from the step where \(x = 2y\) was substituted into the equation \(x + 4y = 300\), we simplified to \(2y + 4y = 300\), which boiled down to \(2y = 300\). Here, coefficient simplification and basic operations are applied:

• Combine like terms (e.g., \(-2y + 4y\) simplifies to \(2y\)).
• Perform operations to isolate and solve for \(y\) (here, dividing both sides by 2).

The manipulation ensures that complex systems can be tackled in an organized manner to yield exact variable values. Proficiency with these techniques is crucial for solving a wide range of algebraic problems effectively.